0
$\begingroup$

What is wrong with the following statement:

Let $f(x)$ be Riemann-integrable, $$F(x)=\int_a^x f(t)dt.$$ Then $F(x)$ is differentiable, and $F'(x)=f(x)$ almost everywhere.

I think the statement is true. Because $f(x)$ be Riemann-integrable, it's continuous almost everywhere. Then by the Second Fundamental Theorem of Calculus, $F(x)$ is is differentiable almost everywhere, and $F'(x)=f(x)$ almost everywhere.

$\endgroup$
3
  • $\begingroup$ I think you should add something like this: "if $f:I \to \mathbb{R}$ is Riemann integrable on the interval $[a,x] \in I$..." $\endgroup$
    – Botond
    Mar 22 '18 at 15:14
  • $\begingroup$ The way it is written, it could be interpreted that $F$ is differentiable everywhere. $\endgroup$ Mar 22 '18 at 15:15
  • $\begingroup$ @JuliánAguirre yes, maybe. $\endgroup$
    – Hongyan
    Mar 22 '18 at 15:27
0
$\begingroup$

Quick answer: $f$ is Riemann integrable iff $f$ is bounded and continuous almost everywhere (a characterization due to Lebesgue).

In elementary analysis, it is well-known that if $f$ is continuous at $x$, then $F'(x)$ exists and $F'(x)=f(x)$.

$\endgroup$
1
  • $\begingroup$ Sorry, forget to mention that the domain of $f$ and $F$ under consideration is a closed and bounded interval. $\endgroup$ Mar 22 '18 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.