0

Let $\mathcal{f}:[a,b]\rightarrow\mathbb{R}$ be Riemann integrable, and define $F(t)=\int_a^tf$ for $t\in[a,b]$. Recall that $F$ must be continuous but need not be differentiable. Prove that $F$ is differentiable (with derivative $f$) almost everywhere.

Andrés E. Caicedo
  • 79,201
user123220
  • 19

1 Answers1

4

This follows immediately from two facts:

  • If $f$ is Riemann integrable, it is continuous almost everywhere. (This is part of Lebesgue's characterization of Riemann integrability: A function is Riemann integrable iff it is both continuous almost everywhere, and bounded. See for instance Theorem 7.6.5 in Abbott's Understanding analysis.)
  • At any point $x$ of continuity of $f$, $F'(x)$ exists and equals $f(x)$. (This is part of the fundamental theorem of calculus. See for instance Theorem 7.5.1.(ii) in Abbott's book.)
Andrés E. Caicedo
  • 79,201
  • Riemann integerable implies Lebesgue interable but not other way around. Now instead I think we need to prove F is lebesgue differentiable(with derivative f) almost everywhere. How to approach this one any idea? – user123220 Apr 26 '14 at 19:24
  • I do not understand. Is this a different question? – Andrés E. Caicedo Apr 26 '14 at 19:44
  • Sorry it is same question. I am just trying to understand more deeply. So, when f is Riemann integrable, then it implies Lebesgue integrable too? How do we prove the differentiablity of F a.e. from measure theory prespective. – user123220 Apr 26 '14 at 21:44
  • @user123220 You prove it using the argument I outlined. The measure theory that appears in the argument, such as it is, is confined to verifying that Riemann integrable functions are continuous almost everywhere. – Andrés E. Caicedo Apr 26 '14 at 21:53
  • If your question is not about Riemann integration, you should really ask it separately. For what is worth, if $g$ is Lebesgue integrable in $[a,b]$, and $F$ is defined by $F(x)=\int^x_a g(t),dt$, where the integral is in the sense of Lebesgue, then it is also true that $f'=g$ a.e., but the argument is entirely different. – Andrés E. Caicedo Apr 26 '14 at 22:25