I have a really general question. Luzin's theorem implies that

ScommaMaruj 2022-07-07 Answered
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?
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Answers (1)

Salma Bradley
Answered 2022-07-08 Author has 13 answers
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 R as f ( x ) = 1 if x 0, 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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