Let $\mu$ be an unsigned measure which is differentiable with respect to the Lebesgue measaure $m$. Let $f$ be its Radon-Nikodym derivative w.r.t. $m$. i.e. $f=\frac{d\mu}{dm}$, or alternatively $\mu(E)=\int_E fdm$. Further assume that $f$ is continuous. I want to show that the function $x\mapsto\mu([0,x])$ is classically differentiable and moreover that $\frac{d}{dx}\mu([0,x])=f(x)$. To show that this function is differentiable, we need to show $$\lim\limits_{h\to 0}\frac{\mu([0,x+h])-\mu([0,x])}{h}$$ exists. If we replace $\mu$ with $\int fdm$, this reduces to showing $$\lim\limits_{h\to 0}\frac{\int_{[0,x+h]}fdm-\int_{[0,x]fdm}}{h}$$ exists. I'm just not sure where to go from here. I guess we need to use the continuity of $f$ somehow but I'm not too sure how to do this.
Edit: For indicator and simple functions, I think I can obtain the result by brutely computing the integrals. I just don't know what to do in the case where $f$ is some arbitrary non-negative measurable function.
Let $h>0$ (the other case is done in the same way), then
$$|\frac{1}{h}(\mu[0,x+h] - \mu[0,x] -hf(x))| = $$ $$\frac{1}{h}|\int_{(x,x+h]} f(t)-f(x) d \lambda(t)| \leq 1/h (h M_h) = M_h$$ where $$M_h = \text{sup}_{t \in (x,x+h]} |f(t)-f(x)| \rightarrow_{h \rightarrow 0} 0$$.
Hence the difference quotient tends to $f(x)$.
It may be worth noting the same holds almost everywhere by the Lebesgue differentiation theorem, in any dimension and for more general sets that shrink in a controlled way, only requiring local integrability of $f$.