Differentiable almost everywhere of antiderivative function Give an integrable function $h$ on $[a,\,b]$. Let $f(x) = \int_{a}^{x} h$ for all $x\in [a,\,b]$. Prove that $f$ is differentiable almost everywhere on $(a,\,b)$\
My attempt: I tried to show that $f$ is differentiable on any subinterval $[c,d]\subset (a,b)$, but I could not see how to use the assumption $h$ is integrable despite spending several hours. Can anyone please help me with this problem?
 A: Note that
$$h = h^+ - h^-,$$
where
$$h^+ = \max(h,0) \geqslant 0, \\ h^- = \max(-h,0) \geqslant 0. $$
Then
$$f(x) = \int_a^x h^+ - \int_a^xh^-.$$
Since $f$ is the difference of two monotone increasing functions, it is of bounded variation and is differentiable almost everywhere.
A: Royden and Fitzpatrick's 4th ed put this as an exercise before they discuss absolute continuity, so I assume we can do this by Lebesgue's Differentiation Theorem which is proved via Vitali's Covering Lemma. Here's my take:
Assume $g \geq 0$. Then $f$ is increasing so by Lebesgue's theorem, we are done. In the general case, we can write $f$ as a difference of two increasing functions. We will be done if we can show that the sum of two functions, each of which is differentiable almost everywhere, is differentiable almost everywhere. This follows since for any such $f_1,f_2$, by the properties of infimum and supremum, $$\underline{D}(f_1)+\underline{D}(f_2)\leq \underline{D}(f_1+f_2) \leq \bar{D}(f_1+f_2) \leq \bar{D}(f_1)+\bar{D}(f_2)$$ and so $$\{x:\bar{D}(f+g)(x)>\underline{D}(f+g)(x)\}\subset \{x:\bar{D}(f)(x)>\underline{D}(f)(x)\} \cup \{x:\bar{D}(g)(x)>\underline{D}(g)(x)\}$$ The latter two sets are null.
$\textbf{Lebesgue's Theorem}$
If $f$ is monotonic, then $E = \{x:\bar{D}f(x)>\underline{D}f(x)\}$ is a null set.
$\textbf{Proof}$
Consider the collection $E_{\alpha,\beta} = \{x:\bar{D}f(x)>\alpha > \beta >\underline{D}f(x)\}$ where $\alpha,\beta \in \mathbb{Q}$. For a fixed $\beta$, let $\mathcal{F}$ be the collection of intervals $[c,d]$ such that $\frac{f(d)-f(c)}{d-c} < \beta$. This is a Vitali cover of $E_{\alpha,\beta}$ so for each $\epsilon>0$, we can find a finite disjoint subcollection $\{[c_k,d_k]\}_{k=1}^n$ such that $m(E_{\alpha,\beta}-\cup_{i=1}^n[c_i,d_i])=\sum_{i=1}^nm(E_{\alpha,\beta}\cap[c_i,d_i])<\epsilon$. Moreover, by Lemma 3 of $\S 6.2$ of the same book, $m(E_{\alpha,\beta}\cap[c_i,d_i])<\frac{f(d_i)-f(c_i)}{\alpha}$. Now, since $$E_{\alpha,\beta}=(E_{\alpha,\beta} \cap (\bigcup_{i=1}^n[c_i,d_i]))\cup (E_{\alpha,\beta} - (\bigcup_{i=1}^n[c_i,d_i]))$$ and the above inequality gives us $$m(E_{\alpha,\beta})\leq \sum_{i=1}^n m(E_{\alpha,\beta}\cap [c_i,d_i])+\epsilon\leq \frac{1}{\alpha}\sum_{i=1}^n(f(d_i)-f(c_i)) + \epsilon$$ and so $m(E_{\alpha,\beta}) \leq \frac{\beta}{\alpha}m(E_{\alpha,\beta})+\frac{\epsilon}{\alpha}+\epsilon$. Since $\epsilon, \alpha$ and $\beta$ were arbitrary, we can conclude that $m(E)=0$.
A: Another way to prove it is to note that:
$$m(b-a)\le \int_a^b f(x)\ dx \le M (b-a)$$
Where $m,M$ are inf/sup of $f(x)$. Thus:
$$|f(b)-f(a)|\le C |b-a|$$
So $f$ is Lipschitz and thus differentiable a.e.
