/-- **The Lebesgue decomposition theorem**: Any pair of σ-finite measures `μ` and `ν`
`HaveLebesgueDecomposition`. That is to say, there exist a measure `ξ` and a measurable function
`f`, such that `ξ` is mutually singular with respect to `ν` and `μ = ξ + ν.withDensity f` -/
instance (priority := 100) haveLebesgueDecomposition_of_sigmaFinite (μ ν : Measure α)
    [SigmaFinite μ] [SigmaFinite ν] : HaveLebesgueDecomposition μ ν :=
  ⟨by
    -- Since `μ` and `ν` are both σ-finite, there exists a sequence of pairwise disjoint spanning
    -- sets which are finite with respect to both `μ` and `ν`
    obtain ⟨S, T, h₁, h₂⟩ := exists_eq_disjoint_finiteSpanningSetsIn μ ν
    have h₃ : Pairwise (Disjoint on T.set) := h₁ ▸ h₂
    -- We define `μn` and `νn` as sequences of measures such that `μn n = μ ∩ S n` and
    -- `νn n = ν ∩ S n` where `S` is the aforementioned finite spanning set sequence.
    -- Since `S` is spanning, it is clear that `sum μn = μ` and `sum νn = ν`
    set μn : ℕ → Measure α := fun n => μ.restrict (S.set n) with hμn
    have hμ : μ = sum μn := by rw [hμn, ← restrict_iUnion h₂ S.set_mem, S.spanning, restrict_univ]
    set νn : ℕ → Measure α := fun n => ν.restrict (T.set n) with hνn
    have hν : ν = sum νn := by rw [hνn, ← restrict_iUnion h₃ T.set_mem, T.spanning, restrict_univ]
    -- As `S` is finite with respect to both `μ` and `ν`, it is clear that `μn n` and `νn n` are
    -- finite measures for all `n : ℕ`. Thus, we may apply the finite Lebesgue decomposition theorem
    -- to `μn n` and `νn n`. We define `ξ` as the sum of the singular part of the Lebesgue
    -- decompositions of `μn n` and `νn n`, and `f` as the sum of the Radon-Nikodym derivatives of
    -- `μn n` and `νn n` restricted on `S n`
    set ξ := sum fun n => singularPart (μn n) (νn n) with hξ
    set f := ∑' n, (S.set n).indicator (rnDeriv (μn n) (νn n))
    -- I claim `ξ` and `f` form a Lebesgue decomposition of `μ` and `ν`
    refine' ⟨⟨ξ, f⟩, _, _, _⟩
    · exact
        Measurable.ennreal_tsum' fun n =>
          Measurable.indicator (measurable_rnDeriv (μn n) (νn n)) (S.set_mem n)
    -- We show that `ξ` is mutually singular with respect to `ν`
    · choose A hA₁ hA₂ hA₃ using fun n => mutuallySingular_singularPart (μn n) (νn n)
      simp only [hξ]
      -- We use the set `B := ⋃ j, (S.set j) ∩ A j` where `A n` is the set provided as
      -- `singularPart (μn n) (νn n) ⟂ₘ νn n`
      refine' ⟨⋃ j, S.set j ∩ A j, MeasurableSet.iUnion fun n => (S.set_mem n).inter (hA₁ n), _, _⟩
      -- `ξ B = 0` since `ξ B = ∑ i j, singularPart (μn j) (νn j) (S.set i ∩ A i)`
      --                     `= ∑ i, singularPart (μn i) (νn i) (S.set i ∩ A i)`
      --                     `≤ ∑ i, singularPart (μn i) (νn i) (A i) = 0`
      -- where the second equality follows as `singularPart (μn j) (νn j) (S.set i ∩ A i)` vanishes
      -- for all `i ≠ j`
      · rw [measure_iUnion]
        · have :
            ∀ i,
              (sum fun n => (μn n).singularPart (νn n)) (S.set i ∩ A i) =
                (μn i).singularPart (νn i) (S.set i ∩ A i) := by
            intro i; rw [sum_apply _ ((S.set_mem i).inter (hA₁ i)), tsum_eq_single i]
            · intro j hij
              rw [hμn, ← nonpos_iff_eq_zero]
              refine' le_trans ((singularPart_le _ _) _ ((S.set_mem i).inter (hA₁ i))) (le_of_eq _)
              rw [restrict_apply ((S.set_mem i).inter (hA₁ i)), inter_comm, ← inter_assoc]
              have : Disjoint (S.set j) (S.set i) := h₂ hij
              rw [disjoint_iff_inter_eq_empty] at this
              rw [this, empty_inter, measure_empty]
          simp_rw [this, tsum_eq_zero_iff ENNReal.summable]
          intro n; exact measure_mono_null (inter_subset_right _ _) (hA₂ n)
        · exact h₂.mono fun i j => Disjoint.mono inf_le_left inf_le_left
        · exact fun n => (S.set_mem n).inter (hA₁ n)
      -- We will now show `ν Bᶜ = 0`. This follows since `Bᶜ = ⋃ n, S.set n ∩ (A n)ᶜ` and thus,
      -- `ν Bᶜ = ∑ i, ν (S.set i ∩ (A i)ᶜ) = ∑ i, (νn i) (A i)ᶜ = 0`
      · have hcompl : IsCompl (⋃ n, S.set n ∩ A n) (⋃ n, S.set n ∩ (A n)ᶜ) := by
          constructor
          · rw [disjoint_iff_inf_le]
            rintro x ⟨hx₁, hx₂⟩; rw [mem_iUnion] at hx₁ hx₂
            obtain ⟨⟨i, hi₁, hi₂⟩, ⟨j, hj₁, hj₂⟩⟩ := hx₁, hx₂
            have : i = j := by by_contra hij; exact (h₂ hij).le_bot ⟨hi₁, hj₁⟩
            exact hj₂ (this ▸ hi₂)
          · rw [codisjoint_iff_le_sup]
            intro x hx
            simp only [mem_iUnion, sup_eq_union, mem_inter_iff, mem_union, mem_compl_iff,
              or_iff_not_imp_left]
            intro h; push_neg at h
            rw [top_eq_univ, ← S.spanning, mem_iUnion] at hx
            obtain ⟨i, hi⟩ := hx
            exact ⟨i, hi, h i hi⟩
        rw [hcompl.compl_eq, measure_iUnion, tsum_eq_zero_iff ENNReal.summable]
        · intro n; rw [inter_comm, ← restrict_apply (hA₁ n).compl, ← hA₃ n, hνn, h₁]
        · exact h₂.mono fun i j => Disjoint.mono inf_le_left inf_le_left
        · exact fun n => (S.set_mem n).inter (hA₁ n).compl
    -- Finally, it remains to show `μ = ξ + ν.withDensity f`. Since `μ = sum μn`, and
    -- `ξ + ν.withDensity f = ∑ n, singularPart (μn n) (νn n)`
    --                        `+ ν.withDensity (rnDeriv (μn n) (νn n)) ∩ (S.set n)`,
    -- it suffices to show that the individual summands are equal. This follows by the
    -- Lebesgue decomposition properties on the individual `μn n` and `νn n`
    · simp only
      nth_rw 1 [hμ]
      rw [withDensity_tsum _, sum_add_sum]
      · refine' sum_congr fun n => _
        nth_rw 1 [haveLebesgueDecomposition_add (μn n) (νn n)]
        suffices heq :
          (νn n).withDensity ((μn n).rnDeriv (νn n)) =
            ν.withDensity ((S.set n).indicator ((μn n).rnDeriv (νn n)))
        · rw [heq]
        rw [hν, withDensity_indicator (S.set_mem n), restrict_sum _ (S.set_mem n)]
        

Proof state:
case heq
α : Type u_1
β : Type u_2
m : MeasurableSpace α
μ✝ ν✝ μ ν : Measure α
inst✝¹ : SigmaFinite μ
inst✝ : SigmaFinite ν
S : FiniteSpanningSetsIn μ {s | MeasurableSet s}
T : FiniteSpanningSetsIn ν {s | MeasurableSet s}
h₁ : S.set = T.set
h₂ : Pairwise (Disjoint on S.set)
h₃ : Pairwise (Disjoint on T.set)
μn : ℕ → Measure α := fun n => restrict μ (FiniteSpanningSetsIn.set S n)
hμn : μn = fun n => restrict μ (FiniteSpanningSetsIn.set S n)
hμ : μ = sum μn
νn : ℕ → Measure α := fun n => restrict ν (FiniteSpanningSetsIn.set T n)
hνn : νn = fun n => restrict ν (FiniteSpanningSetsIn.set T n)
hν : ν = sum νn
ξ : Measure α := sum fun n => singularPart (μn n) (νn n)
hξ : ξ = sum fun n => singularPart (μn n) (νn n)
f : α → ℝ≥0∞ := ∑' (n : ℕ), indicator (FiniteSpanningSetsIn.set S n) (rnDeriv (μn n) (νn n))
n : ℕ
⊢ withDensity (νn n) (rnDeriv (μn n) (νn n)) =
    withDensity (sum fun i => restrict (νn i) (FiniteSpanningSetsIn.set S n)) (rnDeriv (μn n) (νn n))

Next tactic:
suffices hsumeq : (sum fun i : ℕ => (νn i).restrict (S.set n)) = νn n