Main derivation
S01 · definition
Before
u:(X,\Sigma_X)\to(Y,\Sigma_Y)\ \text{measurable}After
\forall B\in\Sigma_Y:\ u^{-1}(B)\in\Sigma_XConditions: none
Rule: mko_measurable_def · Uses: mko_measurable_def
Leaf basis: mko_measurable_def + def_unfold
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.
S02 · implies
Before
\Sigma_X,\ \Sigma_Y,\ \Sigma_Z
After
\text{each is a}\ \sigma\text{-algebra on its carrier}Conditions: none
Rule: mko_sigma_def · Uses: G1, G2, G4, mko_sigma_def
Leaf basis: mko_sigma_def + set_operations_def
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.
S03 · implies
Before
f:\ 0\mapsto0,\ 1\mapsto0,\ 2\mapsto1,\ 3\mapsto2
After
f:X\to Y\ \text{total}Conditions: none
Rule: function_def · Uses: G3, function_def
Leaf basis: function_def
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).
S04 · implies
Before
u:X\to Y\ \text{total}After
u^{-1}(\emptyset)=\emptyset\in\Sigma_X,\quad u^{-1}(Y)=X\in\Sigma_XConditions: none
Rule: image_preimage_def · Uses: image_preimage_def, S03
Leaf basis: image_preimage_def + function_def
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.
S05 · equivalent
Before
f^{-1}(\{0\})After
\{x\mid f(x)=0\}=\{0,1\}\in\Sigma_XConditions: none
Rule: image_preimage_def · Uses: G3, image_preimage_def, G1
Leaf basis: image_preimage_def + set_builder
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.
S06 · equivalent
Before
f^{-1}(\{1,2\})After
\{x\mid f(x)\in\{1,2\}\}=\{2,3\}\in\Sigma_XConditions: none
Rule: image_preimage_def · Uses: G3, image_preimage_def, G1
Leaf basis: image_preimage_def + set_builder
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}.
S07 · implies
Before
B\in\{\emptyset,\{0\},\{1,2\},Y\}After
f^{-1}(B)\in\Sigma_X\ \text{for every}\ B\in\Sigma_Y:\ f\ \text{measurable}Conditions: none
Rule: case_analysis · Uses: S01, S04, S05, S06, G2
Leaf basis: case_analysis + forall_intro
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.
S08 · implies
Before
g^{-1}(\{0\})After
\{0\}\notin\Sigma_X:\ g\ \text{not measurable}Conditions: none
Rule: quantifier_negation · Uses: G3, G1, S01, image_preimage_def
Leaf basis: quantifier_negation + image_preimage_def
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.
S09 · theorem_application
Before
u,\ \psi\ \text{measurable}After
(\psi\circ u)^{-1}(D)=u^{-1}(\psi^{-1}(D)),\ \ \psi\circ u\ \text{measurable}Conditions: none
Rule: L_comp_meas · Uses: L_comp_meas
Leaf basis: L_comp_meas + composition_def
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).
S10 · implies
Before
\varphi:\ 0\mapsto0,\ 1\mapsto1,\ 2\mapsto1
After
\varphi^{-1}(\{0\})=\{0\},\ \varphi^{-1}(\{1\})=\{1,2\}\in\Sigma_Y:\ \varphi\ \text{measurable}Conditions: none
Rule: case_analysis · Uses: G4, G2, image_preimage_def
Leaf basis: case_analysis + image_preimage_def
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.
S11 · theorem_application
Before
\varphi\circ f
After
\varphi\circ f:\ 0\mapsto0,\ 1\mapsto0,\ 2\mapsto1,\ 3\mapsto1,\ \ \text{measurable}Conditions: none
Rule: L_comp_meas · Uses: S07, S10, S09, G3, G4
Leaf basis: L_comp_meas + composition_def
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.
S12 · theorem_application
Before
\mathbf{1}_E:(X,\Sigma)\to(\{0,1\},\mathcal{P}(\{0,1\}))After
\mathbf{1}_E\ \text{measurable}\iff E\in\SigmaConditions: none
Rule: L_indicator · Uses: L_indicator, p_indicator_def
Leaf basis: L_indicator + p_indicator_def
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).
S13 · implies
Before
\varphi\circ f=\mathbf{1}_{\{2,3\}},\ k=\mathbf{1}_{\{1,2,3\}}After
\mathbf{1}_{\{2,3\}}\ \text{measurable};\ \mathbf{1}_{\{0\}},\ k\ \text{not measurable}Conditions: none
Rule: L_indicator · Uses: S12, S11, G1, G5
Leaf basis: L_indicator + set_extensionality
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.
S14 · implies
Before
m:\ 0\mapsto0,\ 1\mapsto0,\ 2\mapsto1,\ 3\mapsto1
After
m\ \text{measurable, yet}\ m(\{2,3\})=\{1\}\notin\Sigma_YConditions: none
Rule: image_preimage_def · Uses: G5, G2, image_preimage_def, S04
Leaf basis: image_preimage_def + case_analysis
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.
S15 · implies
Before
k:X\to Z,\ \Sigma_Z=\mathcal{P}(Z)After
k(A)\in\Sigma_Z\ \text{for every}\ A\in\Sigma_X,\ \text{yet}\ k^{-1}(\{0\})=\{0\}\notin\Sigma_XConditions: none
Rule: image_preimage_def · Uses: G5, G4, S13, image_preimage_def
Leaf basis: image_preimage_def + quantifier_negation
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.
S16 · implies
Before
\text{S07 to S15}After
f,\ \varphi,\ \varphi\circ f,\ m\ \text{measurable};\ g,\ k\ \text{not}Conditions: none
Rule: and_intro_elim · Uses: S07, S08, S11, S12, S14, S15
Leaf basis: and_intro_elim
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.
Substeps and lemmas
S02.1 · implies
Before
\Sigma_X=\{\emptyset,\{0,1\},\{2,3\},X\}After
\emptyset\in\Sigma_X;\ \emptyset^c=X,\ \{0,1\}^c=\{2,3\};\ \{0,1\}\cup\{2,3\}=XConditions: none
Rule: set_operations_def · Uses: G1, set_operations_def
Leaf basis: set_operations_def + case_analysis
∅∈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.
S02.2 · implies
Before
\Sigma_Y=\{\emptyset,\{0\},\{1,2\},Y\},\ \Sigma_Z=\mathcal{P}(Z)After
\Sigma_Y,\ \Sigma_Z\ \text{are}\ \sigma\text{-algebras}Conditions: none
Rule: set_operations_def · Uses: G2, G4, set_operations_def
Leaf basis: set_operations_def + case_analysis
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.
S09.1 · equivalent
Before
x\in(\psi\circ u)^{-1}(D)After
x\in u^{-1}(\psi^{-1}(D))Conditions: none
Rule: image_preimage_def · Uses: composition_def, image_preimage_def
Leaf basis: image_preimage_def + composition_def + set_extensionality
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.
S09.2 · implies
Before
D\in\mathcal{C}After
\psi^{-1}(D)\in\mathcal{B}\ \Rightarrow\ u^{-1}(\psi^{-1}(D))\in\mathcal{A}Conditions: none
Rule: forall_elim · Uses: S09.1, S01
Leaf basis: forall_elim + forall_intro + def_unfold
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.
S12.1 · equivalent
Before
\mathbf{1}_E^{-1}(C),\ C\subseteq\{0,1\}After
\emptyset\mapsto\emptyset,\ \{1\}\mapsto E,\ \{0\}\mapsto X\setminus E,\ \{0,1\}\mapsto XConditions: none
Rule: image_preimage_def · Uses: p_indicator_def, image_preimage_def
Leaf basis: image_preimage_def + p_indicator_def + set_operations_def
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.
S12.2 · equivalent
Before
\{\emptyset,E,X\setminus E,X\}\subseteq\SigmaAfter
\iff E\in\Sigma
Conditions: none
Rule: iff_intro · Uses: S12.1, mko_sigma_def
Leaf basis: iff_intro + mko_sigma_def
⇒: 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.
Error controls and negative controls
Check preimages, not images
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.
Measurable is not measure preserving
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.
Check the sigma-algebras before measurability
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.
- m({2,3})={1} is not in Σ_Y, so m is not measurable →
reject — Measurability concerns preimages: m⁻¹({0})={0,1} and m⁻¹({1,2})={2,3} are both in Σ_X (S14). - Every image of k lies in Σ_Z, so k is measurable →
reject — k⁻¹({0})={0}∉Σ_X (S15). - g differs from f at only one point, so g is measurable too →
reject — g⁻¹({0})={0}∉Σ_X (S08); measurability is not stable under changing a single point. - f is measurable, so μ_X(f⁻¹(B))=μ_Y(B) for any measures μ_X, μ_Y →
reject — The measures of error control 2 give 1/2≠1/3; measurability does not imply measure preservation.