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Let $f$ be a strictly increasing continuous function defined on $(0, 1)$. Prove that $f' > 0$ almost everywhere if and only if $f^{-1}$ is absolutely continuous.

What I did

$(\Leftarrow)$ for $[a,b] \subset (0,1)$ $f$ is of bounded variation and then $f'$ exists a.e., as $f$ is strictly increasing then $f'\geq 0$. $f: [a,b] \rightarrow f([a,b])$ is bijection and $f([a,b])$ is closed interval.

$f^{-1}:f([a,b])\rightarrow [a,b]$ is of absolutely continuous then $(f^{-1})'$ exists a.e. Therefore $f' > 0$ a.e. (in this case $(f^{-1})(y)' = \dfrac{1}{f'(x)}$ where $f(x) = y$)

As $[a,b]$ is arbritary then $f'> 0$ a. e. on $(0,1)$

$\blacksquare$

I would be very grateful for some help to prove $(\Rightarrow)$

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    Could you please clarify which definition of absolute continuity you are using? – 2017-02-23
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    A function $f:[a,b]\rightarrow \mathbb{C}$ is absolutely continuous if for all $\epsilon > 0$ there exists $\delta > 0$ such that $\sum_{i = 1}^{k} |f(a_i) - f(b_i)| < \epsilon$ whenever a finite family $\{[a_i,b_i]\}_{i = 1}^{k}$ of disjoints subintervals of $[a,b]$ is such that $\sum_{i = 1}^{k}|a_i - b_i| < \delta$ – 2017-02-23
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    @BigbearZzz: Do you have any tip for this problem? – 2017-02-24

1 Answers 1

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Let $(a,b)=f((0,1))$. If $f^{-1}:(a,b)\rightarrow(0,1)$ is not absolutely continuous, then there exists a set $E\subset(a,b)$ of Lebesgue measure zero such that $F:=f^{-1}(E)$ has positive measure (have you proved this? It's called the (N) property). Since $f^{\prime}>0$ for $\mathcal{L}^{1}$ a.e. $x\in F$, we can write $$ F=F_{0}\cup\bigcup_{n=1}^{\infty}F_{n}, $$ where $F_{n}:=\{x\in F:\,f^{\prime}(x)\geq\frac{1}{n}\}$ and $F_{0}$ has measure zero. It follows that% $$ \sum_{n=1}^{\infty}\mathcal{L}^{1}(F_{n})\geq\mathcal{L}^{1}(F)>0, $$ and so for some $n$, $\mathcal{L}^{1}(F_{n})>0$. Consider an open set $U$ that covers $f(F_{n})$. Without loss of generality we can assume that $U\subseteq(a,b)$. Then $U$ is a disjoint union of intervals. Let $J$ be any such interval. Since $J\subseteq(a,b)=f((0,1))$, we can write $J=(f(c),f(d))$ where $0\leq c0$, but $f(F_{n})\subseteq f(F)=E$ which has Lebesgue measure zero and so we can find an open set $U$ covering $F_{n}$ with measure less than $\frac{1} {n}\mathcal{L}^{1}(F_{n})$, which gives a contradiction.

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    What means $\mathcal{L^{1}}$? – 2017-02-25
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    It's the Lebesgue measure – 2017-02-25
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    I don't understand the two final lines. You showed that $\mathcal{L^1}(U) \geq \dfrac{1}{n}\mathcal{L^1}(F_n)$ but says that $U$ have measure less than $\dfrac{1}{n}\mathcal{L^1}(F_n)$. – 2017-02-25
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    $\mathcal{L^1}(F_n) = 0$? How then $\dfrac{1}{n}\mathcal{L^1}(F_n) > 0$? – 2017-02-25
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    Sorry, I meant that I showed that $\mathcal{L}^{1}(U)\geq\frac{1}{n}\mathcal{L}^{1}(F_{n})>0$ . On the other hand, since $\mathcal{L}^{1}(f(F_{n}))=0$, I can cover $f(F_n)$ with an open set of measure as close to zero as I want. In particular I can find a small open set covering $f(F_n)$ with measure less than $\frac{1}{n}\mathcal{L}^{1}(F_{n})>0$. This is a contradiction, which shows that $f^{-1}$ send sets of measure zero into sets of measure zero. Together with the continuity it implies that $f^{-1}$ is absolutely continuous. – 2017-02-25
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    Ok, now I see the contradiction. Thanks. – 2017-02-25
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    @Gio67 if (for example) $c=0$, how could you evaluate $f(c)$? – 2017-02-26
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    Filburt, you are right. The function $f(x)=\frac{1}{x}$ would have this property. You either need to assume that $f$ is defined in $[0,1]$ or alternatively you can only prove that $f^{-1}$ is locally absolutely continuous and so apply my proof in a closed interval $I\subset (0,1)$. – 2017-02-27