Does the converse of Lusin's theorem hold? Lusin's theorem says that if $f:[a,b]\to\mathbb{C}$ is a measurable function, then for any given $\varepsilon>0$ there is a continuous function such that $\mu(\{x\in[a,b]:f(x)\neq g(x)\})<\varepsilon$. But I wonder if the converse is true, that is, if for any $\varepsilon>0$ there is a continuous function such that $\mu(\{x\in[a,b]:f(x)\neq g(x)\})<\varepsilon$, is $f$ a measurable function?
 A: Assume that for every $\varepsilon > 0$ there exists a continuous $g_\varepsilon: [a,b] \to \Bbb{C}$ such that the set $\{x \in [a,b]: f(x) \neq g_\varepsilon (x)\}$ is measurable and its measure is less than $\varepsilon$.
Define
$$
A_n := \bigcup_{k \geq n} \{ x \in [a,b] : f(x) \neq g_{2^{-k}}(x) \}\,.
$$
Now $A_n \supset A_{n+1} \supset A_{n+2}...$ and
$$
\mu \left( A_n \right) \leq \sum_{k=n}^\infty \mu \left( \{ x \in [a,b] : f(x) \neq g_{2^{-k}}(x) \}\right) \leq \sum_{k=1}^\infty \frac{1}{2^k} = 1\,.
$$
Also
$$
\lim_{n \to \infty} \mu(A_n) \leq \lim_{n \to \infty} \sum_{k=n}^\infty \frac{1}{2^k} = 0\,.
$$
Now by the convergence of measure
$$
\mu \left( \bigcap_{n \in \Bbb{N}} A_n \right) = \lim_{n \to \infty} \mu(A_n) = 0\,.
$$
Define $g(x) := \lim_{n \to \infty} g_{2^{-n}}(x)$ (we don't know yet for which $x$ this limit even exists). Assume that $x \notin \bigcap_{n \in \Bbb{N}} A_n$. Then there exists $j$ s.t. $x \notin A_j$. So
$$
x \notin \bigcup_{k \geq j} \{ x \in [a,b] : f(x) \neq g_{2^{-k}} (x) \}.
$$
So $g_{2^{-k}}(x) = f(x)$ for all $k \geq j$. So $g(x) = f(x)$. So $g = \lim g_{2^{-n}} =  f$ a.e. So $f$ is a pointwise limit of continuous functions a.e. so $f$ is measurable.
