# real analysis proof question

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 $[0,1$]?
As I mentioned in my answer, this results from the compactness of $[0,1]$. – Isaac Solomon Nov 23 '12 at 7:26
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