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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?

Maclaurin
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    Then $f$ is a almost everywhere pointwise limits of sequence of continuous (thus measureabl) functions. Thus $f$ itself is also measureable. –  Mar 29 '15 at 09:24
  • @John could you flesh this out -- how does one know that $f$ is an almost everywhere pointwise limit of measurable functions? One has no control over the set where $g_\epsilon$ and $g_{\epsilon'}$ agree with $f$, for $\epsilon$ and $\epsilon'$ different. – hunter Mar 29 '15 at 09:25
  • A generalization of this result to Radon measures, together with more comments and references, can be found in V. I. Bogachev's Measure Theory, Volume II, p. 137. – Yakov Shklarov Feb 15 '18 at 07:41

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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.

desos
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