LeanMachineLearning

ProbabilityTheory.indepFun_cond_of_indepFun🔗

Lemma

No docstring.

🔗theorem
ProbabilityTheory.indepFun_cond_of_indepFun.{u_6, u_7, u_8} {α : Type u_6} {β : Type u_7} {γ : Type u_8} { : MeasurableSpace α} { : MeasurableSpace β} { : MeasurableSpace γ} {μ : MeasureTheory.Measure α} {X : α β} {Y : α γ} (hXY : IndepFun X Y μ) (hY : Measurable Y) {s : Set γ} (hs : MeasurableSet s) : IndepFun X Y μ[|Y ⁻¹' s]
ProbabilityTheory.indepFun_cond_of_indepFun.{u_6, u_7, u_8} {α : Type u_6} {β : Type u_7} {γ : Type u_8} { : MeasurableSpace α} { : MeasurableSpace β} { : MeasurableSpace γ} {μ : MeasureTheory.Measure α} {X : α β} {Y : α γ} (hXY : IndepFun X Y μ) (hY : Measurable Y) {s : Set γ} (hs : MeasurableSet s) : IndepFun X Y μ[|Y ⁻¹' s]

Code

lemma indepFun_cond_of_indepFun {α β γ : Type*} {mα : MeasurableSpace α} {mβ : MeasurableSpace β}
    {mγ : MeasurableSpace γ} {μ : Measure α}
    {X : α → β} {Y : α → γ} (hXY : X ⟂ᵢ[μ] Y) (hY : Measurable Y) {s : Set γ}
    (hs : MeasurableSet s) :
    X ⟂ᵢ[μ[|Y ⁻¹' s]] Y
Proof
by
  by_cases h_zero : μ[|Y ⁻¹' s] = 0
  · simp [h_zero]
  rw [cond_eq_zero] at h_zero
  push Not at h_zero -- `h_zero : μ (Y ⁻¹' s) ≠ ⊤ ∧ μ (Y ⁻¹' s) ≠ 0`
  rw [indepFun_iff_measure_inter_preimage_eq_mul] at hXY ⊢
  intro u t hu ht
  rw [cond_apply (hs.preimage hY), cond_apply (hs.preimage hY), cond_apply (hs.preimage hY)]
  have h_eq : Y ⁻¹' s ∩ (X ⁻¹' u ∩ Y ⁻¹' t) = X ⁻¹' u ∩ Y ⁻¹' (s ∩ t) := by grind
  have hsu : μ (X ⁻¹' u ∩ Y ⁻¹' s) = μ (X ⁻¹' u) * μ (Y ⁻¹' s) := hXY u s hu hs
  rw [Set.inter_comm] at hsu
  have hust : μ (X ⁻¹' u ∩ Y ⁻¹' (s ∩ t)) = μ (X ⁻¹' u) * μ (Y ⁻¹' (s ∩ t)) :=
    hXY u (s ∩ t) hu (hs.inter ht)
  rw [hsu, h_eq, hust]
  simp_rw [mul_assoc]
  congr 1
  rw [← mul_assoc (μ (Y ⁻¹' s)), ENNReal.mul_inv_cancel h_zero.2 h_zero.1, one_mul]
  congr

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Meaning last changed in v4.34.0-rc2-1-g439785b (2026-08-23), the 3th recorded change.

Self-contained, with its dependencies inlined and proofs replaced by sorry: download the raw file · open it in the Lean web editor.

Dependency graph

Nothing to draw. Its statement rests on no other declaration in this project, and names nothing from a package left unaudited — so the graph is this declaration alone. That is the answer, not a missing picture.

Audit surface: 0 project declarations, 8 external constants

✓ Proved: no sorry anywhere in its closure

This is the tool's own reading of one build's recorded axioms, and it is not robust against an author who wants it to pass. Checking meant to be relied on should go through Comparator, which replays the proof through the kernel from an export against an explicit list of permitted axioms.