absolutely continuous increasing function on $\mathbb{R}$

Question

Question: Let $f$ be an absolutely continuous increasing function on $[a,b]$. If $E \subseteq (a,b)$ is a Lebesgue measurable set, then $f(E)$ is a Lebesgue measurable set.

Notation: For two sets $A$ and $B$, $$ A \Delta B = (A \setminus B) \cup (B \setminus A) \; . $$ $m^*$ is Lebesgue outer measure.

My attempt: Fix an arbitrary $\varepsilon>0$, then there exists $\delta > 0$, s.t. $$ \sum_{i=1}^{N} \mid f(x_i)-f(y_i) \mid < \varepsilon $$ whenever $\{ (x_i,y_i) \}_{i=1}^N$ is a collection of finitely many disjoint nonempety subintervals of $(a,b)$ satisifying $$ \sum_{i=1}^{N} \mid x_i- y_i \mid < \delta \; . $$ Since $E$ is measurable, then there exists disjoint nonempety subintervals $(a_i,b_i)$($i=1,2,...,n$), s.t. $$ m^* \left(F \Delta \bigcup_{i=1}^{n} \left(a_i,b_i \right) \right) < \delta \; . $$ WLG, we can assume that $$ b_1 \leq a_2, b_2 \leq a_3, \ldots, b_{n-1} \leq a_n \; . $$ Then $$ f \left(\bigcup_{i=1}^{n} \left(a_i,b_i \right) \right) = \bigcup_{i=1}^{n} \left(f(a_i),f(b_i) \right) \; . $$ I want to show that $$ m^* \left(f(E) \Delta \bigcup_{i=1}^{n}(f(a_i),f(b_i)) \right) < \varepsilon \; , $$ but I don't know how to do.

Answer 1

I know a proof of this, but it's more sophisticated than just working directly with outer measures. This proof makes use of two key properties of the Lebesgue measure.

• Regularity. If $E$ is Lebesgue measurable, then for every $\varepsilon > 0$, there exists a closed set $F$ and an open set $U$ such that $F \subset E \subset U$ and $m(U \setminus F) < \epsilon$.
• Completeness. If there exist Lebesgue measurable sets $A$ and $B$ such that $A \subset E \subset B$ and $m(B \setminus A) = 0$, then $E$ is Lebesgue measurable.

From here, the steps are:

• Let $E$ be Lebesgue measurable. Use the regularity property to show that $E$ can be written as a union $E = E_1 \cup E_2$, where $E_1$ is a countable union of closed sets, and $E_2$ is a null set.
• Use the continuity of $f$ to show that $f(E_1)$ is measurable. (Note that a closed subset of $[0, 1]$ is also compact, and continuous functions map compact sets to compact sets.)
• Use the regularity property to show that for any $\delta > 0$, there exists an open set $U$ such that $E_2 \subset U$ and $m(U) < \delta$. Viewing $U$ as a countable union of disjoint open intervals, use the absolute continuity of $f$ to show that for any $\varepsilon > 0$, $f(E_2)$ can be covered by a countable union of intervals whose total length is less than or equal to $\varepsilon$. Use the completeness property to infer that $f(E_2)$ is measurable (with measure zero).
• Conclude that $f(E)$, being the union of the measurable sets $f(E_1)$ and $f(E_2)$, is a measurable set.