Show that if $f: [a, b] \to \Bbb{R}$ is Riemann integrable, then $\int_a^xf(s)ds$ is uniformly continuous on $[a, b]$

Question

Really don't know where to start with this one. Maybe use the fact that integrable functions are bounded? Any help would be appreciated.

Answer 1

Let $\epsilon>0$ and let $M$ be a bound for $f$ on $[a,b]$. Then for any $x,y\in[a,b]$ satisfying $|y-x|<\epsilon/M$, we have

\begin{align} \int_a^y f(s)ds-\int_a^xf(s)ds&=\int_a^y f(s)ds-\int_a^xf(s)ds\\ \\ &=\int_x^yf(s)ds\\ \\ &\le M|y-x|\\ \\ &<\epsilon \end{align}

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This uses two important results. First, we need to know that $f$ is bounded. If you were being picky, you also might prove the following: if $g\colon[a,b]\to \Bbb R$ is integrable on $[a,b]$ and bounded by $M$, then $\int_a^b g\le M(b-a)$. To go about that proof, you would note that the upper Riemann sums are always less than $M(b-a)$. Then the sequence you are looking for is strictly bounded by $M(b-a)$, so we can apply Bolzano Weierstrass.