English
For AEMeasurable f, lintegral μ f = 0 if and only if f =ᵐ μ 0.
Русский
Для AEMеасенABLE f: линегральный интеграл равен нулю тогда и только тогда, когда f =ᵐ μ 0.
LaTeX
$$Iff (lintegral μ f = 0) (f =ᵐ μ 0)$$
Lean4
/-- The Lebesgue integral is zero iff the function is a.e. zero.
The measurability assumption is necessary, otherwise there are counterexamples: for instance, the
conclusion fails if `f` is the characteristic function of a Vitali set. -/
@[simp]
theorem lintegral_eq_zero_iff' {f : α → ℝ≥0∞} (hf : AEMeasurable f μ) : ∫⁻ a, f a ∂μ = 0 ↔ f =ᵐ[μ] 0 := by
-- The proof implicitly uses Markov's inequality,
-- but it has been inlined for the sake of imports
refine ⟨fun h ↦ ?_, lintegral_eq_zero_of_ae_eq_zero⟩
have meas_levels_0 : ∀ ε > 0, μ {x | ε ≤ f x} = 0 := fun ε εpos ↦
by
by_contra! h'; rw [← zero_lt_iff] at h'
refine ((mul_pos_iff.mpr ⟨εpos, h'⟩).trans_le ?_).ne' h
calc
_ ≥ ∫⁻ a in {x | ε ≤ f x}, f a ∂μ := setLIntegral_le_lintegral _ _
_ ≥ ∫⁻ _ in {x | ε ≤ f x}, ε ∂μ := (setLIntegral_mono_ae hf.restrict (ae_of_all μ fun _ ↦ id))
_ = _ := setLIntegral_const _ _
obtain ⟨u, -, bu, tu⟩ := exists_seq_strictAnti_tendsto' (α := ℝ≥0∞) zero_lt_one
have u_union : {x | f x ≠ 0} = ⋃ n, {x | u n ≤ f x} := by
ext x; rw [mem_iUnion, mem_setOf_eq, ← zero_lt_iff]
rw [ENNReal.tendsto_atTop_zero] at tu
constructor <;> intro h'
· obtain ⟨n, hn⟩ := tu _ h'; use n, hn _ le_rfl
· obtain ⟨n, hn⟩ := h'; exact (bu n).1.trans_le hn
have res := measure_iUnion_null_iff.mpr fun n ↦ meas_levels_0 _ (bu n).1
rwa [← u_union] at res
