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I see questions on the topic that are more specific, or in Math Overflow, and a quick search online does yield some results. However, I am after a simple explanation about how the FTC can be adapted to Lebesgue integration.

Clearly the FTC relies on derivatives (and anti-derivatives), with the essential equation being

$$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

with $\Delta x$ implying partitions in the domain, as opposed to the range as in Lebesgue integration. In this regard, statements about the FTC being the same when a function is both Riemann and Lebesgue integrable, such as in here

When they both exist, Lebesgue and Riemann integration give the same thing. In particular, the fundamental theorem of calculus, substitution theorems, etc, are just as true for the Lebesgue integral as for the Riemann integral.

are a bit disconcerting.

Antoni Parellada
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    The phrase "both Riemann and Lebesgue integrable" is a bit odd: If $f$ is a Riemann integrable function on $[a,b]$, then $f$ is Lebesgue integrable on $[a,b]$. The Lebesgue integral generalizes the Riemann integral. In fact, that's essentially the point: Lebesgue integration allows you to integrate functions whose Riemann integral doesn't exist. It's not to provide some sort of alternative way of computing elementary integrals like $\int_0^1 x^2,dx$. – Jesse Madnick Dec 28 '19 at 01:29
  • Also, your "essential equation" is false. The left side should be $f'(x)$ or $\frac{df}{dx}$, not $\displaystyle \lim_{\Delta x \to 0} f(x)$. – Jesse Madnick Dec 28 '19 at 01:31
  • @JesseMadnick Thank you for both comments - the second one, a careless mistake, now corrected. – Antoni Parellada Dec 28 '19 at 01:55
  • See also here. – Arturo Magidin Dec 28 '19 at 02:00
  • A Garden of Integrals by Burk is a good source to look at the different kinds of integrals, from Cauchy to Riemann to Lebesgue and beyond. – Arturo Magidin Dec 28 '19 at 02:01

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The result you might be interested in looking at is the Lebesgue differentiation theorem. Simply put, the value of $f(x)$ is approximately the average of the points that surround it. If $f$ is Riemann integrable than we can apply standard analysis results to conclude FTC. It becomes more complicated when $f$ is only Lebesgue integrable, relying on certain results such as the covering lemma and the Hardy-Littlewood maximal function. So, if we consider $F(x) = \int_a^x f(y) dy$ then for a sufficiently small ball $B(x,h)$ we have $$\frac{F(x+h) - F(x)}{h} = \frac{2}{m(B(x,h))}\int_x^{x+h}f(y) \ dy \approx \frac{1}{m(B(x,h))}\int_{x-h}^{x+h}f(y) \approx f(x)$$ and taking the limit makes this exact. There is a very nice proof in Bass or Folland.