English
Let f and g be strongly measurable with respect to a sub-sigma-algebra m, and hm be a relation m ≤ m0. Then the inequality f ≤ a.e. with respect to the trimmed measure μ.trim hm is equivalent to f ≤ a.e. with respect to μ.
Русский
Пусть f и g разложимы через подмножество модуля, связанные с подпроизвольной сигма-алгеброй, и hm задаёт отношение m ≤ m0. Тогда неравенство f ≤ a.e. по обрезанной мере μ.trim hm эквивалентно неравенству f ≤ a.e. по мере μ.
LaTeX
$$$ f \leq_{\mathrm{ae}}^{\mu_{\mathrm{trim}(hm)}} g \;\Longleftrightarrow\; f \leq_{\mathrm{ae}}^{\mu} g $$$
Lean4
theorem ae_le_trim_iff (hm : m ≤ m₀) (hf : StronglyMeasurable[m] f) (hg : StronglyMeasurable[m] g) :
f ≤ᵐ[μ.trim hm] g ↔ f ≤ᵐ[μ] g :=
⟨ae_le_of_ae_le_trim, ae_le_trim_of_stronglyMeasurable hm hf hg⟩