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I need to show that $f:[0,1] \rightarrow \mathbb{R}$ is Riemann integrable on $[0,1]$ if for all $x \in [0,1]$, $\lim\limits_{t \rightarrow x} f(t)$ exists.

I need to show first $f$ is bounded. Using the compactness of $[0,1]$, I tried using a finite subcover of $[0,1]$ generated from the definition of the limit existing. Now to show that $U(P,f) - L(P,f) < \varepsilon$, I have no idea how to start that part of the proof. Any help is welcome, although help with an elementary approach is preferred since chapter 6 of baby Rudin is mostly what I can work with.

J. W. Tanner
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  • I was comparing the question to Theorem 6.8 in Rudin (where he proves that if a function $f$ on $[a,b]$ is continuous then it's Riemann integrable on $[a,b]$), the only difference here is that there may be finitely many isolated discontinuities on $[0,1]$ here (because of compactness and the limit existing everywhere). So you could try and partition $[0,1]$ using the discontinuous points, and integrate the piecewise continuous components (as a sketch). – BenCWBrown Feb 26 '19 at 00:25
  • The function can have infinitely many discontinuities - for example take $f =1$ except at $1-\frac{1}{n}$ where you take it $1-\frac{1}{n}$ and you can complicate this example quite a lot – Conrad Feb 26 '19 at 01:39

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