{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-measurable-function-01",
  "mko_id": "mko-measurable-function",
  "title": {
    "zh-Hant": "可測函數：以完整列舉逐一核逆像，並以合成、指示函數與像集陷阱對照",
    "en": "Measurable function: checking every preimage by complete enumeration, with composition, indicators and the image-set trap"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-measurable-function-01",
    "source": {
      "mko_id": "mko-measurable-function",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "0c21c3c35f2f6b0451adf573f920b3dd8e797c631bcae8c14bf6bd959dc1571b",
      "english_sha256": "92b1662eff5366dc439172033e87ad4948ab0cbef9b4f22c08d2efd3e2310267"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "image_preimage_def",
      "set_operations_def",
      "function_def",
      "composition_def",
      "set_extensionality"
    ],
    "goal": {
      "zh-Hant": "X={0,1,2,3}、Σ_X={∅,{0,1},{2,3},X}；Y={0,1,2}、Σ_Y={∅,{0},{1,2},Y}；Z={0,1}、Σ_Z 為冪集。(1) 以完整列舉證明 f=(0↦0,1↦0,2↦1,3↦2) 可測，而 g=(0↦0,1↦1,2↦1,3↦2) 不可測；(2) 證明可測映射的合成可測，並把 φ∘f 讀回為指示函數；(3) 證明指示函數 1_E 可測若且唯若 E∈Σ_X；(4) 用 m 與 k 說明「檢查像集」既會誤拒也會誤收。",
      "en": "X={0,1,2,3}, Σ_X={∅,{0,1},{2,3},X}; Y={0,1,2}, Σ_Y={∅,{0},{1,2},Y}; Z={0,1} with Σ_Z the power set. (1) Prove by complete enumeration that f=(0↦0,1↦0,2↦1,3↦2) is measurable and g=(0↦0,1↦1,2↦1,3↦2) is not; (2) prove that a composite of measurable maps is measurable, and read φ∘f back as an indicator function; (3) prove that an indicator 1_E is measurable exactly when E∈Σ_X; (4) use m and k to show that 'checking images' can both wrongly reject and wrongly accept."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={0,1,2,3}，Σ_X={∅,{0,1},{2,3},X}。",
          "en": "X={0,1,2,3}, Σ_X={∅,{0,1},{2,3},X}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "Y={0,1,2}，Σ_Y={∅,{0},{1,2},Y}。",
          "en": "Y={0,1,2}, Σ_Y={∅,{0},{1,2},Y}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "完整映射表：f(0)=0、f(1)=0、f(2)=1、f(3)=2；g(0)=0、g(1)=1、g(2)=1、g(3)=2。",
          "en": "Complete map tables: f(0)=0, f(1)=0, f(2)=1, f(3)=2; g(0)=0, g(1)=1, g(2)=1, g(3)=2."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "Z={0,1}，Σ_Z={∅,{0},{1},Z}；φ:Y→Z，φ(0)=0、φ(1)=1、φ(2)=1。",
          "en": "Z={0,1}, Σ_Z={∅,{0},{1},Z}; φ:Y→Z with φ(0)=0, φ(1)=1, φ(2)=1."
        }
      },
      {
        "id": "G5",
        "statement": {
          "zh-Hant": "像集陷阱例子：m:X→Y，m(0)=0、m(1)=0、m(2)=1、m(3)=1；k:X→Z，k(0)=0、k(1)=1、k(2)=1、k(3)=1。",
          "en": "Image-trap examples: m:X→Y with m(0)=0, m(1)=0, m(2)=1, m(3)=1; k:X→Z with k(0)=0, k(1)=1, k(2)=1, k(3)=1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_measurable_def"
        ],
        "before_tex": "u:(X,\\Sigma_X)\\to(Y,\\Sigma_Y)\\ \\text{measurable}",
        "after_tex": "\\forall B\\in\\Sigma_Y:\\ u^{-1}(B)\\in\\Sigma_X",
        "relation": "definition",
        "rule": "mko_measurable_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開：可測的責任方向是「目標可測集取逆像，回到來源 σ-代數」（MKO 段落 0）。因為 Σ_Y 有限，這個全稱命題可以完整列舉 Σ_Y 的每個成員來核對。",
          "en": "Unfold by the MKO: the direction of the obligation is 'take the preimage of each measurable target set and land in the source sigma-algebra' (MKO paragraph 0). Since Σ_Y is finite, this universal statement can be checked by enumerating every member of Σ_Y."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_measurable_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；方向是逆像，不是像（見 S14、S15）。",
            "en": "Only unfolds the definition; the direction is preimages, not images (see S14, S15)."
          }
        },
        "leaf_basis": "mko_measurable_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2",
          "G4",
          "mko_sigma_def"
        ],
        "before_tex": "\\Sigma_X,\\ \\Sigma_Y,\\ \\Sigma_Z",
        "after_tex": "\\text{each is a}\\ \\sigma\\text{-algebra on its carrier}",
        "relation": "implies",
        "rule": "mko_sigma_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "可測性的定義預設兩邊都是 σ-代數，所以先核對（S02.1、S02.2）。三個族都有限，可數聯集化為有限聯集（MKO 有限歸約），所以只需核 ∅、補集與二元聯集。",
          "en": "The definition of measurability presupposes sigma-algebras on both sides, so these are checked first (S02.1, S02.2). All three families are finite, so countable unions reduce to finite unions (the MKO's finite reduction), and only ∅, complements and binary unions need checking."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_sigma_def"
          ],
          "explanation": {
            "zh-Hant": "有限歸約只對有限族有效；這裡三個族各有 4 個成員。",
            "en": "The finite reduction is valid only for finite families; each family here has 4 members."
          }
        },
        "leaf_basis": "mko_sigma_def + set_operations_def"
      },
      {
        "id": "S02.1",
        "uses": [
          "G1",
          "set_operations_def"
        ],
        "before_tex": "\\Sigma_X=\\{\\emptyset,\\{0,1\\},\\{2,3\\},X\\}",
        "after_tex": "\\emptyset\\in\\Sigma_X;\\ \\emptyset^c=X,\\ \\{0,1\\}^c=\\{2,3\\};\\ \\{0,1\\}\\cup\\{2,3\\}=X",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "∅∈Σ_X。補集：∅ 與 X 互補，{0,1} 與 {2,3} 互補，都在族中。二元聯集：與 ∅ 聯集得另一個集合本身，與 X 聯集得 X，而 {0,1}∪{2,3}=X；所有 6 對都留在族中。所以 Σ_X 是 σ-代數。",
          "en": "∅∈Σ_X. Complements: ∅ and X are complementary, {0,1} and {2,3} are complementary, all in the family. Binary unions: a union with ∅ gives the other set, a union with X gives X, and {0,1}∪{2,3}=X; all 6 pairs stay in the family. So Σ_X is a sigma-algebra."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "C(4,2)=6 對聯集全部涵蓋（N1）；有限族的可數聯集等於有限聯集。",
            "en": "All C(4,2)=6 pairwise unions are covered (N1); a countable union from a finite family equals a finite union."
          }
        },
        "leaf_basis": "set_operations_def + case_analysis"
      },
      {
        "id": "S02.2",
        "uses": [
          "G2",
          "G4",
          "set_operations_def"
        ],
        "before_tex": "\\Sigma_Y=\\{\\emptyset,\\{0\\},\\{1,2\\},Y\\},\\ \\Sigma_Z=\\mathcal{P}(Z)",
        "after_tex": "\\Sigma_Y,\\ \\Sigma_Z\\ \\text{are}\\ \\sigma\\text{-algebras}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Σ_Y 的結構與 Σ_X 相同：{0} 與 {1,2} 互補且聯集為 Y，∅ 與 Y 互補。Σ_Z 是 Z 的全部 2²=4 個子集（N2），任何補集與聯集仍是 Z 的子集，自然封閉。",
          "en": "Σ_Y has the same structure as Σ_X: {0} and {1,2} are complementary with union Y, and ∅ and Y are complementary. Σ_Z consists of all 2²=4 subsets of Z (N2), and any complement or union is again a subset of Z, so closure is automatic."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "G2",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "冪集永遠是 σ-代數；Σ_Y 的核對與 S02.1 逐項對應。",
            "en": "A power set is always a sigma-algebra; the check of Σ_Y corresponds item by item to S02.1."
          }
        },
        "leaf_basis": "set_operations_def + case_analysis"
      },
      {
        "id": "S03",
        "uses": [
          "G3",
          "function_def"
        ],
        "before_tex": "f:\\ 0\\mapsto0,\\ 1\\mapsto0,\\ 2\\mapsto1,\\ 3\\mapsto2",
        "after_tex": "f:X\\to Y\\ \\text{total}",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "映射表對 X 的四個點各列出恰好一個值，且值都在 Y 中，所以 f 是總函數。g、m、φ、k 同理（MKO 假設 a1：完整總映射表）。",
          "en": "The table lists exactly one value for each of the four points of X, all in Y, so f is a total function. The same holds for g, m, φ and k (MKO assumption a1: complete total-map tables)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "function_def",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "若表缺一個點，逆像就無法完整計算；全部 4 個點都列出了。",
            "en": "If the table missed a point, preimages could not be computed completely; all 4 points are listed."
          }
        },
        "leaf_basis": "function_def"
      },
      {
        "id": "S04",
        "uses": [
          "image_preimage_def",
          "S03"
        ],
        "before_tex": "u:X\\to Y\\ \\text{total}",
        "after_tex": "u^{-1}(\\emptyset)=\\emptyset\\in\\Sigma_X,\\quad u^{-1}(Y)=X\\in\\Sigma_X",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任何總函數 u：沒有 x 能使 u(x)∈∅，所以 u⁻¹(∅)=∅；每個 x 都有 u(x)∈Y，所以 u⁻¹(Y)=X。兩者都在 Σ_X 中。所以 Σ_Y 的四個成員中，只剩 {0} 與 {1,2} 需要針對各函數核對。",
          "en": "For any total function u: no x has u(x)∈∅, so u⁻¹(∅)=∅; every x has u(x)∈Y, so u⁻¹(Y)=X. Both are in Σ_X. So of the four members of Σ_Y, only {0} and {1,2} need checking for each function."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "image_preimage_def",
            "function_def"
          ],
          "explanation": {
            "zh-Hant": "u⁻¹(Y)=X 用到 u 是總函數；部分函數會使這一步失敗。",
            "en": "u⁻¹(Y)=X uses that u is total; a partial function would break this step."
          }
        },
        "leaf_basis": "image_preimage_def + function_def"
      },
      {
        "id": "S05",
        "uses": [
          "G3",
          "image_preimage_def",
          "G1"
        ],
        "before_tex": "f^{-1}(\\{0\\})",
        "after_tex": "\\{x\\mid f(x)=0\\}=\\{0,1\\}\\in\\Sigma_X",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐點列舉：f(0)=0 ✓、f(1)=0 ✓、f(2)=1 ✗、f(3)=2 ✗。所以 f⁻¹({0})={0,1}，在 Σ_X 中。",
          "en": "Enumerate point by point: f(0)=0 yes, f(1)=0 yes, f(2)=1 no, f(3)=2 no. So f⁻¹({0})={0,1}, which is in Σ_X."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "四個點全部檢查，逆像集合由定義雙向確定。",
            "en": "All four points are checked, and the preimage set is determined both ways by the definition."
          }
        },
        "leaf_basis": "image_preimage_def + set_builder"
      },
      {
        "id": "S06",
        "uses": [
          "G3",
          "image_preimage_def",
          "G1"
        ],
        "before_tex": "f^{-1}(\\{1,2\\})",
        "after_tex": "\\{x\\mid f(x)\\in\\{1,2\\}\\}=\\{2,3\\}\\in\\Sigma_X",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐點：f(0)=0 ✗、f(1)=0 ✗、f(2)=1 ✓、f(3)=2 ✓。所以 f⁻¹({1,2})={2,3}，在 Σ_X 中。也可用 S05 與補集：f⁻¹(Y∖{0})=X∖f⁻¹({0})={2,3}。",
          "en": "Point by point: f(0)=0 no, f(1)=0 no, f(2)=1 yes, f(3)=2 yes. So f⁻¹({1,2})={2,3}, which is in Σ_X. Alternatively, by S05 and complements: f⁻¹(Y∖{0})=X∖f⁻¹({0})={2,3}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "完整列舉；兩種算法結果一致，可互相核對。",
            "en": "Complete enumeration; the two computations agree and check each other."
          }
        },
        "leaf_basis": "image_preimage_def + set_builder"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S04",
          "S05",
          "S06",
          "G2"
        ],
        "before_tex": "B\\in\\{\\emptyset,\\{0\\},\\{1,2\\},Y\\}",
        "after_tex": "f^{-1}(B)\\in\\Sigma_X\\ \\text{for every}\\ B\\in\\Sigma_Y:\\ f\\ \\text{measurable}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Σ_Y 恰有四個成員（G2），S04、S05、S06 分別處理了它們，每個逆像都在 Σ_X 中。完整列舉涵蓋全部情況，所以 S01 的全稱命題成立，f 可測。",
          "en": "Σ_Y has exactly four members (G2), handled by S04, S05 and S06, and every preimage lies in Σ_X. The complete enumeration covers every case, so the universal statement of S01 holds and f is measurable."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S06",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "這是有限全稱的完整驗證，只證這一個 f（MKO 假設 a2）。",
            "en": "This is a complete verification of a finite universal statement, for this f only (MKO assumption a2)."
          }
        },
        "leaf_basis": "case_analysis + forall_intro"
      },
      {
        "id": "S08",
        "uses": [
          "G3",
          "G1",
          "S01",
          "image_preimage_def"
        ],
        "before_tex": "g^{-1}(\\{0\\})",
        "after_tex": "\\{0\\}\\notin\\Sigma_X:\\ g\\ \\text{not measurable}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐點：g(0)=0 ✓，g(1)=g(2)=1、g(3)=2 都 ✗。所以 g⁻¹({0})={0}，但 {0} 不在 Σ_X 中（Σ_X 只把 0 與 1 一起處理）。一個失敗的 B 就否定全稱命題，g 不可測。另外 g⁻¹({1,2})={1,2,3} 也不在 Σ_X 中。",
          "en": "Point by point: g(0)=0 yes; g(1)=g(2)=1 and g(3)=2 no. So g⁻¹({0})={0}, but {0} is not in Σ_X (Σ_X only handles 0 and 1 together). One failing B refutes the universal statement, so g is not measurable. Also g⁻¹({1,2})={1,2,3} is not in Σ_X."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "quantifier_negation",
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "否定全稱只需一個反例 B={0}；g 與 f 只差 g(1)，但這就足以破壞可測性。",
            "en": "Refuting a universal statement needs one counterexample B={0}; g differs from f only at 1, and that is enough to break measurability."
          }
        },
        "leaf_basis": "quantifier_negation + image_preimage_def"
      },
      {
        "id": "S09",
        "uses": [
          "L_comp_meas"
        ],
        "before_tex": "u,\\ \\psi\\ \\text{measurable}",
        "after_tex": "(\\psi\\circ u)^{-1}(D)=u^{-1}(\\psi^{-1}(D)),\\ \\ \\psi\\circ u\\ \\text{measurable}",
        "relation": "theorem_application",
        "rule": "L_comp_meas",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_comp_meas（S09.1、S09.2 證明，對任意可測空間成立）：合成的逆像是逆像的合成（順序相反），所以兩個可測映射的合成可測（MKO 段落 2）。",
          "en": "Lemma L_comp_meas (proved in S09.1, S09.2, valid for arbitrary measurable spaces): the preimage under a composite is the composite of preimages (in reverse order), so a composite of two measurable maps is measurable (MKO paragraph 2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_comp_meas"
          ],
          "explanation": {
            "zh-Hant": "需要兩個映射各自可測；只有其中一個可測時不宣稱。",
            "en": "Needs each map to be measurable; nothing is claimed when only one of them is."
          }
        },
        "leaf_basis": "L_comp_meas + composition_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "composition_def",
          "image_preimage_def"
        ],
        "before_tex": "x\\in(\\psi\\circ u)^{-1}(D)",
        "after_tex": "x\\in u^{-1}(\\psi^{-1}(D))",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈(ψ∘u)⁻¹(D) ⇔ ψ(u(x))∈D ⇔ u(x)∈ψ⁻¹(D) ⇔ x∈u⁻¹(ψ⁻¹(D))。每一環都是逆像或合成定義的展開，依外延性得集合相等。",
          "en": "x∈(ψ∘u)⁻¹(D) ⇔ ψ(u(x))∈D ⇔ u(x)∈ψ⁻¹(D) ⇔ x∈u⁻¹(ψ⁻¹(D)). Each link unfolds the definition of preimage or composition, and extensionality gives equality of sets."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "composition_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "整條鏈雙向成立；不需要任何映射可逆。",
            "en": "The whole chain holds both ways; no map needs to be invertible."
          }
        },
        "leaf_basis": "image_preimage_def + composition_def + set_extensionality"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "S01"
        ],
        "before_tex": "D\\in\\mathcal{C}",
        "after_tex": "\\psi^{-1}(D)\\in\\mathcal{B}\\ \\Rightarrow\\ u^{-1}(\\psi^{-1}(D))\\in\\mathcal{A}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任取 D∈𝒞。ψ 可測給 ψ⁻¹(D)∈ℬ；把 u 的可測性用在這個集合上給 u⁻¹(ψ⁻¹(D))∈𝒜。由 S09.1 這就是 (ψ∘u)⁻¹(D)。D 任意，所以 ψ∘u 可測。S09.1 與 S09.2 證得引理 L_comp_meas。",
          "en": "Take any D∈𝒞. Measurability of ψ gives ψ⁻¹(D)∈ℬ; measurability of u applied to that set gives u⁻¹(ψ⁻¹(D))∈𝒜. By S09.1 this is (ψ∘u)⁻¹(D). D was arbitrary, so ψ∘u is measurable. S09.1 and S09.2 prove Lemma L_comp_meas."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "forall_elim",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "兩次代入特例；中間集合 ψ⁻¹(D) 必須在 ℬ 中，u 的可測性才能用上。",
            "en": "Two instantiations; the intermediate set ψ⁻¹(D) must lie in ℬ for the measurability of u to apply."
          }
        },
        "leaf_basis": "forall_elim + forall_intro + def_unfold"
      },
      {
        "id": "S10",
        "uses": [
          "G4",
          "G2",
          "image_preimage_def"
        ],
        "before_tex": "\\varphi:\\ 0\\mapsto0,\\ 1\\mapsto1,\\ 2\\mapsto1",
        "after_tex": "\\varphi^{-1}(\\{0\\})=\\{0\\},\\ \\varphi^{-1}(\\{1\\})=\\{1,2\\}\\in\\Sigma_Y:\\ \\varphi\\ \\text{measurable}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Σ_Z 的四個成員：∅ 與 Z 由 S04 的一般事實處理（逆像為 ∅、Y）；φ⁻¹({0})={0}∈Σ_Y；φ⁻¹({1})={1,2}∈Σ_Y。全部通過，φ 可測。",
          "en": "The four members of Σ_Z: ∅ and Z are handled by the general fact of S04 (preimages ∅ and Y); φ⁻¹({0})={0}∈Σ_Y; φ⁻¹({1})={1,2}∈Σ_Y. All pass, so φ is measurable."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "image_preimage_def",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "完整列舉 Σ_Z 的四個成員。",
            "en": "All four members of Σ_Z are enumerated."
          }
        },
        "leaf_basis": "case_analysis + image_preimage_def"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "S10",
          "S09",
          "G3",
          "G4"
        ],
        "before_tex": "\\varphi\\circ f",
        "after_tex": "\\varphi\\circ f:\\ 0\\mapsto0,\\ 1\\mapsto0,\\ 2\\mapsto1,\\ 3\\mapsto1,\\ \\ \\text{measurable}",
        "relation": "theorem_application",
        "rule": "L_comp_meas",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f（S07）與 φ（S10）都可測，L_comp_meas 給 φ∘f 可測。讀回數值：φ(f(0))=φ(0)=0、φ(f(1))=0、φ(f(2))=φ(1)=1、φ(f(3))=φ(2)=1。逐一核對逆像：(φ∘f)⁻¹({1})=f⁻¹(φ⁻¹({1}))=f⁻¹({1,2})={2,3}∈Σ_X，與引理一致。",
          "en": "f (S07) and φ (S10) are both measurable, so L_comp_meas makes φ∘f measurable. Reading back the values: φ(f(0))=φ(0)=0, φ(f(1))=0, φ(f(2))=φ(1)=1, φ(f(3))=φ(2)=1. Checking a preimage directly: (φ∘f)⁻¹({1})=f⁻¹(φ⁻¹({1}))=f⁻¹({1,2})={2,3}∈Σ_X, in agreement with the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_comp_meas",
            "S07",
            "S10"
          ],
          "explanation": {
            "zh-Hant": "引理的兩個前提都已核；直接讀回是獨立的交叉核對。",
            "en": "Both hypotheses of the lemma have been checked; the direct read-back is an independent cross-check."
          }
        },
        "leaf_basis": "L_comp_meas + composition_def"
      },
      {
        "id": "S12",
        "uses": [
          "L_indicator",
          "p_indicator_def"
        ],
        "before_tex": "\\mathbf{1}_E:(X,\\Sigma)\\to(\\{0,1\\},\\mathcal{P}(\\{0,1\\}))",
        "after_tex": "\\mathbf{1}_E\\ \\text{measurable}\\iff E\\in\\Sigma",
        "relation": "theorem_application",
        "rule": "L_indicator",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_indicator（S12.1、S12.2 證明，對任意可測空間成立）：指示函數的可測性恰由事件 E（及其補集）是否在來源 σ-代數中決定（MKO 段落 1）。",
          "en": "Lemma L_indicator (proved in S12.1, S12.2, valid for any measurable space): measurability of an indicator is controlled exactly by whether the event E (and its complement) lies in the source sigma-algebra (MKO paragraph 1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_indicator"
          ],
          "explanation": {
            "zh-Hant": "目標 σ-代數取冪集；若目標只取 {∅,{0,1}}，任何函數都可測，引理的形式會不同。",
            "en": "The target sigma-algebra is the power set; with only {∅,{0,1}} as target every function would be measurable, and the lemma would read differently."
          }
        },
        "leaf_basis": "L_indicator + p_indicator_def"
      },
      {
        "id": "S12.1",
        "uses": [
          "p_indicator_def",
          "image_preimage_def"
        ],
        "before_tex": "\\mathbf{1}_E^{-1}(C),\\ C\\subseteq\\{0,1\\}",
        "after_tex": "\\emptyset\\mapsto\\emptyset,\\ \\{1\\}\\mapsto E,\\ \\{0\\}\\mapsto X\\setminus E,\\ \\{0,1\\}\\mapsto X",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依指示函數的定義，1_E(x)=1 ⇔ x∈E，1_E(x)=0 ⇔ x∉E。所以四個子集的逆像依次是 ∅、E、X∖E、X。",
          "en": "By the definition of the indicator, 1_E(x)=1 ⇔ x∈E and 1_E(x)=0 ⇔ x∉E. So the preimages of the four subsets are ∅, E, X∖E and X."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "p_indicator_def",
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "四個逆像都由定義雙向確定。",
            "en": "All four preimages are determined both ways by the definitions."
          }
        },
        "leaf_basis": "image_preimage_def + p_indicator_def + set_operations_def"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "mko_sigma_def"
        ],
        "before_tex": "\\{\\emptyset,E,X\\setminus E,X\\}\\subseteq\\Sigma",
        "after_tex": "\\iff E\\in\\Sigma",
        "relation": "equivalent",
        "rule": "iff_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇒：若 1_E 可測，則 E=1_E⁻¹({1})∈Σ。⇐：若 E∈Σ，σ-代數含 ∅ 與 X（X=∅ 的補集），並對補集封閉給 X∖E∈Σ，所以四個逆像都在 Σ 中。兩個方向合起來證得引理 L_indicator。",
          "en": "⇒: if 1_E is measurable then E=1_E⁻¹({1})∈Σ. ⇐: if E∈Σ, the sigma-algebra contains ∅ and X (the complement of ∅) and, being closed under complements, X∖E∈Σ, so all four preimages lie in Σ. The two directions together prove Lemma L_indicator."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "iff_intro",
            "mko_sigma_def"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都有理由：⇒ 用一個逆像，⇐ 用 σ-代數的補集封閉。",
            "en": "Both directions have reasons: ⇒ uses one preimage, ⇐ uses closure of the sigma-algebra under complements."
          }
        },
        "leaf_basis": "iff_intro + mko_sigma_def"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "S11",
          "G1",
          "G5"
        ],
        "before_tex": "\\varphi\\circ f=\\mathbf{1}_{\\{2,3\\}},\\ k=\\mathbf{1}_{\\{1,2,3\\}}",
        "after_tex": "\\mathbf{1}_{\\{2,3\\}}\\ \\text{measurable};\\ \\mathbf{1}_{\\{0\\}},\\ k\\ \\text{not measurable}",
        "relation": "implies",
        "rule": "L_indicator",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S11 的 φ∘f 在 {2,3} 上取 1、其餘取 0，正是 1_{{2,3}}；{2,3}∈Σ_X，由引理可測，與 S11 一致。{0}∉Σ_X，所以 1_{{0}} 不可測。G5 的 k 在 {1,2,3} 上取 1，即 k=1_{{1,2,3}}，而 {1,2,3}∉Σ_X，所以 k 不可測。",
          "en": "The map φ∘f of S11 is 1 on {2,3} and 0 elsewhere, exactly 1_{{2,3}}; {2,3}∈Σ_X, so the lemma makes it measurable, consistent with S11. {0}∉Σ_X, so 1_{{0}} is not measurable. The map k of G5 is 1 on {1,2,3}, so k=1_{{1,2,3}}, and {1,2,3}∉Σ_X, so k is not measurable."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_indicator",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "判定 E 是否在 Σ_X 中，只需與 Σ_X 的四個成員逐一比較。",
            "en": "Deciding whether E lies in Σ_X needs only a comparison with the four members of Σ_X."
          }
        },
        "leaf_basis": "L_indicator + set_extensionality"
      },
      {
        "id": "S14",
        "uses": [
          "G5",
          "G2",
          "image_preimage_def",
          "S04"
        ],
        "before_tex": "m:\\ 0\\mapsto0,\\ 1\\mapsto0,\\ 2\\mapsto1,\\ 3\\mapsto1",
        "after_tex": "m\\ \\text{measurable, yet}\\ m(\\{2,3\\})=\\{1\\}\\notin\\Sigma_Y",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逆像：m⁻¹({0})={0,1}∈Σ_X，m⁻¹({1,2})={2,3}∈Σ_X，加上 S04 的 ∅ 與 Y，m 可測。但像集 m({2,3})={1} 不在 Σ_Y 中。所以若誤用「來源可測集的像要在 Σ_Y 中」作檢查，會錯誤地拒絕一個可測函數。",
          "en": "Preimages: m⁻¹({0})={0,1}∈Σ_X and m⁻¹({1,2})={2,3}∈Σ_X, together with ∅ and Y from S04, so m is measurable. But the image m({2,3})={1} is not in Σ_Y. So the mistaken test 'images of measurable source sets must lie in Σ_Y' would wrongly reject a measurable function."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "image_preimage_def",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "可測性由逆像決定；像集不在 Σ_Y 中與可測性無關。",
            "en": "Measurability is decided by preimages; an image outside Σ_Y is irrelevant to it."
          }
        },
        "leaf_basis": "image_preimage_def + case_analysis"
      },
      {
        "id": "S15",
        "uses": [
          "G5",
          "G4",
          "S13",
          "image_preimage_def"
        ],
        "before_tex": "k:X\\to Z,\\ \\Sigma_Z=\\mathcal{P}(Z)",
        "after_tex": "k(A)\\in\\Sigma_Z\\ \\text{for every}\\ A\\in\\Sigma_X,\\ \\text{yet}\\ k^{-1}(\\{0\\})=\\{0\\}\\notin\\Sigma_X",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Σ_Z 是冪集，所以任何像集 k(A)⊆Z 都在 Σ_Z 中，「像集檢查」全部通過。但 k⁻¹({0})={0}∉Σ_X，k 不可測（S13）。所以像集檢查也會錯誤地接受不可測函數。S14 與 S15 合起來說明：像集檢查與可測性互不蘊含，MKO 指出的誤解兩個方向都會出錯。",
          "en": "Σ_Z is the power set, so every image k(A)⊆Z lies in Σ_Z and the 'image test' passes everywhere. But k⁻¹({0})={0}∉Σ_X, so k is not measurable (S13). So the image test can also wrongly accept a non-measurable function. Together S14 and S15 show that neither the image test nor measurability implies the other: the misconception named by the MKO errs in both directions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "image_preimage_def",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "兩個反例各否定一個方向的蘊含。",
            "en": "Each counterexample refutes one direction of implication."
          }
        },
        "leaf_basis": "image_preimage_def + quantifier_negation"
      },
      {
        "id": "S16",
        "uses": [
          "S07",
          "S08",
          "S11",
          "S12",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S07 to S15}",
        "after_tex": "f,\\ \\varphi,\\ \\varphi\\circ f,\\ m\\ \\text{measurable};\\ g,\\ k\\ \\text{not}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：完整列舉逆像得 f、φ、m 可測，g 不可測；合成 φ∘f=1_{{2,3}} 可測；指示函數可測若且唯若事件在 Σ_X 中，所以 k 不可測；像集檢查兩個方向都不可靠。",
          "en": "Summary: complete enumeration of preimages shows f, φ and m measurable and g not; the composite φ∘f=1_{{2,3}} is measurable; an indicator is measurable exactly when its event lies in Σ_X, so k is not; the image test is unreliable in both directions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "S11",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "只合取已證結論；這些結論只針對 G1 到 G5 的實例，兩個引理則對一般可測空間成立。",
            "en": "Only conjoins conclusions already proved; they concern the instances of G1 to G5, while the two lemmas hold for general measurable spaces."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "f 可測（f⁻¹({0})={0,1}、f⁻¹({1,2})={2,3}）；g 不可測（g⁻¹({0})={0}∉Σ_X）。合成可測映射可測，φ∘f=1_{{2,3}} 可測；1_E 可測 ⇔ E∈Σ_X，故 1_{{0}} 與 k=1_{{1,2,3}} 不可測。m 可測但 m({2,3})={1}∉Σ_Y，k 的像全在 Σ_Z 中卻不可測：像集檢查不能代替逆像檢查。",
      "en": "f is measurable (f⁻¹({0})={0,1}, f⁻¹({1,2})={2,3}); g is not (g⁻¹({0})={0}∉Σ_X). Composites of measurable maps are measurable, and φ∘f=1_{{2,3}} is measurable; 1_E is measurable ⇔ E∈Σ_X, so 1_{{0}} and k=1_{{1,2,3}} are not. m is measurable although m({2,3})={1}∉Σ_Y, and k has all images in Σ_Z but is not measurable: an image check cannot replace the preimage check."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "m({2,3})={1} 不在 Σ_Y 中，所以 m 不可測",
          "en": "m({2,3})={1} is not in Σ_Y, so m is not measurable"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "可測性看逆像：m⁻¹({0})={0,1}、m⁻¹({1,2})={2,3} 都在 Σ_X 中（S14）。",
          "en": "Measurability concerns preimages: m⁻¹({0})={0,1} and m⁻¹({1,2})={2,3} are both in Σ_X (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "k 的每個像集都在 Σ_Z 中，所以 k 可測",
          "en": "Every image of k lies in Σ_Z, so k is measurable"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "k⁻¹({0})={0}∉Σ_X（S15）。",
          "en": "k⁻¹({0})={0}∉Σ_X (S15)."
        }
      },
      {
        "claim": {
          "zh-Hant": "g 與 f 只在一個點不同，所以 g 也可測",
          "en": "g differs from f at only one point, so g is measurable too"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "g⁻¹({0})={0}∉Σ_X（S08）；可測性對單點改動並不穩定。",
          "en": "g⁻¹({0})={0}∉Σ_X (S08); measurability is not stable under changing a single point."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 可測，所以對任何測度 μ_X、μ_Y 都有 μ_X(f⁻¹(B))=μ_Y(B)",
          "en": "f is measurable, so μ_X(f⁻¹(B))=μ_Y(B) for any measures μ_X, μ_Y"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "錯誤控制 2 的測度給 1/2≠1/3；可測不推出保測。",
          "en": "The measures of error control 2 give 1/2≠1/3; measurability does not imply measure preservation."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "composition_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合成：對 f:A→B 與 g:B→C，g∘f:A→C 定義為 (g∘f)(a)=g(f(a))。",
          "en": "Composition: for f:A→B and g:B→C, g∘f:A→C is defined by (g∘f)(a)=g(f(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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "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": "mko_measurable_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-measurable-function",
          "sha256": "0c21c3c35f2f6b0451adf573f920b3dd8e797c631bcae8c14bf6bd959dc1571b",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給有限可測空間 (X,Σ_X) 與 (Y,Σ_Y)，總函數 f:X→Y 可測，若每個 B∈Σ_Y 的逆像 f⁻¹(B) 都屬 Σ_X。本輪用完整有限列舉逐一核逆像。",
          "en": "Given finite measurable spaces (X,Σ_X) and (Y,Σ_Y), a total function f:X→Y is measurable if f⁻¹(B) belongs to Σ_X for every B in Σ_Y. This slice checks each preimage by complete finite enumeration."
        }
      },
      {
        "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；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 complete finite carrier permits reduction of the countable obligations to finitely many distinct sets."
        }
      },
      {
        "id": "p_indicator_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對 E⊆X，指示函數 1_E:X→{0,1}：x∈E 時 1_E(x)=1，否則 1_E(x)=0。",
          "en": "For E⊆X, the indicator 1_E:X→{0,1} has 1_E(x)=1 if x∈E and 1_E(x)=0 otherwise."
        }
      },
      {
        "id": "L_comp_meas",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "對任意可測空間，若 u:(A,𝒜)→(B,ℬ) 與 ψ:(B,ℬ)→(C,𝒞) 可測，則 (ψ∘u)⁻¹(D)=u⁻¹(ψ⁻¹(D)) 對每個 D⊆C 成立，且 ψ∘u 可測。",
          "en": "For arbitrary measurable spaces, if u:(A,𝒜)→(B,ℬ) and ψ:(B,ℬ)→(C,𝒞) are measurable, then (ψ∘u)⁻¹(D)=u⁻¹(ψ⁻¹(D)) for every D⊆C, and ψ∘u is measurable."
        }
      },
      {
        "id": "L_indicator",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2"
        ],
        "statement": {
          "zh-Hant": "對任意可測空間 (X,Σ) 與 E⊆X，以 {0,1} 的冪集為目標 σ-代數時，1_E 可測 ⇔ E∈Σ。",
          "en": "For any measurable space (X,Σ) and E⊆X, with the power set of {0,1} as target sigma-algebra, 1_E is measurable ⇔ E∈Σ."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02.1",
        "lhs": "comb(4,2)",
        "rhs": "6"
      },
      {
        "id": "N2",
        "step": "S02.2",
        "lhs": "2**2",
        "rhs": "4"
      },
      {
        "id": "N3",
        "lhs": "1/2+1/2",
        "rhs": "1"
      },
      {
        "id": "N4",
        "lhs": "1/3+2/3",
        "rhs": "1"
      },
      {
        "id": "N5",
        "lhs": "1/2",
        "relation": "!=",
        "rhs": "1/3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 到 G5 指定的有限可測空間與完整總映射表（MKO 假設 a1）。完整有限列舉只證這些實例；合成引理與指示函數引理對任意可測空間證明。可測不推出連續、可積或保測（a2）。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the finite measurable spaces and complete total-map tables of G1 to G5 (MKO assumption a1). Complete finite enumeration establishes only these instances; the composition lemma and the indicator lemma are proved for arbitrary measurable spaces. Measurability does not imply continuity, integrability or measure preservation (a2). The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "(X,\\Sigma_X)",
        "來源可測空間，X={0,1,2,3}",
        "the source measurable space, X={0,1,2,3}"
      ],
      [
        "(Y,\\Sigma_Y)",
        "目標可測空間，Y={0,1,2}",
        "the target measurable space, Y={0,1,2}"
      ],
      [
        "(Z,\\Sigma_Z)",
        "Z={0,1}，Σ_Z 為冪集",
        "Z={0,1} with Σ_Z the power set"
      ],
      [
        "f,g,m:X\\to Y",
        "待檢的總映射",
        "the total maps to be checked"
      ],
      [
        "\\varphi:Y\\to Z,\\ k:X\\to Z",
        "合成用映射與像集陷阱例子",
        "the map used for composition and the image-trap example"
      ],
      [
        "u^{-1}(B)",
        "逆像 {x ∣ u(x)∈B}",
        "the preimage {x ∣ u(x)∈B}"
      ],
      [
        "\\mathbf{1}_E",
        "E 的指示函數",
        "the indicator function of E"
      ]
    ],
    "md_error_controls": [
      {
        "title": "檢查逆像，不是像集 / Check preimages, not images",
        "zh": "MKO 指出的誤解之一是檢查像集而非逆像。S14 的 m 可測，但像集 m({2,3})={1} 不在 Σ_Y 中；S15 的 k 所有像集都在 Σ_Z 中，卻因 k⁻¹({0})={0}∉Σ_X 而不可測。可測性的方向是：目標可測集取逆像，回到來源 σ-代數。",
        "en": "One misconception named by the MKO is to check images instead of preimages. The map m of S14 is measurable although its image m({2,3})={1} is not in Σ_Y; the map k of S15 has all its images in Σ_Z yet is not measurable because k⁻¹({0})={0}∉Σ_X. The direction of measurability is: take preimages of measurable target sets and land in the source sigma-algebra."
      },
      {
        "title": "可測不是保測 / Measurable is not measure preserving",
        "zh": "MKO 假設 a2 提醒可測不推出連續、可積或保測。例如在 Σ_X 上取 μ_X({0,1})=μ_X({2,3})=1/2、在 Σ_Y 上取 μ_Y({0})=1/3、μ_Y({1,2})=2/3（兩者都由原子權重給出，總和 1，N3、N4），f 可測，但 μ_X(f⁻¹({0}))=μ_X({0,1})=1/2≠1/3=μ_Y({0})（N5），所以 f 不保測。",
        "en": "MKO assumption a2 warns that measurability does not imply continuity, integrability or measure preservation. For example, take μ_X({0,1})=μ_X({2,3})=1/2 on Σ_X and μ_Y({0})=1/3, μ_Y({1,2})=2/3 on Σ_Y (both given by atom weights with total 1, N3, N4); f is measurable, but μ_X(f⁻¹({0}))=μ_X({0,1})=1/2≠1/3=μ_Y({0}) (N5), so f is not measure preserving."
      },
      {
        "title": "先核 σ-代數，再談可測 / Check the sigma-algebras before measurability",
        "zh": "可測性的定義預設兩邊都是 σ-代數（S02）。若把 Σ_X 誤寫成 {∅,{0,1},X}（缺 {0,1} 的補集 {2,3}），它就不是 σ-代數，S06 的結論「f⁻¹({1,2})={2,3}∈Σ_X」也會失敗；問題出在集合族而不是函數。",
        "en": "The definition of measurability presupposes sigma-algebras on both sides (S02). If Σ_X were miswritten as {∅,{0,1},X} (missing {2,3}, the complement of {0,1}), it would not be a sigma-algebra, and the conclusion of S06 that f⁻¹({1,2})={2,3}∈Σ_X would fail; the fault would lie in the family, not in the function."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "錯誤控制 2 的 μ_X、μ_Y 以原子權重給出；它們是測度的事實沿用 σ-代數與測度空間條目的原子構造，未在本稿重證。",
        "en": "The measures μ_X and μ_Y of error control 2 are given by atom weights; that they are measures follows the atom construction of the sigma-algebra and measure-space entry and is not re-proved here."
      },
      {
        "zh-Hant": "簡單函數、積分與測度收斂（MKO 的 formal next）不在範圍內。",
        "en": "Simple functions, integration and convergence of measures (the MKO's formal next) are out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "b30cc3fb4fd89f6219b80c50db6bfa4bdd2524d76e965a40117637009646bdf0",
    "original_source_sha256": "b30cc3fb4fd89f6219b80c50db6bfa4bdd2524d76e965a40117637009646bdf0",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}