/-- Given two thresholds `p < q`, the sets `{x | v.limRatio ρ x < p}`
and `{x | q < v.limRatio ρ x}` are obviously disjoint. The key to proving that `v.limRatio ρ` is
almost everywhere measurable is to show that these sets have measurable supersets which are also
disjoint, up to zero measure. This is the content of this lemma. -/
theorem exists_measurable_supersets_limRatio {p q : ℝ≥0} (hpq : p < q) :
    ∃ a b, MeasurableSet a ∧ MeasurableSet b ∧
      {x | v.limRatio ρ x < p} ⊆ a ∧ {x | (q : ℝ≥0∞) < v.limRatio ρ x} ⊆ b ∧ μ (a ∩ b) = 0 := by
  /- Here is a rough sketch, assuming that the measure is finite and the limit is well defined
    everywhere. Let `u := {x | v.limRatio ρ x < p}` and `w := {x | q < v.limRatio ρ x}`. They
    have measurable supersets `u'` and `w'` of the same measure. We will show that these satisfy
    the conclusion of the theorem, i.e., `μ (u' ∩ w') = 0`. For this, note that
    `ρ (u' ∩ w') = ρ (u ∩ w')` (as `w'` is measurable, see `measure_toMeasurable_add_inter_left`).
    The latter set is included in the set where the limit of the ratios is `< p`, and therefore
    its measure is `≤ p * μ (u ∩ w')`. Using the same trick in the other direction gives that this
    is `p * μ (u' ∩ w')`. We have shown that `ρ (u' ∩ w') ≤ p * μ (u' ∩ w')`. Arguing in the same
    way but using the `w` part gives `q * μ (u' ∩ w') ≤ ρ (u' ∩ w')`. If `μ (u' ∩ w')` were nonzero,
    this would be a contradiction as `p < q`.

    For the rigorous proof, we need to work on a part of the space where the measure is finite
    (provided by `spanningSets (ρ + μ)`) and to restrict to the set where the limit is well defined
    (called `s` below, of full measure). Otherwise, the argument goes through.
    -/
  let s := {x | ∃ c, Tendsto (fun a => ρ a / μ a) (v.filterAt x) (𝓝 c)}
  let o : ℕ → Set α := spanningSets (ρ + μ)
  let u n := s ∩ {x | v.limRatio ρ x < p} ∩ o n
  let w n := s ∩ {x | (q : ℝ≥0∞) < v.limRatio ρ x} ∩ o n
  -- the supersets are obtained by restricting to the set `s` where the limit is well defined, to
  -- a finite measure part `o n`, taking a measurable superset here, and then taking the union over
  -- `n`.
  refine'
    ⟨toMeasurable μ sᶜ ∪ ⋃ n, toMeasurable (ρ + μ) (u n),
      toMeasurable μ sᶜ ∪ ⋃ n, toMeasurable (ρ + μ) (w n), _, _, _, _, _⟩
  -- check that these sets are measurable supersets as required
  · exact
      (measurableSet_toMeasurable _ _).union
        (MeasurableSet.iUnion fun n => measurableSet_toMeasurable _ _)
  · exact
      (measurableSet_toMeasurable _ _).union
        (MeasurableSet.iUnion fun n => measurableSet_toMeasurable _ _)
  · intro x hx
    by_cases h : x ∈ s
    · refine' Or.inr (mem_iUnion.2 ⟨spanningSetsIndex (ρ + μ) x, _⟩)
      exact subset_toMeasurable _ _ ⟨⟨h, hx⟩, mem_spanningSetsIndex _ _⟩
    · exact Or.inl (subset_toMeasurable μ sᶜ h)
  · intro x hx
    by_cases h : x ∈ s
    · refine' Or.inr (mem_iUnion.2 ⟨spanningSetsIndex (ρ + μ) x, _⟩)
      exact subset_toMeasurable _ _ ⟨⟨h, hx⟩, mem_spanningSetsIndex _ _⟩
    · exact Or.inl (subset_toMeasurable μ sᶜ h)
  -- it remains to check the nontrivial part that these sets have zero measure intersection.
  -- it suffices to do it for fixed `m` and `n`, as one is taking countable unions.
  

Proof state:
case refine'_5
α : Type u_1
inst✝⁵ : MetricSpace α
m0 : MeasurableSpace α
μ : Measure α
v : VitaliFamily μ
E : Type u_2
inst✝⁴ : NormedAddCommGroup E
inst✝³ : SecondCountableTopology α
inst✝² : BorelSpace α
inst✝¹ : IsLocallyFiniteMeasure μ
ρ : Measure α
inst✝ : IsLocallyFiniteMeasure ρ
hρ : ρ ≪ μ
p q : ℝ≥0
hpq : p < q
s : Set α := {x | ∃ c, Tendsto (fun a => ↑↑ρ a / ↑↑μ a) (filterAt v x) (𝓝 c)}
o : ℕ → Set α := spanningSets (ρ + μ)
u : ℕ → Set α := fun n => s ∩ {x | limRatio v ρ x < ↑p} ∩ o n
w : ℕ → Set α := fun n => s ∩ {x | ↑q < limRatio v ρ x} ∩ o n
⊢ ↑↑μ ((toMeasurable μ sᶜ ∪ ⋃ n, toMeasurable (ρ + μ) (u n)) ∩ (toMeasurable μ sᶜ ∪ ⋃ n, toMeasurable (ρ + μ) (w n))) =
    0

Next tactic:
suffices H : ∀ m n : ℕ, μ (toMeasurable (ρ + μ) (u m) ∩ toMeasurable (ρ + μ) (w n)) = 0