Lusins theorem does not say that f is continuous at every

Question

Questions -

Q1. Show that the hypothesis μ(X) < ∞ is necessary in Egoroff's Theorem. Give an example to show that it cannot even be relaxed to 'μ is σ-finite'.

Q2. (Lusin's Theorem, C-version) Let f [a, b] → C be a Lebesgue measurable, complex-valued function on the compact interval [a, b]. Then for every ∈ > 0 there is a measurable subset E ⊆ [a, b] with λ([a, b] - E) < ∈ such that f|_E is continuous (λ = Lebesgue measure).

a) Lusin's Theorem does NOT say that f is continuous at every point of E. Find a counterexample to this incorrect version of Lusin's 'Theorem'.

Complete details below to provide a proof of Lusin's theorem.

b) It is enough to assume f is Borel measurable and maps to R. It is enough to assume f [a, b] → [0, ∞).

c) There is an A ⊆ [a, b] with λ([a, b] - A) < ∈ such that f is uniformly bounded on A.

d) There is sequence of (Borel) simple functions φ_n, on [a, b] increasing pointwise to f and a B ⊆ [a, b] with λ([a, b] - B) < ∈ such that convergence is uniform on B.

e) If φ is a (Borel) simple function there is a continuous function g on [a, b] such that g = φ except on a subset of measure at most ∈.

f) There is sequence of continuous functions g_n on [a, b] and an F ⊆ [a, b] with λ([a, b] - F) < ∈ and such that g_n converges uniformly to f on F.

g) The conclusion of the Theorem follows.