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Luzin's theorem states that: let $f:[a,b]\rightarrow R$ be an a.e. finite function, $f$ is measurable iff $\forall \epsilon \geq 0: \exists \phi_\epsilon$ continuous on $[a,b]$ and $\mu\{x: f(x)\neq \phi_\epsilon (x)\} \leq \epsilon $

Why doesn't it imply that a measurable function $f$ equals a.e. to a continuous function $\phi_\epsilon$?

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    $\phi_\epsilon$ might be different for every $\epsilon$. How would you use them to pick a single function $\phi$? I think you'll find that your technique doesn't preserve continuity.2012-10-08
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    @NateEldredge I don't necessarily mean to pick a single $\phi$.2012-10-08
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    Well, each $\phi_\epsilon$ could be different from $f$ on a set of positive measure (but at most $\epsilon$). I guess I don't understand why you think Luzin's theorem could possibly imply that $f$ is a.e. equal to some continuous function. And of course, you must know of counterexamples.2012-10-08
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    Anna: In addition to the answer Seirios gave, you might find some things of interest in my [essay on Luzin's Theorem](http://math.stackexchange.com/questions/15088/is-every-lebesgue-measurable-function-on-mathbbr-the-pointwise-limit-of-con).2012-10-08

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