Question:
Suppose that $X: \Omega \to \mathbb{R}$ is a random variable and it's distribution function $F(x) = \mathbf{P}(\xi \le x)$ is differentiable for all $x$. Is it true that $F'(x)$ is density?
What do I know:
Remark 1.
If $F'$ is continuous then $F'$ is density – see, e.g., A random variable $X$ with differentiable distribution function has a density
Hence if there's counterexample $F$ then $F'$ is not continuous everywhere.
Remark 2.
Absolutely continuous measures on $\mathbb{R}$ are precisely those that have densities. And if $g$ is differentiable everywhere then it doesn't follow that $g$ is absolutely continuous.
Unfortunately, the example from the link doesn't help.
Remark 3 (close to remark 2).
In order to make a counterexample it's sufficient to find $F$ such that $F'$ is not intergrable. It's easy to find differentiable function with nonintergable derivative. Consider $g(x) = x^2 \sin(\frac{1}{x^2})$ is differentiable for all $x$ and $\int_{0}^1 g(x) dx$ doesn't exist. Unfortunatelly, $g(x)$ is not a counterexample in our problem, because it's not monotone and hence it's not a distribution function. A function $(g(x) + 1000 x) \cdot const$ also doesn't "work".
Addition: I found the similar question here https://math.stackexchange.com/questions/3387905/derivation-of-distribution-function-is-density-function/3387913
but there're no proofs there, so in fact there's still no answer.
Addition2: If $F'$ exists for all $x$ then it doesn't follow that $F'$ is continious. Counterexample: we may consider $\tilde{F}(x) = \frac{g(x) – g(-5)}{g(5)}I_{|x|\le 5} + I_{x > 5}$ where $g(x) = x^2 \sin (\frac{1}x) + 5x + 10$ and make $F(x)$ which is smooth, nondecreasing and which coinsides with $F$ for all $x \in \mathbb{R} \backslash (U_{\frac1{100}}(-5) \cup U_{\frac1{100}}(5))$.
Best Answer
Use $F(x)$ to define a finite measure $\mu$ on the real line via the usual assignment of $\mu((a, b)) = F(b) - F(a)$. By Lebesgue-Radon-Nikodym, we can decompose $\text d\mu = h \text dm + \text d\lambda$ s.t. $h \in L^1(m)$ and $\lambda \perp m$. Then, $m$-almost everywhere (see Rudin's Real and Complex Analysis Chapter 7, theorem 7.14) we have $F' = D\mu = h$, whence $F' \in L^1(m)$. Theorem 7.21 from the same book shows that $F(x) = \int_{-\infty}^x F'(s) \text ds$.