## May 1990 #4

Problem Statement: Using properties of the Riemann Integral show that if $f$ is Riemann Integrable on $[a,b]$ and $x_0\in [a,b]$ then $\phi (x)=\int\limits_{x_0}^{x}f(t)dt$ is uniformly continuous on $[a,b]$.

Proof: We wish to show that for every $\varepsilon>0$ there exists a $\delta>0$ such that for every $|x-y|<\delta$ in $[a,b]$ it follows that $|\phi (x)-\phi (y)|<\varepsilon$.

Since $f$ is Riemann Integrable on $[a,b]$ it follows that $f$ is continuous on a compact set and thus bounded. Let $M\in\mathbb{R}$ such that $|f(x)|\leq M$ for every $x\in [a,b]$.

Let $\varepsilon>0$ and $\delta=\dfrac{\varepsilon}{M}$. Consider $x,y\in [a,b]$ such that $|x-y|<\delta$. Wlog assume that $x.

$|\phi (x)-\phi (y)|=|\int\limits_{x_0}^x f(t)dt-\int\limits_{x_0}^{y}f(t)dt|$

$=|\int\limits_{x}^{y}f(t)dt|$

$\leq |\int\limits_{x}^{y}M dt|$

$=|M(y-x)|$

$

$=M\dfrac{\varepsilon}{M}=\varepsilon$

Therefore, $\phi (x)$ is uniformly continuous on $[a,b]$.

$\Box$

Reflection: Again, another one that really falls out straight by definition. The reason we’re able to combine the integrals is because the $x_0$ is fixed in $[a,b]$. Also, we are given that $f$ is Riemann Integrable and so that tells us that it’s bounded, which is really where the meat of the proof lies.