## September 2001 #4

Problem Statement: Assume $f(x)$ is a bounded Riemann Integrable function on $[a,b]$. Set $F(z)=\int_{a}^z f(x)dx$. Prove that $F(z)$ is uniformly continuous on $[a,b]$.

Proof: Let $M>0$ such that $|f(x)|\leq M$ for every $x\in [a,b]$. Let $\varepsilon>0$ and let $\delta=\dfrac{\varepsilon}{M}$. Then for $|z-y|<\delta$ it follows that $|F(z)-F(y)|=|\int_{a}^zf(x)dx-\int_{a}^yf(x)dx|$. We are given that $f(x)$ is Riemann Integrable on $[a,b]$ so these integrals exist and we may manipulate them. WLOG assume that $z\leq y$, then $|F(z)-F(y)|=|\int_{z}^yf(x)dx|\leq|\int_{z}^yMdx|=|M||y-z|. Thus $F(z)$ is uniformly continuous on $[a,b]$.

Reflection: This proof has reminded me to not over think these problems. When I first approached this problem I still had the idea of a bounded derivative implies uniformly continuous fresh in my mind. This led me to use the Fundamental Theorem of Calculus, parts one and two to show that $F'(z)$ was bounded on $[a,b]$ and using the MVT we get that $F(z)$ is uniformly continuous. This was all fine and dandy until I realized that to use FTC we need $f(x)$ to be continuous. Oh no!! We don’t know anything about $f(x)$ other than that it’s Riemann Integrable. Unlike being differentiable, which tells us that $f(x)$ is continuous, being Riemann Integrable really doesn’t tell us a whole lot…So, after realizing that we are not guaranteed $f(x)$ is continuous we  had to try another simpler, more direct, approach.  So, the moral of the story is, don’t over think!!