`ProbabilityTheory.hasCondDistrib_prod_right_iff`🔗

Lemma

No docstring.

theorem

ProbabilityTheory.hasCondDistrib_prod_right_iff.{u_1, u_2, u_3, u_4}
  {α : Type u_1} {β : Type u_2} {γ : Type u_3} {Ω : Type u_4}
  {mα : MeasurableSpace α} {mβ : MeasurableSpace β}
  {mγ : MeasurableSpace γ} {mΩ : MeasurableSpace Ω}
  {μ : MeasureTheory.Measure α} {κ : Kernel β Ω}
  [MeasureTheory.IsFiniteMeasure μ] [IsFiniteKernel κ] (X : α → β)
  (Y : α → Ω) {f : β → γ} (hf : Measurable f) :
  HasCondDistrib Y (fun a => (X a, f (X a))) (Kernel.prodMkRight γ κ)
      μ ↔
    HasCondDistrib Y X κ μ
ProbabilityTheory.hasCondDistrib_prod_right_iff.{u_1,
    u_2, u_3, u_4}
  {α : Type u_1} {β : Type u_2}
  {γ : Type u_3} {Ω : Type u_4}
  {mα : MeasurableSpace α}
  {mβ : MeasurableSpace β}
  {mγ : MeasurableSpace γ}
  {mΩ : MeasurableSpace Ω}
  {μ : MeasureTheory.Measure α}
  {κ : Kernel β Ω}
  [MeasureTheory.IsFiniteMeasure μ]
  [IsFiniteKernel κ] (X : α → β)
  (Y : α → Ω) {f : β → γ}
  (hf : Measurable f) :
  HasCondDistrib Y
      (fun a => (X a, f (X a)))
      (Kernel.prodMkRight γ κ) μ ↔
    HasCondDistrib Y X κ μ

Code

lemma hasCondDistrib_prod_right_iff [IsFiniteMeasure μ] [IsFiniteKernel κ] (X : α → β) (Y : α → Ω)
    {f : β → γ} (hf : Measurable f) :
    HasCondDistrib Y (fun a ↦ (X a, f (X a))) (κ.prodMkRight _) μ ↔ HasCondDistrib Y X κ μ

Proof

by
  refine ⟨fun h ↦ ?_, fun h ↦ h.prod_right hf⟩
  have hX : AEMeasurable X μ := by
    have := h.aemeasurable_snd
    have h_eq : X = (fun p ↦ p.1) ∘ (fun a ↦ (X a, f (X a))) := by ext; simp
    rw [h_eq]
    exact Measurable.comp_aemeasurable (by fun_prop) (by fun_prop)
  refine ⟨by fun_prop, ?_⟩
  have h_eq := h.map_eq
  calc μ.map (fun x ↦ (X x, Y x))
  _ = (μ.map (fun ω ↦ ((X ω, f (X ω)), Y ω))).map (fun p ↦ (p.1.1, p.2)) := by
    rw [AEMeasurable.map_map_of_aemeasurable (by fun_prop) (by fun_prop)]
    congr
  _ = (μ.map (fun a ↦ (X a, f (X a))) ⊗ₘ κ.prodMkRight γ).map (fun p ↦ (p.1.1, p.2)) := by rw [h_eq]
  _ = ((μ.map X).map (fun a ↦ (a, f a)) ⊗ₘ κ.prodMkRight γ).map (fun p ↦ (p.1.1, p.2)) := by
    rw [AEMeasurable.map_map_of_aemeasurable (by fun_prop) (by fun_prop)]
    congr
  _ = μ.map X ⊗ₘ κ := by
    simp_rw [Measure.compProd_eq_comp_prod,
      ← Measure.deterministic_comp_eq_map (f := fun a ↦ (a, f a)) (by fun_prop),
      ← Measure.deterministic_comp_eq_map (f := fun p : (β × γ) × Ω ↦ (p.1.1, p.2)) (by fun_prop),
      Measure.comp_assoc]
    congr 1
    ext b : 1
    rw [Kernel.comp_apply, Kernel.comp_apply, Kernel.prod_apply, Kernel.id_apply,
      Kernel.deterministic_apply, Measure.dirac_bind (Kernel.measurable _),
      Kernel.prod_apply, Measure.deterministic_comp_eq_map, Kernel.prodMkRight_apply,
      Kernel.id_apply]
    change Measure.map (Prod.map (fun x ↦ x.1) id) ((Measure.dirac (b, f b)).prod (κ b)) = _
    rw [← Measure.map_prod_map _ _ (by fun_prop) (by fun_prop), Measure.map_id,
      Measure.map_dirac' (by fun_prop)]

Actions: Source · Open Issue

Meaning last changed in v4.34.0-rc2-1-g439785b (2026-08-23), the 4th 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, 12 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.

ProbabilityTheory.hasCondDistrib_prod_right_iff🔗

`ProbabilityTheory.hasCondDistrib_prod_right_iff`🔗