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If $ f$ is uniformly continuous on a bounded open interval $ (a,b) $, then $ \lim_{x \to a^+} f(x) $ and $ \lim_{x \to b^-} f(x) $ exist and are finite.

I was reading this book: http://classicalrealanalysis.info/documents/T-CalculusIntegral-AllChapters-Portrait.pdf

the above is one direction of theorem 1.12, and is proven in exercise 65. I was having some trouble understanding the proof, and was hoping that someone might be able to explain it more in-depth.

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What parts you don't understand? Try to be specific. –  Pedro Tamaroff Sep 16 '13 at 1:42
    
I suppose that I don't quite understand the point of the oscillation of a function, and its use in the proof. –  user95072 Sep 16 '13 at 1:55
    
Can you at least see, intuitively, why the oscillation is zero if and only if the function is continuous there, say? –  Pedro Tamaroff Sep 16 '13 at 1:58
    
Yes, that makes sense –  user95072 Sep 16 '13 at 2:06
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1 Answer

Extension of uniformly continuous functions is discussed in detail in $\S$ 10.11 of these notes.

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(I guess this is the way to go, the other notes are strange.) –  Pedro Tamaroff Sep 16 '13 at 2:17
    
@Peter: For some reason this basic result is hard to find in most elementary analysis courses. I remember thinking that when I taught undergrad analysis at McGill almost ten years ago and making a point of putting it in the notes. IIRC though, the course text used at the time (by Russell Gordon; it is really quite a good book) did treat this material, and I took my discussion more or less from there. –  Pete L. Clark Sep 16 '13 at 2:23
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