TITLE: Measure is continuous and difference quotient is indicator function
QUESTION [1 upvotes]: Let $E \subseteq [a,b]$ be a measurable set with positive measure. For $a \leq x \leq b$, define $h(x) = m(E \cap [a,x])$. Prove that $H$ is continous on $[a,b]$. Further, prove
$$\lim_{h \to 0} \frac{m(E \cap [x,x+h])}{h} = \chi_E(x) a.e.$$
on $[a,b]$.
Attempt
Let $x,y \in [a,b]$. WLOG, let $x<y$
$$E \cap [a,x] \setminus E \cap [a,y] = E \cap [x,y]$$
Choosing $|x-y| < \varepsilon$ we have
$$|m(E \cap [a,y])- E \cap [a,y]|=|m(E \cap [x,y])| \leq m([x,y]) < \varepsilon$$
It is clear that for all $x \in E$ we have $\lim_{h \to 0} \frac{m(E \cap [x,x+h])}{h} = \chi_E(x)$. Now, we may presume $E$ is the disjoint union of open sets (I think?) and also that $h$ is strictly positive. So the only places outside $E$ that we might have a non-zero value is the closure of $E$? Maybe?

REPLY [1 votes]: It looks like your proof that $H$ is continuous is correct apart from some typographical errors - the key point is that $|H(y) - H(x)| = m(E \cap [x,y]) \leq m([x,y])$.
For the bit about the limit, it is certainly not the case that $E$ is the union of open sets (disjoint or otherwise) - $E$ could be the complement of $\mathbb{Q}$ in $[a,b]$, for instance.  Also, the equation need not hold for all $x \in E$: for instance if $x$ is an isolated point of $E$ then $m(E \cap [x, x+h]) = 0$ for $h$ sufficiently small and hence the limit is $0$.
Unfortunately you aren't going to find a straightforward characterization of the points where the limit takes the desired value; no matter how you cut it you're going to have to use some tools.  Perhaps the simplest approach is to apply the Lebesgue differentiation theorem to $H$ by expressing it as $H(x) = \int_a^x \chi_E(x)\, dm$.  The conclusion of the theorem is that $H'(x) = \chi_E$ almost everywhere, and this proves what you want since
$$H'(x) = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h} = \lim_{h \to 0} \frac{m(E \cap [x,x+h])}{h}$$
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