Let $f$ be an asbolutely continuous, non-decreasing function on a finite, closed interval $[a,b]$. Let $E$ be a set of Lebesgue measure zero in $[a,b]$. Prove that the measurable set $f(E)$ has Lebesgue measure zero.


Fix $\varepsilon >0$. Then there exists $\delta$ such that $\sum|f(b_i)-f(b_i) < \varepsilon$ when $\sum|b_i-a_i|<\delta$ is a finite collection of intervals. Clearly, since $E$ is of measure zero we can parition in $E$ into finitely many disjoint sub-intervals of measure $ < \delta$ for all $\delta > 0$. Thus, $\lambda([f(a),f(b)]) \leq \sum|f(b_i)-f(a_i)| < \varepsilon$.


Fix $\varepsilon >0$. Then there exists $\delta$ such that $\sum|f(b_i)-f(b_i) < \varepsilon$ when $\sum|b_i-a_i|<\delta$ is a finite collection of intervals.

Since $E$ is of measure zero we can write $E=\cup_{i=1}^\infty E_i$ where $\sum \lambda (E_i) < \delta$. However, since $[a,b]$ is compact we need only finitely many $E_i$ andso $E = \cup^k E_i$ and $\sum^k \lambda (E_i) <\delta \implies$ giving the result. Note: we may assume $E_i = (a_i,b_i)$.

  • $\begingroup$ You have the right ideas, but your proof is not correct. You cannot "partition in $E$ into finitely many disjoint sub-intervals of measure $< \delta$" (partially because I'm not sure what you mean by this...) because that's not what measure zero means. It means that you can find a collection of intervals $\{E_i\}_{i = 1}^{\infty}$ such that $\sum |E_i| < \delta$ and $E \subseteq \bigcup_i E_i$. $\endgroup$ – user296602 Jan 7 '18 at 19:21
  • $\begingroup$ … and this may happen if $E$ itself has no non-degenerate subintervals, for instance if $E = ℚ ∩ [0..1]$ in $[0..1]$. $\endgroup$ – k.stm Jan 7 '18 at 19:22
  • $\begingroup$ Can you provide correction? $\endgroup$ – RhythmInk Jan 7 '18 at 19:42

You may modify your reasoning slightly like:

Fix an $\epsilon>0$, then there exists a $\delta>0$ such that $\displaystyle\sum|f(b_{i})-f(a_{i})|<\epsilon$ whenever $\displaystyle\sum|b_{i}-a_{i}|<\delta$, where $\{[a_{i},b_{i}]\}$ is a collection of non-overlapping intervals in $[a,b]$. Here the indexes $i$ may be countably infinite.

Now, as $E$ is of measure zero, there exists a collection $\{I_{i}\}$ of intervals such that $E\subseteq\displaystyle\bigcup_{i}I_{i}$ and $\displaystyle\sum_{i}|I_{i}|<\delta$. By absorbing the overlapping intervals, we know that the resulting collection is non-overlapping and must have measure no more than $\displaystyle\sum_{i}|I_{i}|$, so we may assume that $\{I_{i}\}$ is non-overlapping. Express each $I_{i}$ as $[a_{i},b_{i}]$, then $\displaystyle\sum_{i}|b_{i}-a_{i}|<\delta$. Since $f$ is non-decreasing, then $J_{i}:=[f(a_{i}),f(b_{i})]$ is an interval and $\displaystyle\sum_{i}|J_{i}|=\sum_{i}|f(b_{i})-f(a_{i})|<\epsilon$.

The result follows if we are able to show that the collection $\{J_{i}\}$ is a cover for $f(E)$. For any $x\in E$, choose an $i$ with $x\in I_{i}$, then $a_{i}\leq x\leq b_{i}$, so $f(a_{i})\leq f(x)\leq f(b_{i})$, and hence $f(x)\in[f(a_{i}),f(b_{i})]=J_{i}$, we are done.

  • $\begingroup$ In your last paragraph, do you mean that $\{f(I_i)\}$ is a cover for $f(E)$? $\endgroup$ – Antoine Love Jun 12 '20 at 15:25
  • $\begingroup$ The $\{f(J_{i})\}$ is a cover for $f(E)$. $\endgroup$ – user284331 Jun 12 '20 at 15:26
  • $\begingroup$ Sure, but with your definition of $J_i$ we have $f(J_i) = f([f(a_i),f(b_i)]) = [f(f(a_i)),f(f(b_i))]$, right? I could be wrong, but I think $\{J_i\}$ is a cover for $f(E)$ is what you meant. $\endgroup$ – Antoine Love Jun 12 '20 at 15:36
  • $\begingroup$ Sorry sorry, this was long ago so I missed it. Actually it should be $\{J_{i}\}$ being the cover, no $f$ instead. $\endgroup$ – user284331 Jun 12 '20 at 15:38
  • $\begingroup$ Totally get it. I'm surprised that you commented on it! I'll edit it for you if you don't mind. $\endgroup$ – Antoine Love Jun 12 '20 at 15:38

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