Let's consider $f:~]-\infty,\infty[~\to\mathbb{R}$ a continuous function. Our professor nonchalantly said that if we assume that $\int\limits_{-\infty}^xf(t)dt$ exists, then differentiating by using Fundamental Theorem of Calculus (FTC) yields $f(x)$. The meaning is clear but it is not 100% clear to me why we can state this with no further justification. So I wonder if the following proves this claim.
Let be $(\alpha_n)_{n\in\mathbb{N}}$ a real valued sequence with $\lim\limits_{n\to\infty}\alpha_n=-\infty$ and define $F_n(x):=\int\limits_{\alpha_n}^xf(t)dt$. Then, by applying FTC we get $$ F'_n(x):=\left(\int\limits_{\alpha_n}^xf(t)dt\right)'=f(x) $$ for all $n\in\mathbb{N}$. So we immediatley see that the sequence of functions $F'_n$ converges uniformly to $f$. This allows us to swap $\lim\limits_{n\to\infty}$ and differentiating, so $$ f(x)=\lim\limits_{n\to\infty}F'_n(x)=\left(\lim\limits_{n\to\infty}F_n(x)\right)'=\left(\int\limits_{-\infty}^xf(t)\right)'. $$
This proves the claim.
Is this correct? Maybe there is a more obvious reason why the claim from our professor is true?
It's actually easier. For any $x\in (-\infty, \infty)$, let $I$ be an open interval containing $x$ and $c< x$, $c\notin I$ be fixed. Then
$$ \int_{-\infty} ^x f(t) dt = \int_c^x f(t)dx+ \int_{-\infty}^c f(t) dt.$$
Since $\int_{-\infty}^c f(t) dt$ is just a constant, we have
$$ \left( \int_{-\infty} ^x f(t) dt \right)'= \left( \int_c^x f(t)dx\right)'= f(x).$$