{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-sigma-algebra-measure-space-01",
  "mko_id": "mko-sigma-algebra-measure-space",
  "title": {
    "zh-Hant": "σ-代數與測度空間：四點載體上由原子生成的 σ-代數與一個機率測度",
    "en": "Sigma-algebras and measure spaces: an atom-generated sigma-algebra and a probability measure on a four-point carrier"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-sigma-algebra-measure-space-01",
    "source": {
      "mko_id": "mko-sigma-algebra-measure-space",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "36f29056738113842fd5fc437c89ae6f234815fa3abaea4cc68e968a358d6650",
      "english_sha256": "2148bebcaf753b4b69f42f056cf1cca5abc628d871406923f55af3da86c876b6"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "set_operations_def",
      "set_extensionality",
      "finite_sum_rules",
      "finite_counting_basics",
      "nat_induction"
    ],
    "goal": {
      "zh-Hant": "在 X={0,1,2,3} 上，以原子 A_1={0}、A_2={1,2}、A_3={3} 生成的 8 個集合構成 Σ，並以 m_1=1/2、m_2=1/3、m_3=1/6 定義 μ(∪_{i∈I}A_i)=Σ_{i∈I}m_i。證明 Σ 是 σ-代數（含可數聯集的有限歸約），證明 μ 是測度（含可數可加性的有限歸約）且為機率測度；再說明 𝒢={∅,{0},{1},X} 不是 σ-代數，以及只改一個值的 ν 不是測度。",
      "en": "On X={0,1,2,3}, the 8 sets generated by the atoms A_1={0}, A_2={1,2}, A_3={3} form Σ, and μ(∪_{i∈I}A_i)=Σ_{i∈I}m_i with m_1=1/2, m_2=1/3, m_3=1/6. Prove that Σ is a sigma-algebra (with the finite reduction of countable unions), that μ is a measure (with the finite reduction of countable additivity) and a probability measure; then show that 𝒢={∅,{0},{1},X} is not a sigma-algebra and that ν, which changes a single value, is not a measure."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "載體 X={0,1,2,3}；原子 A_1={0}、A_2={1,2}、A_3={3}。",
          "en": "Carrier X={0,1,2,3}; atoms A_1={0}, A_2={1,2}, A_3={3}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "完整 canonical 集合族 Σ={∅,{0},{1,2},{3},{0,1,2},{0,3},{1,2,3},X}，描述為 Σ={E_I ∣ I⊆{1,2,3}}，其中 E_I=∪_{i∈I}A_i（MKO 假設 a1）。",
          "en": "Complete canonical set family Σ={∅,{0},{1,2},{3},{0,1,2},{0,3},{1,2,3},X}, described as Σ={E_I ∣ I⊆{1,2,3}} with E_I=∪_{i∈I}A_i (MKO assumption a1)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "精確非負有理權重 m_1=1/2、m_2=1/3、m_3=1/6；定義 μ(E_I)=Σ_{i∈I}m_i。",
          "en": "Exact nonnegative rational weights m_1=1/2, m_2=1/3, m_3=1/6; define μ(E_I)=Σ_{i∈I}m_i."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "反例：𝒢={∅,{0},{1},X}；ν 定義在 Σ 上，除了 ν({0,1,2})=1 之外處處等於 μ。",
          "en": "Counterexamples: 𝒢={∅,{0},{1},X}; ν is defined on Σ and equals μ everywhere except ν({0,1,2})=1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_sigma_def"
        ],
        "before_tex": "(X,\\Sigma,\\mu)\\ \\text{is a measure space}",
        "after_tex": "\\emptyset\\in\\Sigma;\\ E\\in\\Sigma\\Rightarrow X\\setminus E\\in\\Sigma;\\ E_k\\in\\Sigma\\Rightarrow\\bigcup_{k\\ge1}E_k\\in\\Sigma;\\ \\mu\\ge0,\\ \\mu(\\emptyset)=0,\\ \\mu\\Big(\\bigcup_k E_k\\Big)=\\sum_k\\mu(E_k)\\ \\text{(pairwise disjoint)}",
        "relation": "definition",
        "rule": "mko_sigma_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開「測度空間」：Σ 要含 ∅、對補集封閉、對可數聯集封閉；μ 在 Σ 上取非負值、μ(∅)=0，並對兩兩互斥的可數族可數可加。之後每一項都分別核對，任何一項失敗就不是測度空間。",
          "en": "Unfold 'measure space' by the MKO: Σ must contain ∅, be closed under complements and under countable unions; μ takes nonnegative values on Σ, μ(∅)=0, and μ is countably additive on pairwise disjoint countable families. Each item is then checked separately; if any one fails, it is not a measure space."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_sigma_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義，不產生新事實；這張清單就是後面要逐項完成的責任表。",
            "en": "Only unfolds the definition and produces no new fact; this list is the set of obligations to be discharged item by item."
          }
        },
        "leaf_basis": "mko_sigma_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "A_1=\\{0\\},\\ A_2=\\{1,2\\},\\ A_3=\\{3\\}",
        "after_tex": "A_i\\neq\\emptyset,\\ \\ A_i\\cap A_j=\\emptyset\\ (i\\neq j),\\ \\ A_1\\cup A_2\\cup A_3=X",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一核對：三個原子都非空；{0}∩{1,2}=∅、{0}∩{3}=∅、{1,2}∩{3}=∅；三者聯集是 {0,1,2,3}=X。所以原子構成 X 的一個分割：X 的每個點恰在一個原子中。",
          "en": "Check one by one: all three atoms are nonempty; {0}∩{1,2}=∅, {0}∩{3}=∅, {1,2}∩{3}=∅; their union is {0,1,2,3}=X. So the atoms form a partition of X: every point of X lies in exactly one atom."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "三對交集與一個聯集是完整的有限核對；非空性在 S03.1 與 S12.1 要用到。",
            "en": "Three pairwise intersections and one union form a complete finite check; nonemptiness is needed in S03.1 and S12.1."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G2",
          "L_rep_unique"
        ],
        "before_tex": "\\Sigma=\\{E_I\\mid I\\subseteq\\{1,2,3\\}\\}",
        "after_tex": "|\\Sigma|=8:\\ \\emptyset,\\{0\\},\\{1,2\\},\\{3\\},\\{0,1,2\\},\\{0,3\\},\\{1,2,3\\},X",
        "relation": "implies",
        "rule": "L_rep_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "I↦E_I 把 {1,2,3} 的子集送到原子聯集。引理 L_rep_unique（S03.1）說這個對應是單射，所以每個 E∈Σ 有唯一的指標集 I；S03.2 數出 8 個相異集合，正好是 G2 的清單。",
          "en": "I↦E_I sends subsets of {1,2,3} to unions of atoms. Lemma L_rep_unique (S03.1) says this map is injective, so every E∈Σ has a unique index set I; S03.2 counts 8 distinct sets, exactly the list of G2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "L_rep_unique",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "單射性依賴原子非空且兩兩不交（S02）；若原子重疊，同一集合可能有兩個指標集。",
            "en": "Injectivity depends on the atoms being nonempty and pairwise disjoint (S02); with overlapping atoms one set could have two index sets."
          }
        },
        "leaf_basis": "L_rep_unique + finite_counting_basics"
      },
      {
        "id": "S03.1",
        "uses": [
          "S02",
          "contradiction"
        ],
        "before_tex": "E_I=E_J",
        "after_tex": "I=J",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "A_i\\neq\\emptyset",
          "A_i\\cap A_j=\\emptyset\\ (i\\neq j)"
        ],
        "explanation": {
          "zh-Hant": "假設 I≠J，不妨設有 i∈I 但 i∉J。A_i 非空，取 a∈A_i；則 a∈E_I=E_J，所以 a∈A_j 對某個 j∈J。a 同時在 A_i 與 A_j 中，兩兩不交迫使 j=i，於是 i∈J，矛盾。另一方向（j∈J 而 j∉I）對稱。所以 I=J，這證得引理 L_rep_unique。",
          "en": "Suppose I≠J; without loss of generality some i∈I has i∉J. A_i is nonempty; take a∈A_i. Then a∈E_I=E_J, so a∈A_j for some j∈J. Since a lies in both A_i and A_j, pairwise disjointness forces j=i, so i∈J, a contradiction. The other direction (j∈J with j∉I) is symmetric. Hence I=J, which proves Lemma L_rep_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "contradiction",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "非空性給出點 a，不交性把 j 鎖定為 i；兩個條件缺一不可。",
            "en": "Nonemptiness supplies the point a and disjointness pins j down to i; neither condition can be dropped."
          }
        },
        "leaf_basis": "contradiction + set_operations_def + set_extensionality"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "finite_counting_basics",
          "G2"
        ],
        "before_tex": "I\\mapsto E_I\\ \\text{injective on}\\ \\mathcal{P}(\\{1,2,3\\})",
        "after_tex": "|\\Sigma|=2^3=8",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "{1,2,3} 有 2^3=8 個子集（N1），單射把它們送到 8 個相異集合：I=∅→∅、{1}→{0}、{2}→{1,2}、{3}→{3}、{1,2}→{0,1,2}、{1,3}→{0,3}、{2,3}→{1,2,3}、{1,2,3}→X。這正是 G2 列出的 8 個集合，所以 G2 的清單是完整的。",
          "en": "{1,2,3} has 2^3=8 subsets (N1), and the injective map sends them to 8 distinct sets: I=∅→∅, {1}→{0}, {2}→{1,2}, {3}→{3}, {1,2}→{0,1,2}, {1,3}→{0,3}, {2,3}→{1,2,3}, {1,2,3}→X. These are exactly the 8 sets listed in G2, so the list of G2 is complete."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "單射加上逐一列出才得到 |Σ|=8 與清單吻合。",
            "en": "Injectivity together with the explicit listing gives |Σ|=8 and agreement with the list."
          }
        },
        "leaf_basis": "finite_counting_basics + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "I=\\emptyset",
        "after_tex": "E_{\\emptyset}=\\emptyset\\in\\Sigma",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "空指標集的聯集不含任何點，所以 E_∅=∅，而 ∅ 在 G2 的清單中。σ-代數的第一條責任完成。",
          "en": "The union over the empty index set contains no point, so E_∅=∅, and ∅ is on the list of G2. The first sigma-algebra obligation is met."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "只用到空聯集的定義。",
            "en": "Uses only the definition of the empty union."
          }
        },
        "leaf_basis": "def_unfold + set_operations_def"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "L_complement"
        ],
        "before_tex": "E_I\\in\\Sigma",
        "after_tex": "X\\setminus E_I=E_{I'}\\in\\Sigma,\\ \\ I'=\\{1,2,3\\}\\setminus I",
        "relation": "theorem_application",
        "rule": "L_complement",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_complement（S05.1、S05.2 證明）：E_I 的補集是其餘原子的聯集，而 E_{I'} 仍是原子聯集，屬於 Σ。例：X∖{0,3}：I={1,3}、I'={2}，補集是 {1,2}∈Σ。這完成補集封閉性。",
          "en": "Lemma L_complement (proved in S05.1, S05.2): the complement of E_I is the union of the remaining atoms, and E_{I'} is again a union of atoms, so it lies in Σ. Example: X∖{0,3}: I={1,3}, I'={2}, and the complement is {1,2}∈Σ. This gives closure under complements."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_complement",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "依賴原子構成分割（S02）；對任意集合族，補集一般不會自動留在族內（見 S14）。",
            "en": "Depends on the atoms forming a partition (S02); for an arbitrary family, complements need not stay in the family (see S14)."
          }
        },
        "leaf_basis": "L_complement + set_extensionality"
      },
      {
        "id": "S05.1",
        "uses": [
          "S02",
          "set_operations_def"
        ],
        "before_tex": "x\\in X\\setminus E_I",
        "after_tex": "x\\in E_{I'}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 x∈X 且 x∉E_I。由分割（S02），x 恰在一個原子 A_j 中。若 j∈I，則 x∈E_I，矛盾；所以 j∉I，即 j∈I'，因此 x∈E_{I'}。這證得 X∖E_I⊆E_{I'}。",
          "en": "Let x∈X with x∉E_I. By the partition (S02), x lies in exactly one atom A_j. If j∈I then x∈E_I, a contradiction; so j∉I, that is j∈I', hence x∈E_{I'}. This proves X∖E_I⊆E_{I'}."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "用到「每點至少在一個原子中」（聯集為 X）。",
            "en": "Uses that every point lies in at least one atom (the union is X)."
          }
        },
        "leaf_basis": "set_operations_def + subset_def"
      },
      {
        "id": "S05.2",
        "uses": [
          "S02",
          "S05.1",
          "set_extensionality"
        ],
        "before_tex": "x\\in E_{I'}",
        "after_tex": "x\\in X\\setminus E_I,\\ \\text{hence}\\ X\\setminus E_I=E_{I'}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 x∈A_j 且 j∉I。若 x∈E_I，則 x∈A_i 對某個 i∈I，而原子不交迫使 i=j∈I，矛盾。所以 x∉E_I，即 x∈X∖E_I。連同 S05.1 的反向包含，依外延性兩集合相等，證得引理 L_complement。",
          "en": "Let x∈A_j with j∉I. If x∈E_I, then x∈A_i for some i∈I, and disjointness of atoms forces i=j∈I, a contradiction. So x∉E_I, that is x∈X∖E_I. With the reverse inclusion of S05.1, extensionality gives equality, proving Lemma L_complement."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S05.1",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "用到「每點至多在一個原子中」（兩兩不交）；兩個包含合起來才是相等。",
            "en": "Uses that every point lies in at most one atom (pairwise disjoint); only the two inclusions together give equality."
          }
        },
        "leaf_basis": "set_extensionality + set_operations_def + subset_def"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "L_union"
        ],
        "before_tex": "E_I,\\ E_J\\in\\Sigma",
        "after_tex": "E_I\\cup E_J=E_{I\\cup J}\\in\\Sigma",
        "relation": "theorem_application",
        "rule": "L_union",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_union（S06.1 證明）：兩個原子聯集的聯集，就是指標集取聯集後的原子聯集，仍在 Σ 中。例：{0}∪{3}=E_{{1}}∪E_{{3}}=E_{{1,3}}={0,3}。這是二元聯集封閉性。",
          "en": "Lemma L_union (proved in S06.1): the union of two unions of atoms is the union of atoms over the union of the index sets, so it stays in Σ. Example: {0}∪{3}=E_{{1}}∪E_{{3}}=E_{{1,3}}={0,3}. This is closure under binary unions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_union",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "二元聯集封閉是可數聯集封閉的必要部分，但本身還不夠；S07 處理可數情形。",
            "en": "Closure under binary unions is a necessary part of closure under countable unions but is not enough on its own; S07 handles the countable case."
          }
        },
        "leaf_basis": "L_union + set_extensionality"
      },
      {
        "id": "S06.1",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "x\\in E_I\\cup E_J",
        "after_tex": "x\\in E_{I\\cup J}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈E_I∪E_J ⇔（存在 i∈I 使 x∈A_i）或（存在 j∈J 使 x∈A_j）⇔ 存在 k∈I∪J 使 x∈A_k ⇔ x∈E_{I∪J}。每個 ⇔ 都是聯集定義的展開，依外延性得到集合相等，證得引理 L_union。",
          "en": "x∈E_I∪E_J ⇔ (some i∈I has x∈A_i) or (some j∈J has x∈A_j) ⇔ some k∈I∪J has x∈A_k ⇔ x∈E_{I∪J}. Each ⇔ unfolds the definition of a union, and extensionality gives equality of sets, proving Lemma L_union."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "每一環都是定義的雙向改寫，所以整條鏈雙向成立；這一步不需要原子不交。",
            "en": "Each link is a two-way rewriting of a definition, so the whole chain holds both ways; this step does not need the atoms to be disjoint."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S06",
          "L_countable_reduce"
        ],
        "before_tex": "(E_k)_{k\\ge1},\\ E_k\\in\\Sigma",
        "after_tex": "\\bigcup_{k\\ge1}E_k=\\bigcup_{E\\in D}E\\in\\Sigma,\\ \\ D=\\{E_k\\mid k\\ge1\\}",
        "relation": "theorem_application",
        "rule": "L_countable_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "這是 MKO 段落 0 的有限歸約：引理 L_countable_reduce（S07.1 到 S07.3）說，Σ 中任一可數序列只含有限多個相異集合，它們的聯集等於整個序列的聯集，並且由二元聯集逐次組成，所以留在 Σ 中。",
          "en": "This is the finite reduction of MKO paragraph 0: Lemma L_countable_reduce (S07.1 to S07.3) says that any countable sequence in Σ contains only finitely many distinct sets, their union equals the union of the whole sequence, and it is built by repeated binary unions, so it stays in Σ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_countable_reduce",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "關鍵前提是 Σ 有限（S03）；在無限 σ-代數上，可數聯集不能這樣化為有限聯集。",
            "en": "The key premise is that Σ is finite (S03); in an infinite sigma-algebra a countable union cannot be reduced to a finite union this way."
          }
        },
        "leaf_basis": "L_countable_reduce + L_union"
      },
      {
        "id": "S07.1",
        "uses": [
          "S03",
          "subset_def",
          "finite_counting_basics"
        ],
        "before_tex": "D=\\{E_k\\mid k\\ge1\\}",
        "after_tex": "D\\subseteq\\Sigma,\\ \\ |D|\\le 8",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "序列可以有無限多項，但每一項都屬於 Σ，所以相異值的集合 D 是 Σ 的子集；Σ 只有 8 個元素（S03），所以 D 有限，|D|≤8。重複出現的集合在 D 中只算一次。",
          "en": "The sequence may have infinitely many terms, but each term belongs to Σ, so the set D of distinct values is a subset of Σ; Σ has only 8 elements (S03), so D is finite with |D|≤8. A set that occurs repeatedly counts once in D."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "有限集的子集有限；這裡沒有宣稱序列本身有限。",
            "en": "A subset of a finite set is finite; no claim is made that the sequence itself is finite."
          }
        },
        "leaf_basis": "subset_def + finite_counting_basics"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "set_extensionality"
        ],
        "before_tex": "x\\in\\bigcup_{k\\ge1}E_k",
        "after_tex": "x\\in\\bigcup_{E\\in D}E",
        "relation": "equivalent",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x 在序列聯集中 ⇔ 存在 k 使 x∈E_k ⇔ 存在 E∈D 使 x∈E（因為 D 正是所有 E_k 的值）。依外延性，序列聯集等於 D 的有限聯集。",
          "en": "x is in the union of the sequence ⇔ some k has x∈E_k ⇔ some E∈D has x∈E (since D is exactly the set of values E_k). By extensionality, the union of the sequence equals the finite union over D."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_extensionality",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "⇒：x∈E_k 時取 E=E_k∈D；⇐：E∈D 意味 E=E_k 對某個 k。雙向都成立。",
            "en": "⇒: if x∈E_k take E=E_k∈D; ⇐: E∈D means E=E_k for some k. Both directions hold."
          }
        },
        "leaf_basis": "set_extensionality + set_operations_def"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.2",
          "S04",
          "S06",
          "nat_induction"
        ],
        "before_tex": "D\\subseteq\\Sigma\\ \\text{finite}",
        "after_tex": "\\bigcup_{E\\in D}E\\in\\Sigma",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 |D| 歸納：|D|=0 時聯集為 ∅∈Σ（S04）；若任何 r 個 Σ 成員的聯集在 Σ 中，加入第 r+1 個成員 F 時，聯集 =（前 r 個的聯集）∪F，由 S06 的二元聯集封閉仍在 Σ 中。所以 D 的有限聯集屬於 Σ；連同 S07.2，序列的聯集屬於 Σ。S07.1 到 S07.3 證得引理 L_countable_reduce。",
          "en": "Induct on |D|: for |D|=0 the union is ∅∈Σ (S04); if the union of any r members of Σ is in Σ, then adding an (r+1)-st member F gives (the union of the first r) ∪ F, which is in Σ by binary closure (S06). So the finite union over D belongs to Σ; with S07.2 the union of the sequence belongs to Σ. S07.1 to S07.3 prove Lemma L_countable_reduce."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "歸納只對有限的 |D| 進行；它不證明任意無限聯集封閉，那一步由 S07.1 的有限性取代。",
            "en": "The induction runs only over finite |D|; it does not prove closure under arbitrary infinite unions, a step replaced by the finiteness of S07.1."
          }
        },
        "leaf_basis": "nat_induction + L_union"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S05",
          "S07",
          "S01"
        ],
        "before_tex": "\\emptyset\\in\\Sigma,\\ \\text{complements},\\ \\text{countable unions}",
        "after_tex": "\\Sigma\\ \\text{is a}\\ \\sigma\\text{-algebra on}\\ X",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S04、S05、S07 分別完成 S01 中 σ-代數的三條責任，合取得到 Σ 是 X 上的 σ-代數。有限的可數族可以用 ∅ 補成序列，所以也被 S07 涵蓋。",
          "en": "S04, S05 and S07 discharge the three sigma-algebra obligations of S01; their conjunction says Σ is a sigma-algebra on X. A finite family can be padded with ∅ to a sequence, so it is covered by S07 as well."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "三條責任缺一不可；只含 ∅ 與 X 並不足夠（S14）。",
            "en": "None of the three obligations may be omitted; containing ∅ and X is not enough (S14)."
          }
        },
        "leaf_basis": "and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "S03",
          "G3",
          "L_rep_unique"
        ],
        "before_tex": "\\mu(E_I)=\\sum_{i\\in I}m_i",
        "after_tex": "\\mu:\\ 0,\\tfrac12,\\tfrac13,\\tfrac16,\\tfrac56,\\tfrac23,\\tfrac12,1\\ \\text{on}\\ \\emptyset,\\{0\\},\\{1,2\\},\\{3\\},\\{0,1,2\\},\\{0,3\\},\\{1,2,3\\},X",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "μ 由指標集定義；因為每個 E∈Σ 的指標集唯一（L_rep_unique），μ(E) 只有一個值，μ 是 Σ 上定義良好的函數。逐一計算：μ({0,1,2})=1/2+1/3=5/6（N3）、μ({0,3})=1/2+1/6=2/3（N4）、μ({1,2,3})=1/3+1/6=1/2（N5）、μ(X)=1（N2）。",
          "en": "μ is defined through index sets; since the index set of each E∈Σ is unique (L_rep_unique), μ(E) has a single value, so μ is a well-defined function on Σ. Computing each value: μ({0,1,2})=1/2+1/3=5/6 (N3), μ({0,3})=1/2+1/6=2/3 (N4), μ({1,2,3})=1/3+1/6=1/2 (N5), μ(X)=1 (N2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_rep_unique",
            "G3",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "本構造以表示唯一保證定義良好；一般只需不同表示給相同值。8 個值都可由 N2 到 N5 獨立重算。",
            "en": "Unique representation guarantees well-definedness for this construction; in general it is enough that different representations give the same value. The 8 values can be recomputed independently through N2 to N5."
          }
        },
        "leaf_basis": "def_intro + finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "finite_sum_rules"
        ],
        "before_tex": "m_i\\ge 0",
        "after_tex": "\\mu(E)\\ge 0\\ \\ (E\\in\\Sigma),\\quad \\mu(\\emptyset)=\\sum_{i\\in\\emptyset}m_i=0",
        "relation": "implies",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非負數的有限和非負，所以每個 μ(E)≥0；空和等於 0，所以 μ(∅)=0。測度的前兩條責任完成。",
          "en": "A finite sum of nonnegative numbers is nonnegative, so every μ(E)≥0; the empty sum is 0, so μ(∅)=0. The first two measure obligations are met."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_sum_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要每個權重 m_i≥0；若某個權重為負，非負性就可能失敗。",
            "en": "Requires every weight m_i≥0; with a negative weight, nonnegativity could fail."
          }
        },
        "leaf_basis": "finite_sum_rules + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "S09",
          "L_fin_add"
        ],
        "before_tex": "E,F\\in\\Sigma,\\ E\\cap F=\\emptyset",
        "after_tex": "\\mu(E\\cup F)=\\mu(E)+\\mu(F)",
        "relation": "theorem_application",
        "rule": "L_fin_add",
        "conditions": [
          "E\\cap F=\\emptyset"
        ],
        "explanation": {
          "zh-Hant": "引理 L_fin_add（S11.1、S11.2 證明）：兩個不交的 Σ 成員的指標集不交，因此權重和可以拆開。這是有限可加性，也是 S12 可數可加性的基本步。",
          "en": "Lemma L_fin_add (proved in S11.1, S11.2): two disjoint members of Σ have disjoint index sets, so the weight sum splits. This is finite additivity, the basic step for countable additivity in S12."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_fin_add"
          ],
          "explanation": {
            "zh-Hant": "條件 E∩F=∅ 不可省：{0,1,2} 與 {0,3} 相交，μ 的和 5/6+2/3 大於 μ({0,1,2,3})=1。",
            "en": "The condition E∩F=∅ cannot be dropped: {0,1,2} and {0,3} intersect, and 5/6+2/3 exceeds μ({0,1,2,3})=1."
          }
        },
        "leaf_basis": "L_fin_add + finite_sum_rules"
      },
      {
        "id": "S11.1",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "E=E_I,\\ F=E_J,\\ E\\cap F=\\emptyset",
        "after_tex": "I\\cap J=\\emptyset",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [
          "A_i\\neq\\emptyset"
        ],
        "explanation": {
          "zh-Hant": "若有 i∈I∩J，則 A_i⊆E 且 A_i⊆F，而 A_i 非空（S02），所以 E∩F 含 A_i 的點，與 E∩F=∅ 矛盾。所以 I∩J=∅。",
          "en": "If some i∈I∩J, then A_i⊆E and A_i⊆F, and A_i is nonempty (S02), so E∩F contains a point of A_i, contradicting E∩F=∅. Hence I∩J=∅."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "用到原子非空；反向（I∩J=∅ 推出 E∩F=∅）也成立但此處不需要。",
            "en": "Uses that atoms are nonempty; the converse (I∩J=∅ implies E∩F=∅) also holds but is not needed here."
          }
        },
        "leaf_basis": "set_operations_def + contradiction"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "S06",
          "S09",
          "finite_sum_rules"
        ],
        "before_tex": "I\\cap J=\\emptyset",
        "after_tex": "\\mu(E_I\\cup E_J)=\\mu(E_{I\\cup J})=\\sum_{i\\in I\\cup J}m_i=\\sum_{i\\in I}m_i+\\sum_{j\\in J}m_j",
        "relation": "implies",
        "rule": "finite_sum_rules",
        "conditions": [
          "I\\cap J=\\emptyset"
        ],
        "explanation": {
          "zh-Hant": "由 S06，E_I∪E_J=E_{I∪J}；由定義（S09），μ 是 I∪J 上的權重和。指標集不交時，有限和可拆成 I 上的和加 J 上的和，即 μ(E)+μ(F)。S11.1 與 S11.2 證得引理 L_fin_add。例：μ({0,3})+μ({1,2})=2/3+1/3=1=μ(X)（N6）。",
          "en": "By S06, E_I∪E_J=E_{I∪J}; by the definition (S09), μ is the weight sum over I∪J. Since the index sets are disjoint, the finite sum splits into the sum over I plus the sum over J, which is μ(E)+μ(F). S11.1 and S11.2 prove Lemma L_fin_add. Example: μ({0,3})+μ({1,2})=2/3+1/3=1=μ(X) (N6)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11.1",
            "S06",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "拆和需要指標集不交；若有重疊，重疊的權重會被算兩次。",
            "en": "Splitting the sum needs disjoint index sets; with overlap the shared weights would be counted twice."
          }
        },
        "leaf_basis": "finite_sum_rules + L_union"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "S11",
          "L_ctbl_add"
        ],
        "before_tex": "(E_k)_{k\\ge1}\\subseteq\\Sigma\\ \\text{pairwise disjoint}",
        "after_tex": "\\mu\\Big(\\bigcup_{k\\ge1}E_k\\Big)=\\sum_{k\\ge1}\\mu(E_k)",
        "relation": "theorem_application",
        "rule": "L_ctbl_add",
        "conditions": [
          "E_k\\cap E_l=\\emptyset\\ (k\\neq l)"
        ],
        "explanation": {
          "zh-Hant": "引理 L_ctbl_add（S12.1 到 S12.3 證明）：在這個有限 Σ 中，兩兩互斥的序列至多有 3 個非空成員，其餘都是 ∅；級數只有有限多個非零項，可數可加性化為有限可加性。這就是 MKO 段落 1：有限載體中互斥非空族最多有限個。",
          "en": "Lemma L_ctbl_add (proved in S12.1 to S12.3): in this finite Σ a pairwise disjoint sequence has at most 3 nonempty members and all others are ∅; the series has only finitely many nonzero terms, and countable additivity reduces to finite additivity. This is MKO paragraph 1: on a finite carrier a disjoint family of nonempty sets is finite."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_ctbl_add",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "此處化約依賴原子只有 3 個，因而 Σ 有限；並不要求每個原子或載體本身有限。無限載體若仍只有有限個原子、Σ 為其聯集族，同樣可化約。錯誤控制 2 只說明任意無限事件族的有限可加性不足以保證可數可加性。",
            "en": "The reduction here uses the three-atom bound and hence finiteness of Σ, not finiteness of every atom or of the carrier itself. It also works on an infinite carrier with a finite atom partition and its union family. Error control 2 only shows that finite additivity on arbitrary infinite event families need not imply countable additivity."
          }
        },
        "leaf_basis": "L_ctbl_add + L_fin_add"
      },
      {
        "id": "S12.1",
        "uses": [
          "S11.1",
          "S02",
          "finite_counting_basics"
        ],
        "before_tex": "K=\\{k\\mid E_k\\neq\\emptyset\\},\\ E_k=E_{I_k}",
        "after_tex": "I_k\\neq\\emptyset,\\ I_k\\cap I_l=\\emptyset\\ (k\\neq l\\in K),\\ \\ |K|\\le 3",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 k∈K，E_k 非空，所以 I_k 非空；對相異 k、l∈K，E_k∩E_l=∅，由 S11.1 得 I_k∩I_l=∅。每個 k∈K 至少佔用 {1,2,3} 中一個指標，且不同的 k 不共用指標，所以 K 至多有 3 個元素。",
          "en": "For k∈K, E_k is nonempty, so I_k is nonempty; for distinct k, l∈K, E_k∩E_l=∅, and S11.1 gives I_k∩I_l=∅. Each k∈K uses at least one index from {1,2,3} and distinct k share no index, so K has at most 3 elements."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11.1",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "這是把 K 單射到 {1,2,3}（每個 k 取 I_k 中的一個指標）；單射進 3 元集合推出 |K|≤3。",
            "en": "This injects K into {1,2,3} (each k picks one index from I_k); an injection into a 3-element set gives |K|≤3."
          }
        },
        "leaf_basis": "finite_counting_basics + injective_surjective_def"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "S10",
          "p_finite_support_series"
        ],
        "before_tex": "\\sum_{k\\ge1}\\mu(E_k)",
        "after_tex": "\\sum_{k\\in K}\\mu(E_k)",
        "relation": "implies",
        "rule": "p_finite_support_series",
        "conditions": [
          "|K|\\le 3"
        ],
        "explanation": {
          "zh-Hant": "k∉K 時 E_k=∅，μ(E_k)=0（S10）。所以級數只有 K 中至多 3 個非零項；依前提規則 p_finite_support_series，它收斂，和等於這有限個項的和。",
          "en": "For k∉K, E_k=∅ and μ(E_k)=0 (S10). So the series has at most the 3 nonzero terms indexed by K; by the prerequisite rule p_finite_support_series it converges and its sum is the sum of these finitely many terms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.1",
            "S10",
            "p_finite_support_series"
          ],
          "explanation": {
            "zh-Hant": "收斂性來自「只有有限多個非零項」，不是來自一般級數理論。",
            "en": "Convergence comes from having only finitely many nonzero terms, not from general series theory."
          }
        },
        "leaf_basis": "p_finite_support_series + finite_sum_rules"
      },
      {
        "id": "S12.3",
        "uses": [
          "S12.2",
          "S11",
          "S06",
          "set_extensionality"
        ],
        "before_tex": "\\bigcup_{k\\ge1}E_k",
        "after_tex": "\\mu\\Big(\\bigcup_{k\\in K}E_k\\Big)=\\sum_{k\\in K}\\mu(E_k)",
        "relation": "implies",
        "rule": "L_fin_add",
        "conditions": [],
        "explanation": {
          "zh-Hant": "空集不增加點，所以 ∪_k E_k=∪_{k∈K}E_k（外延性）。|K|≤3：K 為空時兩邊都是 0；否則把 K 的成員逐一加入，每次前面的聯集（由 S06 仍在 Σ 中）與新成員不交，套用 S11 一次，至多兩次後得到 μ(∪_{k∈K}E_k)=Σ_{k∈K}μ(E_k)。連同 S12.2，S12.1 到 S12.3 證得引理 L_ctbl_add。",
          "en": "The empty sets add no points, so ∪_k E_k=∪_{k∈K}E_k (extensionality). |K|≤3: if K is empty both sides are 0; otherwise add the members of K one at a time; each time the previous union (still in Σ by S06) is disjoint from the new member, and S11 applies, so after at most two applications μ(∪_{k∈K}E_k)=Σ_{k∈K}μ(E_k). With S12.2, S12.1 to S12.3 prove Lemma L_ctbl_add."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S06",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "前面的聯集與新成員不交，因為新成員與前面每一個成員都不交；S11 只對兩集合使用。",
            "en": "The previous union is disjoint from the new member because the new member is disjoint from each earlier member; S11 is only ever applied to two sets."
          }
        },
        "leaf_basis": "L_fin_add + set_extensionality + nat_induction"
      },
      {
        "id": "S13",
        "uses": [
          "S08",
          "S10",
          "S12",
          "S09"
        ],
        "before_tex": "\\mu(X)=\\tfrac12+\\tfrac13+\\tfrac16",
        "after_tex": "\\mu(X)=1:\\ (X,\\Sigma,\\mu)\\ \\text{is a probability space}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S08、S10、S12 完成 S01 的全部責任，所以 (X,Σ,μ) 是測度空間。再算 μ(X)=1/2+1/3+1/6=1（N2）；MKO 段落 2 指出總量為 1 時才是機率測度，所以 μ 是機率測度。",
          "en": "S08, S10 and S12 discharge every obligation of S01, so (X,Σ,μ) is a measure space. Further, μ(X)=1/2+1/3+1/6=1 (N2); MKO paragraph 2 says it is a probability measure exactly when the total is 1, so μ is a probability measure."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S10",
            "S12",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "測度空間的結論需要全部責任；機率測度另需 μ(X)=1，這是精確有理數計算，不是近似。",
            "en": "The measure-space conclusion needs every obligation; a probability measure also needs μ(X)=1, which is an exact rational computation, not an approximation."
          }
        },
        "leaf_basis": "numeric_evaluation + and_intro_elim"
      },
      {
        "id": "S14",
        "uses": [
          "G4",
          "S01"
        ],
        "before_tex": "\\mathcal{G}=\\{\\emptyset,\\{0\\},\\{1\\},X\\}",
        "after_tex": "X\\setminus\\{0\\}=\\{1,2,3\\}\\notin\\mathcal{G}:\\ \\mathcal{G}\\ \\text{is not a}\\ \\sigma\\text{-algebra}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "𝒢 含 ∅ 與 X，但 {0}∈𝒢 的補集 {1,2,3} 不在 𝒢 中，補集封閉失敗；另外 {0}∪{1}={0,1} 也不在 𝒢 中，聯集封閉也失敗。只要一條責任失敗就不是 σ-代數。這正是 MKO 指出的誤解：含 ∅ 與全體不等於 σ-代數。",
          "en": "𝒢 contains ∅ and X, but the complement {1,2,3} of {0}∈𝒢 is not in 𝒢, so closure under complements fails; also {0}∪{1}={0,1} is not in 𝒢, so closure under unions fails too. One failed obligation is enough to rule out a sigma-algebra. This is the misconception named by the MKO: containing ∅ and the whole set does not make a sigma-algebra."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "否定全稱的封閉性只需一個反例；這裡給了兩個。",
            "en": "Refuting a universal closure property needs one counterexample; two are given here."
          }
        },
        "leaf_basis": "set_operations_def + quantifier_negation"
      },
      {
        "id": "S15",
        "uses": [
          "G4",
          "S09",
          "S01"
        ],
        "before_tex": "\\nu=\\mu\\ \\text{except}\\ \\nu(\\{0,1,2\\})=1",
        "after_tex": "\\nu(\\{0\\}\\cup\\{1,2\\})=1\\neq\\tfrac56=\\nu(\\{0\\})+\\nu(\\{1,2\\})",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [
          "\\{0\\}\\cap\\{1,2\\}=\\emptyset"
        ],
        "explanation": {
          "zh-Hant": "ν 非負且 ν(∅)=0，但 {0} 與 {1,2} 不交，ν 的值 1 不等於 1/2+1/3=5/6（N7），可加性失敗，所以 ν 不是測度。非負與空集為零不足以構成測度。",
          "en": "ν is nonnegative with ν(∅)=0, but {0} and {1,2} are disjoint and ν's value 1 differs from 1/2+1/3=5/6 (N7), so additivity fails and ν is not a measure. Nonnegativity and vanishing on ∅ do not make a measure."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一組不交集合的失敗就足以否定可加性；兩個集合的情形是可數族的特例（其餘補 ∅）。",
            "en": "One failing pair of disjoint sets refutes additivity; a pair is a special case of a countable family (pad with ∅)."
          }
        },
        "leaf_basis": "numeric_evaluation + quantifier_negation"
      }
    ],
    "result": {
      "zh-Hant": "Σ={E_I ∣ I⊆{1,2,3}} 恰有 8 個集合，含 ∅、對補集封閉（X∖E_I=E_{I'}）、對可數聯集封閉（化為至多 8 個相異集合的有限聯集），所以是 σ-代數；μ(E_I)=Σ_{i∈I}m_i 定義良好、非負、μ(∅)=0，且可數可加（互斥序列至多 3 個非空成員），μ(X)=1，所以 (X,Σ,μ) 是機率空間。𝒢={∅,{0},{1},X} 不是 σ-代數，ν 不是測度。",
      "en": "Σ={E_I ∣ I⊆{1,2,3}} has exactly 8 sets, contains ∅, is closed under complements (X∖E_I=E_{I'}) and under countable unions (reduced to a finite union of at most 8 distinct sets), so it is a sigma-algebra; μ(E_I)=Σ_{i∈I}m_i is well defined, nonnegative, μ(∅)=0 and countably additive (a disjoint sequence has at most 3 nonempty members), with μ(X)=1, so (X,Σ,μ) is a probability space. 𝒢={∅,{0},{1},X} is not a sigma-algebra and ν is not a measure."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "任何含 ∅ 與 X 的集合族都是 σ-代數",
          "en": "Every family containing ∅ and X is a sigma-algebra"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "𝒢={∅,{0},{1},X} 含兩者，但 {0} 的補集 {1,2,3}∉𝒢（S14）。",
          "en": "𝒢={∅,{0},{1},X} contains both, but the complement {1,2,3} of {0} is not in 𝒢 (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "ν 非負且 ν(∅)=0，所以 ν 是測度",
          "en": "ν is nonnegative with ν(∅)=0, so ν is a measure"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "{0} 與 {1,2} 不交，但 ν({0,1,2})=1≠5/6=ν({0})+ν({1,2})（S15）。",
          "en": "{0} and {1,2} are disjoint, but ν({0,1,2})=1≠5/6=ν({0})+ν({1,2}) (S15)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 μ 在這個有限 Σ 上有限可加即可數可加，所以任何載體上有限可加的集函數都可數可加",
          "en": "Since finite additivity gives countable additivity for μ on this finite Σ, every finitely additive set function on any carrier is countably additive"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "ℕ 的有限/餘有限族上的 0-1 值 λ 有限可加，但 λ(ℕ)=1≠0=Σλ({n})（錯誤控制 2）。",
          "en": "The 0-1 valued λ on the finite/cofinite family of ℕ is finitely additive, but λ(ℕ)=1≠0=Σλ({n}) (error control 2)."
        }
      },
      {
        "claim": {
          "zh-Hant": "μ 是 X 的冪集上的機率測度，所以 μ({1}) 有定義",
          "en": "μ is a probability measure on the power set of X, so μ({1}) is defined"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "μ 只定義在 Σ 上；{1} 不是原子聯集（它只含 A_2 的一部分），不在 Σ 中，Σ 不是冪集。",
          "en": "μ is defined only on Σ; {1} is not a union of atoms (it contains only part of A_2), so it is not in Σ, and Σ is not the power set."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_sigma_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-sigma-algebra-measure-space",
          "sha256": "36f29056738113842fd5fc437c89ae6f234815fa3abaea4cc68e968a358d6650",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在集合 X 上，σ-代數含空集，對補集及可數聯集封閉。測度對可測集給非負值、空集為零，並對兩兩互斥可數族可數可加。完整有限載體可把可數責任歸約為有限多個相異集合。",
          "en": "On a set X, a sigma-algebra contains the empty set and is closed under complements and countable unions. A measure assigns nonnegative values to measurable sets, is zero on the empty set and is countably additive on pairwise disjoint countable families. A complete finite carrier permits reduction to finitely many distinct sets."
        }
      },
      {
        "id": "L_rep_unique",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "對 G1 的原子，若 E_I=E_J，則 I=J。",
          "en": "For the atoms of G1, if E_I=E_J then I=J."
        }
      },
      {
        "id": "L_complement",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對每個 I⊆{1,2,3}：X∖E_I=E_{I'}，其中 I'={1,2,3}∖I。",
          "en": "For every I⊆{1,2,3}: X∖E_I=E_{I'}, where I'={1,2,3}∖I."
        }
      },
      {
        "id": "L_union",
        "kind": "lemma",
        "proved_in": [
          "S06.1"
        ],
        "statement": {
          "zh-Hant": "對所有 I、J⊆{1,2,3}：E_I∪E_J=E_{I∪J}。",
          "en": "For all I, J⊆{1,2,3}: E_I∪E_J=E_{I∪J}."
        }
      },
      {
        "id": "L_countable_reduce",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2",
          "S07.3"
        ],
        "statement": {
          "zh-Hant": "對 Σ 中任一序列 (E_k)_{k≥1}，集合 D={E_k ∣ k≥1} 是 Σ 的有限子族，∪_k E_k=∪_{E∈D}E，而且這個聯集屬於 Σ。",
          "en": "For every sequence (E_k)_{k≥1} in Σ, the set D={E_k ∣ k≥1} is a finite subfamily of Σ, ∪_k E_k=∪_{E∈D}E, and this union belongs to Σ."
        }
      },
      {
        "id": "L_fin_add",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "對 E、F∈Σ 且 E∩F=∅：μ(E∪F)=μ(E)+μ(F)。",
          "en": "For E, F∈Σ with E∩F=∅: μ(E∪F)=μ(E)+μ(F)."
        }
      },
      {
        "id": "L_ctbl_add",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2",
          "S12.3"
        ],
        "statement": {
          "zh-Hant": "對 Σ 中兩兩互斥的序列 (E_k)_{k≥1}：級數 Σ_k μ(E_k) 收斂，且 μ(∪_k E_k)=Σ_k μ(E_k)。",
          "en": "For a pairwise disjoint sequence (E_k)_{k≥1} in Σ: the series Σ_k μ(E_k) converges and μ(∪_k E_k)=Σ_k μ(E_k)."
        }
      },
      {
        "id": "p_finite_support_series",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "若一個實數級數除有限多項外每一項都是 0，則它收斂，而且和等於那些非零項（有限個）的和。",
          "en": "If all but finitely many terms of a real series are 0, the series converges and its sum is the (finite) sum of the nonzero terms."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03.2",
        "lhs": "2**3",
        "rhs": "8"
      },
      {
        "id": "N2",
        "step": "S13",
        "lhs": "1/2+1/3+1/6",
        "rhs": "1"
      },
      {
        "id": "N3",
        "step": "S09",
        "lhs": "1/2+1/3",
        "rhs": "5/6"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "1/2+1/6",
        "rhs": "2/3"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "1/3+1/6",
        "rhs": "1/2"
      },
      {
        "id": "N6",
        "step": "S11.2",
        "lhs": "2/3+1/3",
        "rhs": "1"
      },
      {
        "id": "N7",
        "step": "S15",
        "lhs": "1",
        "relation": "!=",
        "rhs": "1/2+1/3"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "5/6+2/3",
        "relation": ">",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 到 G4 指定的有限測度空間與兩個反例（MKO 假設 a1、a2）。可數聯集與可數可加性只在這個有限 Σ 上歸約為有限情形；不外推到任意無限 σ-代數或 extended-real 值測度。錯誤控制中的有限/餘有限例子只作對照說明。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the finite measure space specified by G1 to G4 and two counterexamples (MKO assumptions a1, a2). Countable unions and countable additivity are reduced to finite cases only for this finite Σ; nothing is extended to arbitrary infinite sigma-algebras or extended-real-valued measures. The finite/cofinite example in the error controls is only a contrast. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X=\\{0,1,2,3\\}",
        "有限載體",
        "the finite carrier"
      ],
      [
        "A_1,A_2,A_3",
        "原子 {0}、{1,2}、{3}",
        "the atoms {0}, {1,2}, {3}"
      ],
      [
        "E_I=\\bigcup_{i\\in I}A_i",
        "指標集 I⊆{1,2,3} 對應的原子聯集",
        "the union of atoms indexed by I⊆{1,2,3}"
      ],
      [
        "\\Sigma",
        "G2 的 8 個集合",
        "the 8 sets of G2"
      ],
      [
        "\\mu,\\ m_i",
        "測度與原子權重 1/2、1/3、1/6",
        "the measure and atom weights 1/2, 1/3, 1/6"
      ],
      [
        "\\mathcal{G},\\ \\nu",
        "反例：非 σ-代數的族與非測度的集函數",
        "counterexamples: a family that is not a sigma-algebra and a set function that is not a measure"
      ]
    ],
    "md_error_controls": [
      {
        "title": "含 ∅ 與全體不夠 / Containing ∅ and the whole set is not enough",
        "zh": "MKO 指出的誤解之一是只因集合族含 ∅ 與全體，就直接稱它為 σ-代數。σ-代數還要求補集與可數聯集封閉。S14 的 𝒢 含 ∅ 與 X，卻缺 {0} 的補集 {1,2,3}，也缺 {0}∪{1}。（只含 ∅ 與 X 的族 {∅,X} 本身確實是 σ-代數，但那是因為它通過全部封閉性檢查，不是因為含 ∅ 與 X。）",
        "en": "One misconception named by the MKO is to call a family a sigma-algebra just because it contains ∅ and the whole set. A sigma-algebra must also be closed under complements and countable unions. The family 𝒢 of S14 contains ∅ and X but lacks the complement {1,2,3} of {0} and lacks {0}∪{1}. (The family {∅,X} itself is a sigma-algebra, but that is because it passes every closure check, not because it contains ∅ and X.)"
      },
      {
        "title": "有限加法不能無條件外推 / Finite additivity does not extend unconditionally",
        "zh": "S07 與 S12 的歸約用到 Σ 有限、原子只有 3 個。載體無限本身不阻止化約，例如以偶數與奇數為兩個原子的四事件 σ-代數仍可化約。但不能對任意無限事件族直接這樣做：在 ℕ 的有限或餘有限子集族上定義 λ(A)=0（A 有限）、λ(A)=1（A 餘有限）。兩個不交集合不可能都是餘有限（兩個餘有限集之交仍是餘有限，因而非空），由此可驗證 λ 有限可加；但 ℕ 是單點集 {n} 的可數不交聯集，λ(ℕ)=1 而 Σλ({n})=0，可數可加失敗。這個族也不是 σ-代數（偶數集是單點集的可數聯集，卻既非有限也非餘有限）。",
        "en": "The reductions of S07 and S12 use that Σ is finite with only 3 atoms. An infinite carrier alone does not prevent this reduction: the four-event sigma-algebra generated by the even and odd integers still permits it. The reduction cannot instead be assumed for an arbitrary infinite event family: on the family of finite or cofinite subsets of ℕ define λ(A)=0 for finite A and λ(A)=1 for cofinite A. Two disjoint sets cannot both be cofinite (the intersection of two cofinite sets is cofinite, hence nonempty), from which finite additivity of λ can be checked; but ℕ is the countable disjoint union of the singletons {n}, and λ(ℕ)=1 while Σλ({n})=0, so countable additivity fails. That family is not a sigma-algebra either (the set of even numbers is a countable union of singletons but is neither finite nor cofinite)."
      },
      {
        "title": "唯一表示保證本構造定義良好 / Unique representation makes this construction well defined",
        "zh": "μ 是透過指標集定義的，須核對不同表示是否給同一個值。僅有重疊不必然使表示不唯一；例如另取 B₁={0,1}、B₂={1,2}、B₃={0,1,2} 並各給權重 1，B₁∪B₂=B₃，指標 {1,2} 與 {3} 卻給出總權重 2 與 1，此時該規則就不定義良好。S03.1 的 L_rep_unique 由給定原子非空且兩兩不交證明表示唯一，足以保證 S09 的構造定義良好。",
        "en": "Since μ is defined through index sets, different representations must assign the same value. Overlap alone need not destroy unique representation. For a concrete failure, take B₁={0,1}, B₂={1,2}, B₃={0,1,2}, each with weight 1. Then B₁∪B₂=B₃, but index sets {1,2} and {3} assign totals 2 and 1. That rule is not well defined. Lemma L_rep_unique in S03.1 uses the given atoms' nonemptiness and pairwise disjointness to prove unique representation, which is sufficient for the construction in S09."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "錯誤控制 2 的有限/餘有限例子以文字論證，沒有拆成編號步驟；本稿的結論不依賴它。",
        "en": "The finite/cofinite example in error control 2 is argued in prose and not broken into numbered steps; the conclusions of this guide do not depend on it."
      },
      {
        "zh-Hant": "本稿不證明「有限集上的每個 σ-代數都由某個分割的原子生成」；這裡的 Σ 是由給定原子直接構造的。",
        "en": "This guide does not prove that every sigma-algebra on a finite set is generated by the atoms of some partition; the Σ here is constructed directly from given atoms."
      },
      {
        "zh-Hant": "可測函數、Lebesgue 積分與收斂定理（MKO 的 formal next）不在範圍內。",
        "en": "Measurable functions, the Lebesgue integral and convergence theorems (the MKO's formal next) are out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "bfb135ecb5d6a207da34abed712f1b8ff3396f1a8b27a67641074674be0911ca",
    "original_source_sha256": "8f515e03122f38b50b7ae712b1e7f7a1dbe068f8776f00b0a9a35142bd5feba6",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M058"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}