# Fundamental Theorem of Calculus for Lebesgue Integral

I am trying to prove (96) on pg. 324 of baby Rudin.

ie. That is $f$ is Riemann integrable on $[a,b]$ and if $$F(x)=\int_a ^x f(t)dt$$ then $F'=f(x)$ $a.e$ on $[a,b]$

Anything I do seems to just create a circular argument. But so far I know if $f$ is Riemann integrable on $[a,b]$, then $f$ is Lesbesgue integrable on $[a,b]$. It is also possible to create the $U(P,f)$ and $L(P,f)$ as in the Riemann integral definition and get $L=U\ a.e$ thus $$L(t)=f(t)=U(t)$$ And this is where I am lost.

• What happens when you look at $\frac{F(x+h)-F(x)}{h}$? You're integrating on an increasingly small interval so the partitions should work out nicely... Commented Dec 10, 2012 at 1:56
• I cant believe I didnt even consider looking at the definition differentiation. Thanks! Commented Dec 10, 2012 at 4:58
• Baby Rudin? :D.
– D1X
Commented May 30, 2016 at 11:36
• Funny you commented on this and it's nearly 4 years old!!! Baby Rudin refers to the text: Principles of Mathematical Analysis, because it's the bread and butter of many undergraduate courses in Analysis, and it is comparatively elementary to one of Rudin's other well known text: Real and Complex Analysis. Commented Jun 1, 2016 at 19:46

I think you need the condition that a function is Riemann integrable if and only if it is bounded and the set of discontinuities of $f$ has measure 0. This is the so called "Lesbegue's Criterion". Using this it should not be so difficult to prove the above statement.
Following Alex's comment and the above answer. Let's take $\,h>0\,$ for simplicity:
$$\left|\frac{F(x+h)-F(x)}{h}-f(x)\right|=\frac{1}{h}\left|\int_x^{x+h}\left[f(t)-f(x)\right]\,dt\right|\leq$$
$$\leq\frac{1}{h}(x+h-x)\max_{t\in[x,x+h]}|f(t)-f(x)|$$
Check that the last expression above approaches zero a.e. when $\,h\to 0\,$ as the discontinuities of $\,f\,$ in the integration interval have measure zero.