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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?

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  • $\begingroup$ Are you aware that (good) answers should be accepted. $\endgroup$ – callculus Apr 30 at 15:26
  • $\begingroup$ sorry I do not know what you mean $\endgroup$ – math2020 Apr 30 at 15:27
  • $\begingroup$ You can accept an answer by clicking on the checkmark. $\endgroup$ – callculus Apr 30 at 15:28
  • $\begingroup$ okay thanks will do ! $\endgroup$ – math2020 Apr 30 at 15:29
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Yes, a Riemann integrable function is bounded over closed intervalls. Now consider

$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.

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  • $\begingroup$ Sorry, I don't quite understand how this makes F continuous. $\endgroup$ – math2020 Mar 25 at 12:20
  • $\begingroup$ 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 $\delta< c/M$. $\endgroup$ – Cellardoor Mar 25 at 12:27

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