# 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.

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.

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

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 $F'(x)=f(x)$.
• Sorry, forget to mention that the domain of $f$ and $F$ under consideration is a closed and bounded interval. Mar 22 '18 at 15:30