Fundamental theorem of Calculus for nonincreasing function defined on an open interval

I have a question about the integrability of function on an open interval.

Let $f:(0,1)\rightarrow \mathbb{R}$ be a positive, nonincreasing and integrable (Riemann or Lebesgue) on (0,1). Define

$$F(x):=\int_0^x f(t)dt,\text{ for all }x\in [0,1].$$

Is it possible to prove that $F$ is continuous on $[0,1]?$ It seems that the statement holds if $f(x)=\frac{1}{\sqrt{x}}$ but I do not know the proof in general?

Thank you in advance for your help!

Solution 1:

This follows from the dominated convergence theorem $$ \lim_{x\rightarrow x_0} F(x) =\lim_{x\rightarrow x_0} \int_0^1 1_{[0;x)}(t) f(t) dt = \int_0^1 1_{[0;x_0)}(t) f(t) dt = F(x_0).$$ All we need is that $f$ is Lebesgue-integrable.