ProbabilityTheory.IndepFun.of_measurable_right
of_measurable_right🔗
Lemma
ProbabilityTheory.IndepFun.of_measurable_rightNo docstring.
theorem
ProbabilityTheory.IndepFun.of_measurable_right.{u_1, u_3, u_4, u_6} {α : Type u_1} {γ : Type u_3} {δ : Type u_4} {δ' : Type u_6} {mα : MeasurableSpace α} {mγ : MeasurableSpace γ} {mδ : MeasurableSpace δ} {mδ' : MeasurableSpace δ'} [StandardBorelSpace δ'] [Nonempty δ'] {μ : MeasureTheory.Measure α} {Y : α → γ} {Z : α → δ} {Z' : α → δ'} (h_indep : IndepFun Y Z μ) (hZ_meas : Measurable Z') : IndepFun Y Z' μProbabilityTheory.IndepFun.of_measurable_right.{u_1, u_3, u_4, u_6} {α : Type u_1} {γ : Type u_3} {δ : Type u_4} {δ' : Type u_6} {mα : MeasurableSpace α} {mγ : MeasurableSpace γ} {mδ : MeasurableSpace δ} {mδ' : MeasurableSpace δ'} [StandardBorelSpace δ'] [Nonempty δ'] {μ : MeasureTheory.Measure α} {Y : α → γ} {Z : α → δ} {Z' : α → δ'} (h_indep : IndepFun Y Z μ) (hZ_meas : Measurable Z') : IndepFun Y Z' μ
Code
lemma IndepFun.of_measurable_right
(h_indep : Y ⟂ᵢ[μ] Z) (hZ_meas : Measurable[mδ.comap Z] Z') :
Y ⟂ᵢ[μ] Z'Actions: Source · Open Issue
Proof
by obtain ⟨ψ, hψ_meas, h_eqZ⟩ : ∃ ψ, Measurable ψ ∧ Z' = ψ ∘ Z := hZ_meas.exists_eq_measurable_comp rw [h_eqZ] exact h_indep.comp measurable_id hψ_meas