l3lab/ntp-mathlib huggingface dataset:
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; 

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
n : ℕ
⊢ VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A

Next tactic:
exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A
⊢ ∃ ε,
    0 < ε ∧
      ∃ E,
        MeasurableSet E ∧
          0 < ↑↑ν E ∧
            VectorMeasure.restrict 0 E ≤ VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure (ε • ν)) E

Next tactic:
have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
    exacts [ne_of_lt (measure_lt_top _ _), ne_of_lt (measure_lt_top _ _)]
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A
⊢ ∀ (n : ℕ), ↑↑μ A ≤ ↑(1 / (↑n + 1)) * ↑↑ν A

Next tactic:
intro n
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A
n : ℕ
⊢ ↑↑μ A ≤ ↑(1 / (↑n + 1)) * ↑↑ν A

Next tactic:
have := nonpos_of_restrict_le_zero _ (hA₂ n)
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A
n : ℕ
this : ↑(toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ 0
⊢ ↑↑μ A ≤ ↑(1 / (↑n + 1)) * ↑↑ν A

Next tactic:
rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
    exacts [ne_of_lt (measure_lt_top _ _), ne_of_lt (measure_lt_top _ _)]
  have hμ : μ A = 0 := by
    

Proof state:
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hA₁ : A = ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet A
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) A ≤ VectorMeasure.restrict 0 A
hA₃ : ∀ (n : ℕ), ↑↑μ A ≤ ↑(1 / (↑n + 1)) * ↑↑ν A
⊢ ↑↑μ A = 0

Next tactic:
lift μ A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with μA
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
    exacts [ne_of_lt (measure_lt_top _ _), ne_of_lt (measure_lt_top _ _)]
  have hμ : μ A = 0 := by
    lift μ A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with μA
    

Proof state:
case intro
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet (⋂ n, (f n)ᶜ)
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (⋂ n, (f n)ᶜ) ≤
      VectorMeasure.restrict 0 (⋂ n, (f n)ᶜ)
μA : ℝ≥0
hA₁✝ hA₁ : True
hA₃✝ hA₃ : ∀ (n : ℕ), ↑μA ≤ ↑(1 / (↑n + 1)) * ↑↑ν (⋂ n, (f n)ᶜ)
⊢ ↑μA = 0

Next tactic:
lift ν A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with νA
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
    exacts [ne_of_lt (measure_lt_top _ _), ne_of_lt (measure_lt_top _ _)]
  have hμ : μ A = 0 := by
    lift μ A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with μA
    lift ν A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with νA
    

Proof state:
case intro.intro
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet (⋂ n, (f n)ᶜ)
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (⋂ n, (f n)ᶜ) ≤
      VectorMeasure.restrict 0 (⋂ n, (f n)ᶜ)
μA : ℝ≥0
hA₁✝ hA₁ : True
hA₃✝¹ : ∀ (n : ℕ), ↑μA ≤ ↑(1 / (↑n + 1)) * ↑↑ν (⋂ n, (f n)ᶜ)
νA : ℝ≥0
hA₃✝ hA₃ : ∀ (n : ℕ), ↑μA ≤ ↑(1 / (↑n + 1)) * ↑νA
⊢ ↑μA = 0

Next tactic:
rw [ENNReal.coe_eq_zero]
/-- If two finite measures `μ` and `ν` are not mutually singular, there exists some `ε > 0` and
a measurable set `E`, such that `ν(E) > 0` and `E` is positive with respect to `μ - εν`.

This lemma is useful for the Lebesgue decomposition theorem. -/
theorem exists_positive_of_not_mutuallySingular (μ ν : Measure α) [IsFiniteMeasure μ]
    [IsFiniteMeasure ν] (h : ¬μ ⟂ₘ ν) :
    ∃ ε : ℝ≥0, 0 < ε ∧
      ∃ E : Set α,
        MeasurableSet E ∧ 0 < ν E ∧ 0 ≤[E] μ.toSignedMeasure - (ε • ν).toSignedMeasure := by
  -- for all `n : ℕ`, obtain the Hahn decomposition for `μ - (1 / n) ν`
  have :
    ∀ n : ℕ, ∃ i : Set α,
      MeasurableSet i ∧
        0 ≤[i] μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ∧
          μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[iᶜ] 0 := by
    intro; exact exists_compl_positive_negative _
  choose f hf₁ hf₂ hf₃ using this
  -- set `A` to be the intersection of all the negative parts of obtained Hahn decompositions
  -- and we show that `μ A = 0`
  set A := ⋂ n, (f n)ᶜ with hA₁
  have hAmeas : MeasurableSet A := MeasurableSet.iInter fun n => (hf₁ n).compl
  have hA₂ : ∀ n : ℕ, μ.toSignedMeasure - ((1 / (n + 1) : ℝ≥0) • ν).toSignedMeasure ≤[A] 0 := by
    intro n; exact restrict_le_restrict_subset _ _ (hf₁ n).compl (hf₃ n) (iInter_subset _ _)
  have hA₃ : ∀ n : ℕ, μ A ≤ (1 / (n + 1) : ℝ≥0) * ν A := by
    intro n
    have := nonpos_of_restrict_le_zero _ (hA₂ n)
    rwa [toSignedMeasure_sub_apply hAmeas, sub_nonpos, ENNReal.toReal_le_toReal] at this
    exacts [ne_of_lt (measure_lt_top _ _), ne_of_lt (measure_lt_top _ _)]
  have hμ : μ A = 0 := by
    lift μ A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with μA
    lift ν A to ℝ≥0 using ne_of_lt (measure_lt_top _ _) with νA
    rw [ENNReal.coe_eq_zero]
    

Proof state:
case intro.intro
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : IsFiniteMeasure μ
inst✝ : IsFiniteMeasure ν
h : ¬μ ⟂ₘ ν
f : ℕ → Set α
hf₁ : ∀ (n : ℕ), MeasurableSet (f n)
hf₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict 0 (f n) ≤
      VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)
hf₃ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (f n)ᶜ ≤
      VectorMeasure.restrict 0 (f n)ᶜ
A : Set α := ⋂ n, (f n)ᶜ
hAmeas : MeasurableSet (⋂ n, (f n)ᶜ)
hA₂ :
  ∀ (n : ℕ),
    VectorMeasure.restrict (toSignedMeasure μ - toSignedMeasure ((1 / (↑n + 1)) • ν)) (⋂ n, (f n)ᶜ) ≤
      VectorMeasure.restrict 0 (⋂ n, (f n)ᶜ)
μA : ℝ≥0
hA₁✝ hA₁ : True
hA₃✝¹ : ∀ (n : ℕ), ↑μA ≤ ↑(1 / (↑n + 1)) * ↑↑ν (⋂ n, (f n)ᶜ)
νA : ℝ≥0
hA₃✝ hA₃ : ∀ (n : ℕ), ↑μA ≤ ↑(1 / (↑n + 1)) * ↑νA
⊢ μA = 0

Next tactic:
by_cases hb : 0 < νA