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Suppose $A \subseteq \mathbb{R}$, and let $x \in [a,b]$. Put

$$ F(x) = \int_{[a,x]} \chi_A dm $$

Then of course we know that $F $ is differentiable for almost all $x \in [a.b]$ and $$ F'(x) = 1 \; \; \text{for almost all } x \in [a,b] \cap A $$

Can someone explain to me why does it follow that $F'(x) = 1 $ implies that there exists $\epsilon$ and $\delta >0$ such that

$$ 1 \geq \frac{ m( [x , x +h] \cap A )}{h} > 1 - \epsilon \; \; for\; \; 0 < h < \delta$$

and

$$ 1 \geq \frac{ m( [x , x +k] \cap A )}{k} > 1 - \epsilon \; \; for \; \; 0 < k < \delta$$

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  • $\begingroup$ If $A=\emptyset$, then $F' = 0$. $\endgroup$ – copper.hat Dec 8 '13 at 7:30
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$$F(x+h)-F(x)=\int_{[x,x+h]}\chi_{A}dm=m([x,x+h]\cap A)$$

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  • $\begingroup$ How does this answer the question? If the set $A$ has measure zero, then $F(x+h)-F(x) = 0$, and there is no way it can satisfy the condition above? $\endgroup$ – copper.hat Dec 9 '13 at 2:47
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This is not true. If $A$ is the Cantor set, then $F=0$ and so $F'=0$ everywhere.

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  • $\begingroup$ Can someone explain the downvote? $\endgroup$ – copper.hat Dec 9 '13 at 2:44

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