Let be $X$ an absolutely continuous random variable. Then, we know that there exists a probability density function (pdf) $f_X$ such that its cumulative distribution function (cdf) $F_X(x)$ can be represented by $$ F_X(x)=\int\limits_{-\infty}^xf_X(t)dt. $$
In this context our professor made the statement, that
$F_X(x)$ is continuously differentiable $\iff$ $f_X$ is continuous.
I tried to verify both directions:
"$\impliedby$":
If $f_X$ is continuous, then the statement follows from the Fundamental Theorem of Calculus (applied to improper integrals).
"$\implies$":
Maybe be we can look at $\int\limits_{-\infty}^xF'_X(t)dt$, which also is an antiderivative of $F'_X(x)$ and compare it to $F_X(x)=\int\limits_{-\infty}^x f_X(t)dt$. My guess was that $F'_X(x)$ could be utilized as pdf which would lead to something like $F_X(x)=\int\limits_{-\infty}^x f_X(t)dt=\int\limits_{-\infty}^x F'_X(t)dt$. But I am not sure if this is the way to go...
Any help is welcome!
PS: I know that there already exist similar questions but mostly they are answered wrong, rely on additional assumptions or leave out important details, see for example https://math.stackexchange.com/a/248272/579544
Continuously differentiable functions satisfy FTC so $F_X(y)-F_X(x)=\int_x^{y}F_X'(t)dt$ for $x <y$. $F_X'(t)$ is non-negtaive because $F_X$ is increasing. Letting $x \to -\infty$ and $y \to +\infty$ in $F_X(y)-F_X(x)=\int_x^{y}F_X'(t)dt$ we see that $F_x'$ integrates to $1$. Hence, $F_x'$ is a density function. Since $P(a<X\leq b)=\int_a^{b} F_X'(t)dt$ it follows that $P(X\in A)=\int_A F_X'(t)dt$ for every Borel set $A$ and $F_X'(t)dt$ is indeed the density function of $X$. Thus, $f_X=F_X'$ which is continuous.