/-- The **Lusin-Souslin theorem**: the range of a continuous injective function defined on a Polish
space is Borel-measurable. -/
theorem measurableSet_range_of_continuous_injective {β : Type*} [TopologicalSpace γ]
    [PolishSpace γ] [TopologicalSpace β] [T2Space β] [MeasurableSpace β] [OpensMeasurableSpace β]
    {f : γ → β} (f_cont : Continuous f) (f_inj : Injective f) :
    MeasurableSet (range f) := by
  /- We follow [Fremlin, *Measure Theory* (volume 4, 423I)][fremlin_vol4].
    Let `b = {s i}` be a countable basis for `α`. When `s i` and `s j` are disjoint, their images
    are disjoint analytic sets, hence by the separation theorem one can find a Borel-measurable set
    `q i j` separating them.
    Let `E i = closure (f '' s i) ∩ ⋂ j, q i j \ q j i`. It contains `f '' (s i)` and it is
    measurable. Let `F n = ⋃ E i`, where the union is taken over those `i` for which `diam (s i)`
    is bounded by some number `u n` tending to `0` with `n`.
    We claim that `range f = ⋂ F n`, from which the measurability is obvious. The inclusion `⊆` is
    straightforward. To show `⊇`, consider a point `x` in the intersection. For each `n`, it belongs
    to some `E i` with `diam (s i) ≤ u n`. Pick a point `y i ∈ s i`. We claim that for such `i`
    and `j`, the intersection `s i ∩ s j` is nonempty: if it were empty, then thanks to the
    separating set `q i j` in the definition of `E i` one could not have `x ∈ E i ∩ E j`.
    Since these two sets have small diameter, it follows that `y i` and `y j` are close.
    Thus, `y` is a Cauchy sequence, converging to a limit `z`. We claim that `f z = x`, completing
    the proof.
    Otherwise, one could find open sets `v` and `w` separating `f z` from `x`. Then, for large `n`,
    the image `f '' (s i)` would be included in `v` by continuity of `f`, so its closure would be
    contained in the closure of `v`, and therefore it would be disjoint from `w`. This is a
    contradiction since `x` belongs both to this closure and to `w`. -/
  letI := upgradePolishSpace γ
  obtain ⟨b, b_count, b_nonempty, hb⟩ :
    ∃ b : Set (Set γ), b.Countable ∧ ∅ ∉ b ∧ IsTopologicalBasis b := exists_countable_basis γ
  haveI : Encodable b := b_count.toEncodable
  let A := { p : b × b // Disjoint (p.1 : Set γ) p.2 }
  -- for each pair of disjoint sets in the topological basis `b`, consider Borel sets separating
  -- their images, by injectivity of `f` and the Lusin separation theorem.
  have : ∀ p : A, ∃ q : Set β,
      f '' (p.1.1 : Set γ) ⊆ q ∧ Disjoint (f '' (p.1.2 : Set γ)) q ∧ MeasurableSet q := by
    intro p
    apply
      AnalyticSet.measurablySeparable ((hb.isOpen p.1.1.2).analyticSet_image f_cont)
        ((hb.isOpen p.1.2.2).analyticSet_image f_cont)
    exact Disjoint.image p.2 (f_inj.injOn univ) (subset_univ _) (subset_univ _)
  choose q hq1 hq2 q_meas using this
  -- define sets `E i` and `F n` as in the proof sketch above
  let E : b → Set β := fun s =>
    closure (f '' s) ∩ ⋂ (t : b) (ht : Disjoint s.1 t.1), q ⟨(s, t), ht⟩ \ q ⟨(t, s), ht.symm⟩
  obtain ⟨u, u_anti, u_pos, u_lim⟩ :
    ∃ u : ℕ → ℝ, StrictAnti u ∧ (∀ n : ℕ, 0 < u n) ∧ Tendsto u atTop (𝓝 0) :=
    exists_seq_strictAnti_tendsto (0 : ℝ)
  let F : ℕ → Set β := fun n => ⋃ (s : b) (_ : IsBounded s.1 ∧ diam s.1 ≤ u n), E s
  -- it is enough to show that `range f = ⋂ F n`, as the latter set is obviously measurable.
  suffices range f = ⋂ n, F n by
    have E_meas : ∀ s : b, MeasurableSet (E s) := by
      intro b
      refine' isClosed_closure.measurableSet.inter _
      refine' MeasurableSet.iInter fun s => _
      exact MeasurableSet.iInter fun hs => (q_meas _).diff (q_meas _)
    have F_meas : ∀ n, MeasurableSet (F n) := by
      intro n
      refine' MeasurableSet.iUnion fun s => _
      exact MeasurableSet.iUnion fun _ => E_meas _
    rw [this]
    exact MeasurableSet.iInter fun n => F_meas n
  -- we check both inclusions.
  apply Subset.antisymm
  -- we start with the easy inclusion `range f ⊆ ⋂ F n`. One just needs to unfold the definitions.
  · rintro x ⟨y, rfl⟩
    refine mem_iInter.2 fun n => ?_
    obtain ⟨s, sb, ys, hs⟩ : ∃ (s : Set γ), s ∈ b ∧ y ∈ s ∧ s ⊆ ball y (u n / 2) := by
      apply hb.mem_nhds_iff.1
      exact ball_mem_nhds _ (half_pos (u_pos n))
    have diam_s : diam s ≤ u n := by
      apply (diam_mono hs isBounded_ball).trans
      convert diam_ball (x := y) (half_pos (u_pos n)).le
      ring
    refine' mem_iUnion.2 ⟨⟨s, sb⟩, _⟩
    refine' mem_iUnion.2 ⟨⟨isBounded_ball.subset hs, diam_s⟩, _⟩
    apply mem_inter (subset_closure (mem_image_of_mem _ ys))
    refine' mem_iInter.2 fun t => mem_iInter.2 fun ht => ⟨_, _⟩
    · apply hq1
      exact mem_image_of_mem _ ys
    · apply disjoint_left.1 (hq2 ⟨(t, ⟨s, sb⟩), ht.symm⟩)
      exact mem_image_of_mem _ ys
  -- Now, let us prove the harder inclusion `⋂ F n ⊆ range f`.
  · intro x hx
    -- pick for each `n` a good set `s n` of small diameter for which `x ∈ E (s n)`.
    have C1 : ∀ n, ∃ (s : b) (_ : IsBounded s.1 ∧ diam s.1 ≤ u n), x ∈ E s := fun n => by
      simpa only [mem_iUnion] using mem_iInter.1 hx n
    choose s hs hxs using C1
    have C2 : ∀ n, (s n).1.Nonempty := by
      intro n
      rw [nonempty_iff_ne_empty]
      intro hn
      have := (s n).2
      rw [hn] at this
      exact b_nonempty this
    -- choose a point `y n ∈ s n`.
    choose y hy using C2
    have I : ∀ m n, ((s m).1 ∩ (s n).1).Nonempty := by
      intro m n
      rw [← not_disjoint_iff_nonempty_inter]
      by_contra! h
      have A : x ∈ q ⟨(s m, s n), h⟩ \ q ⟨(s n, s m), h.symm⟩ :=
        haveI := mem_iInter.1 (hxs m).2 (s n)
        (mem_iInter.1 this h : _)
      

Proof state:
α : Type u_1
ι : Type u_2
γ : Type u_3
β : Type u_4
inst✝⁵ : TopologicalSpace γ
inst✝⁴ : PolishSpace γ
inst✝³ : TopologicalSpace β
inst✝² : T2Space β
inst✝¹ : MeasurableSpace β
inst✝ : OpensMeasurableSpace β
f : γ → β
f_cont : Continuous f
f_inj : Injective f
this✝ : UpgradedPolishSpace γ := upgradePolishSpace γ
b : Set (Set γ)
b_count : Set.Countable b
b_nonempty : ∅ ∉ b
hb : IsTopologicalBasis b
this : Encodable ↑b
A✝ : Type u_3 := { p // Disjoint ↑p.1 ↑p.2 }
q : A✝ → Set β
hq1 : ∀ (p : A✝), f '' ↑(↑p).1 ⊆ q p
hq2 : ∀ (p : A✝), Disjoint (f '' ↑(↑p).2) (q p)
q_meas : ∀ (p : A✝), MeasurableSet (q p)
E : ↑b → Set β :=
  fun s =>
    closure (f '' ↑s) ∩
      ⋂ t,
        ⋂ (ht : Disjoint ↑s ↑t),
          q { val := (s, t), property := ht } \ q { val := (t, s), property := (_ : Disjoint ↑t ↑s) }
u : ℕ → ℝ
u_anti : StrictAnti u
u_pos : ∀ (n : ℕ), 0 < u n
u_lim : Tendsto u atTop (𝓝 0)
F : ℕ → Set β := fun n => ⋃ s, ⋃ (_ : Bornology.IsBounded ↑s ∧ diam ↑s ≤ u n), E s
x : β
hx : x ∈ ⋂ n, F n
s : ℕ → ↑b
hs : ∀ (n : ℕ), Bornology.IsBounded ↑(s n) ∧ diam ↑(s n) ≤ u n
hxs : ∀ (n : ℕ), x ∈ E (s n)
y : ℕ → γ
hy : ∀ (n : ℕ), y n ∈ ↑(s n)
m n : ℕ
h : Disjoint ↑(s m) ↑(s n)
A : x ∈ q { val := (s m, s n), property := h } \ q { val := (s n, s m), property := (_ : Disjoint ↑(s n) ↑(s m)) }
⊢ False

Next tactic:
have B : x ∈ q ⟨(s n, s m), h.symm⟩ \ q ⟨(s m, s n), h⟩ :=
        haveI := mem_iInter.1 (hxs n).2 (s m)
        (mem_iInter.1 this h.symm : _)