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I have been trying to solve this problem from Bass.

Let $f$ be a real valued absolutely continuous function defined on $[0,1].$ Denote $f(A)=\{f(x): x \in A\}$ for $A \subset [0,1].$ Prove that if $A$ has Lebesgue-measure zero, then so does $f(A)$.

My attempt: By outer regularity, given an $\varepsilon>0$, I have a collection of disjoint open intervals $\{(a_i,b_i)\}_{i \geq 1}$ such that $A \subset \cup_i (a_i,b_i)$ and $\sum_i (b_i-a_i) < \varepsilon$.

Can we link $f(A)$ and $\cup_i f((a_i,b_i))$ ? We know that $f(a,b)$ is an interval but what about it's length ? Is it related to $f(b)-f(a)?$

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Readers of this problem might want to know some context for it.

A function that maps sets of measure zero to sets of measure zero is said to satisfy Lusin's condition (N). The condition plays a key role in many investigations in real analysis. Absolutely continuous functions (as you have seen here) satisfy the condition. In fact there is a partial converse: a necessary and sufficient condition that a continuous function of bounded variation be also absolutely continuous is that it satisfy Lusin's condition (N).

An accessible introduction to the idea along with techniques for handling questions such as the one posed here is given in Saks, Theory of the Integral, pp. 224-228.

Saks remarks that Lusin first introduced the notion in a 1915 study of trigonometric series. He points out for continuous functions this condition is necessary and sufficient in order that the function map measurable sets to measurable sets (a result he attributes to Rademacher and Hahn).

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Suppose $f:[0,1] \to \mathbb{R}$ is absolutely continuous and $A \subset [0,1]$ is a set of measure zero. Let $\varepsilon>0$ be given. Since $f$ is absolutely continuous there exists a positive number $\delta$ so that \begin{equation} \overset{N}{\underset{i=1}\sum}|f(b_i)-f(a_i)| <\varepsilon \, \, \, \text{whenever} \, \, \overset{N}{\underset{i=1}\sum}(b_i-a_i)<\delta, \end{equation} and the intervals $(a_i, b_i)$, $i=1, 2, \ldots, N$ are disjoint. Now fix a compact subset $K$ of $A$. Since $\lambda$ is outer regular there exists an open set $U$ such that $\lambda(U) \leq \lambda(K) + \delta/2$. Since $U$ is open it can be expressed as the countable union of disjoint intervals. Thus a finite subcollection of these intervals $\{(a_i, b_i)\}_{i=1}^N$ covers $K$, \begin{equation} \overset{N}{\underset{i=1}\bigcup}(a_i, b_i) \supset K . \end{equation} Combining the above, notice that we have \begin{equation} \lambda \Big(\overset{N}{\underset{i=1}\bigcup}(a_i, b_i) \Big) = \overset{N}{\underset{i=1}\sum}(b_i-a_i)<\delta . \end{equation}

Thus \begin{equation} \lambda \Big(f\Big( \overset{N}{\underset{i=1}\bigcup}(a_i, b_i) \Big) \Big) \leq \overset{N}{\underset{i=1}\sum}|f(b_i)-f(a_i)|<\varepsilon , \end{equation} and so by monotonicity of $\lambda$ we have that $\lambda(f(K))< \varepsilon$. Observe that any compact set $K'$ contained in $f(A)$ will be the image under $f$ of some compact set $K \subset A$, $\, K'=f(K)$. Therefore we combine to conclude that $\lambda(f(A))=0$.

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    $\begingroup$ I think your argument works if $f$ is monotone, but it does not cover the general case (last inequality sign). $\endgroup$ – Bananeen Nov 25 '16 at 0:41

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