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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.

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    $\begingroup$ 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... $\endgroup$ – Alex R. Dec 10 '12 at 1:56
  • $\begingroup$ I cant believe I didnt even consider looking at the definition differentiation. Thanks! $\endgroup$ – Andrew Jean Bédard Dec 10 '12 at 4:58
  • $\begingroup$ Baby Rudin? :D. $\endgroup$ – D1X May 30 '16 at 11:36
  • $\begingroup$ 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. $\endgroup$ – Andrew Jean Bédard Jun 1 '16 at 19:46
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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.

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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.

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