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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.
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.
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.
A quite related article can be found at here.
