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Let $f$ fail to be of bounded variation on $[0, 1]$. Show that there is a point $x_0 \in [0, 1]$ such that $f$ fails to be of bounded variation on each nondegenerate closed subinterval of $[0, 1]$ that contains $x_0$.

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You can do this with a proof by contradiction. Suppose this is not the case, so that around each point you can find a nondegenerate closed subinterval where $f$ has bounded variation. The interiors of all these intervals forms an open cover of compact $[0,1]$, so one can cover $[0,1]$ with finitely many such open intervals. Can you see why this implies that $f$ must be of bounded variation on all of $[0,1$]?

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1)why one can cover [0,1] with finitely many such open intervals, not infinitely? 2)why this implies that f must be of bounded variation on all of [0,1]? –  i_a_n Nov 23 '12 at 7:25
    
As I mentioned in my answer, this results from the compactness of $[0,1]$. –  Isaac Solomon Nov 23 '12 at 7:26
    
Sorry what compactness means? and why this implies that f must be of bounded variation on all of [0,1]? –  i_a_n Nov 23 '12 at 7:30
    
Compactness: en.wikipedia.org/wiki/Compact_space –  Isaac Solomon Nov 23 '12 at 7:41
    
If $f$ does not vary much on each finite interval, it cannot vary much on the finite union of such intervals. –  Isaac Solomon Nov 23 '12 at 7:42
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