Main derivation
S01 · definition
Before
(X,\Sigma,\mu)\ \text{is a measure space}After
\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)}Conditions: none
Rule: mko_sigma_def · Uses: mko_sigma_def
Leaf basis: mko_sigma_def + def_unfold
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.
S02 · implies
Before
A_1=\{0\},\ A_2=\{1,2\},\ A_3=\{3\}After
A_i\neq\emptyset,\ \ A_i\cap A_j=\emptyset\ (i\neq j),\ \ A_1\cup A_2\cup A_3=X
Conditions: none
Rule: set_operations_def · Uses: G1
Leaf basis: set_operations_def + set_extensionality
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.
S03 · implies
Before
\Sigma=\{E_I\mid I\subseteq\{1,2,3\}\}After
|\Sigma|=8:\ \emptyset,\{0\},\{1,2\},\{3\},\{0,1,2\},\{0,3\},\{1,2,3\},XConditions: none
Rule: L_rep_unique · Uses: S02, G2, L_rep_unique
Leaf basis: L_rep_unique + finite_counting_basics
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.
S04 · implies
Before
I=\emptyset
After
E_{\emptyset}=\emptyset\in\SigmaConditions: none
Rule: def_unfold · Uses: S03
Leaf basis: def_unfold + set_operations_def
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.
S05 · theorem_application
Before
E_I\in\Sigma
After
X\setminus E_I=E_{I'}\in\Sigma,\ \ I'=\{1,2,3\}\setminus IConditions: none
Rule: L_complement · Uses: S03, L_complement
Leaf basis: L_complement + set_extensionality
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.
S06 · theorem_application
Before
E_I,\ E_J\in\Sigma
After
E_I\cup E_J=E_{I\cup J}\in\SigmaConditions: none
Rule: L_union · Uses: S03, L_union
Leaf basis: L_union + set_extensionality
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.
S07 · theorem_application
Before
(E_k)_{k\ge1},\ E_k\in\SigmaAfter
\bigcup_{k\ge1}E_k=\bigcup_{E\in D}E\in\Sigma,\ \ D=\{E_k\mid k\ge1\}Conditions: none
Rule: L_countable_reduce · Uses: S03, S06, L_countable_reduce
Leaf basis: L_countable_reduce + L_union
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.
S08 · implies
Before
\emptyset\in\Sigma,\ \text{complements},\ \text{countable unions}After
\Sigma\ \text{is a}\ \sigma\text{-algebra on}\ XConditions: none
Rule: and_intro_elim · Uses: S04, S05, S07, S01
Leaf basis: and_intro_elim
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.
S09 · definition
Before
\mu(E_I)=\sum_{i\in I}m_iAfter
\mu:\ 0,\tfrac12,\tfrac13,\tfrac16,\tfrac56,\tfrac23,\tfrac12,1\ \text{on}\ \emptyset,\{0\},\{1,2\},\{3\},\{0,1,2\},\{0,3\},\{1,2,3\},XConditions: none
Rule: def_intro · Uses: S03, G3, L_rep_unique
Leaf basis: def_intro + finite_sum_rules + numeric_evaluation
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).
S10 · implies
Before
m_i\ge 0
After
\mu(E)\ge 0\ \ (E\in\Sigma),\quad \mu(\emptyset)=\sum_{i\in\emptyset}m_i=0Conditions: none
Rule: finite_sum_rules · Uses: S09, finite_sum_rules
Leaf basis: finite_sum_rules + order_basics
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.
S11 · theorem_application
Before
E,F\in\Sigma,\ E\cap F=\emptyset
After
\mu(E\cup F)=\mu(E)+\mu(F)
Conditions: E\cap F=\emptyset
Rule: L_fin_add · Uses: S09, L_fin_add
Leaf basis: L_fin_add + finite_sum_rules
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.
S12 · theorem_application
Before
(E_k)_{k\ge1}\subseteq\Sigma\ \text{pairwise disjoint}After
\mu\Big(\bigcup_{k\ge1}E_k\Big)=\sum_{k\ge1}\mu(E_k)Conditions: E_k\cap E_l=\emptyset\ (k\neq l)
Rule: L_ctbl_add · Uses: S10, S11, L_ctbl_add
Leaf basis: L_ctbl_add + L_fin_add
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.
S13 · implies
Before
\mu(X)=\tfrac12+\tfrac13+\tfrac16
After
\mu(X)=1:\ (X,\Sigma,\mu)\ \text{is a probability space}Conditions: none
Rule: numeric_evaluation · Uses: S08, S10, S12, S09
Leaf basis: numeric_evaluation + and_intro_elim
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.
S14 · implies
Before
\mathcal{G}=\{\emptyset,\{0\},\{1\},X\}After
X\setminus\{0\}=\{1,2,3\}\notin\mathcal{G}:\ \mathcal{G}\ \text{is not a}\ \sigma\text{-algebra}Conditions: none
Rule: set_operations_def · Uses: G4, S01
Leaf basis: set_operations_def + quantifier_negation
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.
S15 · implies
Before
\nu=\mu\ \text{except}\ \nu(\{0,1,2\})=1After
\nu(\{0\}\cup\{1,2\})=1\neq\tfrac56=\nu(\{0\})+\nu(\{1,2\})Conditions: \{0\}\cap\{1,2\}=\emptyset
Rule: numeric_evaluation · Uses: G4, S09, S01
Leaf basis: numeric_evaluation + quantifier_negation
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.
Substeps and lemmas
S03.1 · implies
Before
E_I=E_J
After
I=J
Conditions: A_i\neq\emptyset; A_i\cap A_j=\emptyset\ (i\neq j)
Rule: contradiction · Uses: S02, contradiction
Leaf basis: contradiction + set_operations_def + set_extensionality
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.
S03.2 · implies
Before
I\mapsto E_I\ \text{injective on}\ \mathcal{P}(\{1,2,3\})After
|\Sigma|=2^3=8
Conditions: none
Rule: finite_counting_basics · Uses: S03.1, finite_counting_basics, G2
Leaf basis: finite_counting_basics + numeric_evaluation
{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.
S05.1 · implies
Before
x\in X\setminus E_I
After
x\in E_{I'}Conditions: none
Rule: set_operations_def · Uses: S02, set_operations_def
Leaf basis: set_operations_def + subset_def
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'}.
S05.2 · implies
Before
x\in E_{I'}After
x\in X\setminus E_I,\ \text{hence}\ X\setminus E_I=E_{I'}Conditions: none
Rule: set_extensionality · Uses: S02, S05.1, set_extensionality
Leaf basis: set_extensionality + set_operations_def + subset_def
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.
S06.1 · equivalent
Before
x\in E_I\cup E_J
After
x\in E_{I\cup J}Conditions: none
Rule: set_operations_def · Uses: set_operations_def
Leaf basis: set_operations_def + set_extensionality
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.
S07.1 · implies
Before
D=\{E_k\mid k\ge1\}After
D\subseteq\Sigma,\ \ |D|\le 8
Conditions: none
Rule: finite_counting_basics · Uses: S03, subset_def, finite_counting_basics
Leaf basis: subset_def + finite_counting_basics
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.
S07.2 · equivalent
Before
x\in\bigcup_{k\ge1}E_kAfter
x\in\bigcup_{E\in D}EConditions: none
Rule: set_extensionality · Uses: S07.1, set_extensionality
Leaf basis: set_extensionality + set_operations_def
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.
S07.3 · implies
Before
D\subseteq\Sigma\ \text{finite}After
\bigcup_{E\in D}E\in\SigmaConditions: none
Rule: nat_induction · Uses: S07.2, S04, S06, nat_induction
Leaf basis: nat_induction + L_union
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.
S11.1 · implies
Before
E=E_I,\ F=E_J,\ E\cap F=\emptyset
After
I\cap J=\emptyset
Conditions: A_i\neq\emptyset
Rule: set_operations_def · Uses: S02, S03
Leaf basis: set_operations_def + contradiction
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=∅.
S11.2 · implies
Before
I\cap J=\emptyset
After
\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_jConditions: I\cap J=\emptyset
Rule: finite_sum_rules · Uses: S11.1, S06, S09, finite_sum_rules
Leaf basis: finite_sum_rules + L_union
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).
S12.1 · implies
Before
K=\{k\mid E_k\neq\emptyset\},\ E_k=E_{I_k}After
I_k\neq\emptyset,\ I_k\cap I_l=\emptyset\ (k\neq l\in K),\ \ |K|\le 3
Conditions: none
Rule: finite_counting_basics · Uses: S11.1, S02, finite_counting_basics
Leaf basis: finite_counting_basics + injective_surjective_def
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.
S12.2 · implies
Before
\sum_{k\ge1}\mu(E_k)After
\sum_{k\in K}\mu(E_k)Conditions: |K|\le 3
Rule: p_finite_support_series · Uses: S12.1, S10, p_finite_support_series
Leaf basis: p_finite_support_series + finite_sum_rules
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.
S12.3 · implies
Before
\bigcup_{k\ge1}E_kAfter
\mu\Big(\bigcup_{k\in K}E_k\Big)=\sum_{k\in K}\mu(E_k)Conditions: none
Rule: L_fin_add · Uses: S12.2, S11, S06, set_extensionality
Leaf basis: L_fin_add + set_extensionality + nat_induction
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.