$\int_a^bf'=f(b)-f(a)$ 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.
 A: 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 $[a,b]$.
A: As stated at the moment there is nothing indicating that $f$ is continuous at the endpoints, so the statement is false.   But an easy edit changes it to this:
Theorem.    Suppose that   $f:[a,b]\to \mathbb R$ is continuous and differentiable on  $(a,b)$ excepting an at most countable number of points.   If $f'$ is  Lebesgue integrable then
$$ \int_a^b f'(x)\,dx = f(b)-f(a). \tag{*}$$
I think there is a particularly simple way of thinking about this. But it comes with a warning:  you have to know integration theory one more step past the Lebesgue theory.  As Gerry Edgar pointed out in a comment, many of us exposed to that theory immediately think of the problem in this context.  Under these hypotheses we know that (*) holds in the sense of the Denjoy-Perron-Henstock-Kurzweil integral, but can fail for the Lebesgue integral.  That integral includes the Lebesgue integral so the extra hypothesis that $f'$ is Lebesgue integrable is both necessary and sufficient.


Here is another way of looking at this, but also invokes some material past the usual Lebesgue theory, namely a well-known generalization of absolute continuity.

*

*If  $g$ is Lebesgue integrable on  $[a,b]$  then there is an absolutely continuous function  $G:[a,b]\to \mathbb R$  so that $G'=g$ almost everywhere and
$$ \int_a^b g(t)\,dt = G(b)-G(a).$$

[Well known]



*On the other hand if $f:[a,b]\to \mathbb R$ is continuous and $f'$ exists at all but countably many points then  $f$ is  ACG${}_*$.


[Well known in the 1930s and known now only to students of
nonabsolutely convergent integrals such as the
Denjoy-Perron-Henstock-Kurzweil integral.]

Now put these together assuming $g=f'$.
Thus  $f -G$ satisfies  $(f-G)'=0$ a.e.  and  $f-G$ is ACG${}_*$ simply because $G$  is absolutely continuous and $f$ is ACG${}_*$ .
Every such function is constant.  So  $f(b)-f(a)=G(b)-G(a)$.
$$ \int_a^b f'(t)\,dt =  \int_a^b g(t)\,dt = G(b)-G(a) = f(b)-f(a).$$
