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$f$ is continuous and of bounded variation on $[0,1]$, $f$ is absolutely continuous on any $[c,1]$ with $c \in (0,1]$. Then $f$ is absolutely continuous on $[0,1]$.

How to show this?

Thanks.

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1 Answer 1

up vote 2 down vote accepted

Here are some hints to get you started.

Given $\epsilon>0$, choose $c\in(0,1]$ so that the total variation of $f$ on $[0,c]$ is less than $\epsilon/2$. Then choose $\delta>0$ so that $\sum_{k=1}^n|f(y_k)-f(x_k)|<\epsilon/2$ whenever $\{[x_k,y_k]:k=1,\dots,n\}$ is a finite family of pairwise disjoint intervals in $[c,1]$ such that $\sum_{k=1}^n(y_k-x_k)<\delta$.

Now show that if $\{[x_k,y_k]:k=1,\dots,n\}$ is a finite family of pairwise disjoint intervals in $[0,1]$ such that $\sum_{k=1}^n(y_k-x_k)<\delta$, then $\sum_{k=1}^n|f(y_k)-f(x_k)|<\epsilon$. Bear in mind that $c$ may fall in the interior of one of the intervals $[x_k,y_k]$.

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I have the same in mind BUT is there something like bounded varia –  Salih Ucan Sep 8 '12 at 16:37

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