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I have a really general question. Luzin's theorem implies that Lebesgue measurable functions are continuous almost everywhere. So is there an analogous version of the intermediate value theorem for measurable functions, with maybe some extra conditions?

Thanks!

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Since the heavyside function is measurable, what kind of analogouous statement do you expect? –  Hagen von Eitzen Oct 8 '12 at 15:40
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You have a good point, but I was wondering if it was possible to add some stronger hypothesis on measurable functions to get the intermediate value theorem, without necessarily asking the function to be continuous. –  stefano demineva Oct 8 '12 at 15:43

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I do not know what kind of result you expect, but in my opinion there is no reasonable extension of the mean value theorem to measurable functions. Consider the function defined on $\mathbb{R}$ as $f(x)=1$ if $x\ge0$, $f(x)=0$ if $x<0$. $f$ is measurable and discontinuous only at one point, namely at $x=0$. It takes the values $0$ and $1$, but none of the intermediate values.

On the other hand, the derivative of any differentiable function satisfies the mean value property, even if it is not continuous.

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