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  1. If $S\subseteq \Bbb R$ and $m(S) >0$, then show that for all $c \in (0,1)$ $\exists I$ such that $ {m(S\cap I)} > c\cdot m(I)$

  2. If $f \in BV[a,b]$ and $f\ge 0$, then show $2^f$, $\sqrt{f}$, and $f(\phi)$ for $\phi:[a,b]\rightarrow[a,b]$ continuous is in $BV[a,b]$.

Please any help would be appreciated. I know $f$ can be rewritten as difference of increasing functions $g - h$. But then I get $2^g/2^h$ and $\sqrt{g-h}$.

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$f(\phi)$ doesn't even have the correct domain unless $\phi:[a,b]\to[a,b]$. $f(\phi)$ need not be BV under the stated assumptions. –  Jonas Meyer Feb 10 '13 at 5:19
    
Do you have any work/thoughts to share on the problems? –  Jonas Meyer Feb 10 '13 at 5:20
    
For $\sqrt f$, are you assuming that $f\geq 0$? –  Jonas Meyer Feb 10 '13 at 5:21
    
In 1., what is $I$? Is it an interval? And for 2., I suggest writing down the definition of BV and seeing if you can find upperbounds for the sums. Note that $[a,b]$ is compact, so $\phi$ is uniformly continuous. –  Thomas E. Feb 10 '13 at 6:32
    
@ThomasE: It doesn't help that $\phi$ is uniformly continuous. $f(\phi)$ need not have bounded variation. –  Jonas Meyer Feb 10 '13 at 6:33
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1 Answer

1) It is true, and I suggest first working with the case where $m(S)<\infty$. (Later you can reduce to this case.) If $c>0$ is a constant such that $m(S\cap I)\leq cm(I)$ for each interval $I$, then for each sequence $\{I_n\}_n$ of intervals such that $S\subseteq\cup_n I_n$, we have $m(S)\leq c\sum_n m(I_k)$. Try taking the infimum of the right-hand side of the last equation to see what possible values $c$ can have.

2) Of the three claims here, one is true and two are false.

  • $2^f$ has bounded variation. The image of $f$ is a bounded subset of $\mathbb R$. The function $x\mapsto 2^x$ is Lipschitz on each bounded subset of $\mathbb R$. This allows you to bound the sums used in finding the variation of $2^f$ in terms of the sums for $f$.
  • $\sqrt f$ need not have bounded variation. See here: Composition of two absolute functions
  • $f(\phi)$ need not have bounded variation. E.g., consider cases where $\phi$ is not of bounded variation and $f(x)=x$. Even if $\phi$ also has bounded variation it need not be true; even if $f$ is also continuous. This can be seen in the link above.
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@confused: $S=S\cap\cup_n I_n=\cup_n(S\cap I_n)$. The inequality that results after taking the infimum implies that $c\geq 1$, in the case where $0<m(S)<\infty$. (If $m(S)=\infty$, find a subset $S'\subset S$ such that $0<m(S')<\infty$ and work with $S'$.) –  Jonas Meyer Feb 10 '13 at 8:35
    
@confused: Regarding the second concern in your comment, note that $\forall c>0, P(c)\implies c\geq 1$ is equivalent to $\forall c\in (0,1), \lnot P(c)$. Try using this while replacing "$P(c)$" with "for all intervals $I$, $m(S\cap I)\leq cm(I)$." Regarding the first concern in your comment, note that if $0<a<\infty$ and $a\leq ca$, then $c\geq 1$. –  Jonas Meyer Feb 10 '13 at 23:06
    
@confused: What is the definition of the Lebesgue outer measure of $S$? It is helpful to use this when showing that $m(S)\leq cm(S)$ under the hypothesis that $m(S\cap I)\leq cm(I)$ for each interval $I$. –  Jonas Meyer Feb 10 '13 at 23:16
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