# Absolutely continuous on a set of measure zero

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.

# Attempt

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, $\lambda([f(a),f(b)]) \leq \sum|f(b_i)-f(a_i)| < \varepsilon$.

# Edit

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 $E_i = (a_i,b_i)$.

• 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 $E \subseteq \bigcup_i E_i$. – user296602 Jan 7 '18 at 19:21
• … and this may happen if $E$ itself has no non-degenerate subintervals, for instance if $E = ℚ ∩ [0..1]$ in $[0..1]$. – k.stm Jan 7 '18 at 19:22
• Can you provide correction? – 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.

• In your last paragraph, do you mean that $\{f(I_i)\}$ is a cover for $f(E)$? – Antoine Love Jun 12 '20 at 15:25
• The $\{f(J_{i})\}$ is a cover for $f(E)$. – user284331 Jun 12 '20 at 15:26
• 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. – Antoine Love Jun 12 '20 at 15:36
• Sorry sorry, this was long ago so I missed it. Actually it should be $\{J_{i}\}$ being the cover, no $f$ instead. – user284331 Jun 12 '20 at 15:38
• Totally get it. I'm surprised that you commented on it! I'll edit it for you if you don't mind. – Antoine Love Jun 12 '20 at 15:38