# Given a Riemann Integrable function $F(x) = \int_0^x f$ . Prove F is continuous

Let $$f :\mathbb{R} → \mathbb{R}$$ be Riemann integrable on every interval $$[a,b]$$ and define $$F :\mathbb{R} → \mathbb{R}$$ by $$F(x) = \int_0^x f$$ . Prove that F is continuous.

My thought process behind this proof is that to be Riemann integrable we are assuming f to be bounded, which implies continuity. Then the fundamental theorem of calculus states: Let $$f : [a, b] → \mathbb{R}$$ be continuous and define $$F : [a, b] → \mathbb{R}$$ by $$F(x)= \int_a^x f$$. Then $$F$$ is differentiable. So as we have f a continuous function we can apply the fundamental theorem of calculus. This gives F is differentiable and differentiable functions are continuous so F is continuous.

Is this correct or have I made wrong assumptions at all?

• Are you aware that (good) answers should be accepted. – callculus Apr 30 at 15:26
• sorry I do not know what you mean – math2020 Apr 30 at 15:27
• You can accept an answer by clicking on the checkmark. – callculus Apr 30 at 15:28
• okay thanks will do ! – math2020 Apr 30 at 15:29

$$F(x+h)-F(x)=\int_{x}^{x+h}f(y)dy$$
We have $$|F(x+h)-F(x)|\leq |h|M$$ for some $$M$$ since on $$[x,x+h]$$, $$f$$ is bounded. Clearly $$F$$ is continuous.
• The definition of continuity states that for each $c>0$ we should find a $\delta>0$ such that $|a-b|<\delta$ implies $|F(a)-F(b)|<c$. In our case, for some $c>0$, we let $\delta< c/M$. – Cellardoor Mar 25 at 12:27