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|<M\delta=M\left(\dfrac{\varepsilon}{M}\right)=\varepsilon. 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!! 

This entry was posted in Analysis, Riemann Integrable, Uniform Continuity and tagged , , . Bookmark the permalink.

2 Responses to September 2001 #4

  1. Sir Wadesworth says:

    Yup, sometimes you can make a straightforward problem a hard one by over-thinking. You got this one though.

    Just be careful with your inequalities. The first < in the chain of inequalities should be a =, right?

  2. eekelly2388 says:

    Lol, yes it should. Thanks, Sir Wadesworth 🙂

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