$\int_a^bf'=f(b)-f(a)$ if $f'$ is integrable, but not continuous?

Let $a<b$ and $f:(a,b)\to\mathbb R$ be differentiable (i.e. $f'$ exists, but is NOT necessarily continuous). Assuming that $f'$ is Lebesgue integrable, does it still hold that $$\int_a^bf'=f(b)-f(a)?\tag1$$ The claim is clearly true when $f'$ is continuous, since then $(1)$ is simply the second fundamental theorem of calculus. I wasn't able to come up with a couterexample, but I didn't found a reference of the claim either.

There are a few versions of the FTC, which depend on the hypotheses you decide to impose. All of these can be found in Rudin's RCA chapter $7$. For the version you're asking about, note that Rudin's theorem $7.21$ is the following (apparently this version of the theorem isn't as well-known as the other one involving the equivalence with AC functions):

If $f:[\alpha,\beta]\to\Bbb{R}$ is differentiable at every point of $[\alpha,\beta]$ and $f'$ is Lebesgue integrable on $[\alpha,\beta]$, then for all $x\in [\alpha,\beta]$, we have \begin{align} f(x)-f(\alpha)&=\int_{\alpha}^xf'(t)\,dt \end{align}

Note that one can very easily extend this to functions with values in $\Bbb{C}$, or really any finite-dimensional real/complex vector space (and if you manage to define integrals for arbitrary Banach-space valued mappings, you can deduce this theorem for those maps as well by using Hahn-Banach, and reducing to the scalar-valued case above).

Note also that the hypothesis here is that the function is differentiable at every point of the interval; this is definitely not the weakest possible assumption, though based on the way you phrased your question, I presume this is what you're interested in. However, some restrictions are necessary: if you only assume the function is differentiable at almost every point of the interval, with the a.e defined function $f'$ being Lebesgue integrable, then the statement is false, as mentioned in the comments; the Cantor function provides a counter example.

In your case, you're asking about open intervals, and the answer is still yes, because we can just apply the above theorem to every compact subset of $(a,b)$. More precisely,

If $f:(a,b)\to\Bbb{C}$ is differentiable at every point of $(a,b)$, and $f'$ is Lebesgue integrable on $(a,b)$, then $f$ extends to a continuous function $F$ on $[a,b]$, and for all $x\in [a,b]$, we have \begin{align} F(x)-F(a)&=\int_a^xf'(t)\,dt. \end{align}

$f$ has a continuous extension because if you fix an $x_0\in (a,b)$, then $f'\in L^1((a,b))$ implies that the mapping $x\mapsto \int_{x_0}^xf'(t)\,dt$ is uniformly continuous (it's actually absolutely continuous) on $(a,b)$, hence has a continuous extension to the closure $[a,b]$.