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<y.

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


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




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


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.

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