TITLE: Image of set of measure zero has measure zero if the function is absolutely continuous
QUESTION [13 upvotes]: 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 $A \subset \bigcup_i (a_i,b_i)$ and $\sum_i (b_i-a_i) < \varepsilon$. 
Can we link $f(A)$ and $\bigcup_i f((a_i,b_i))$ ? We know that $f(a,b)$ is an interval but what about it's length ? Is it related to $f(b)-f(a)?$

REPLY [8 votes]: This answer complements the answer by Pelto making  it rigorous. In that answer it may  not be true that $\lambda (f((a_i,b_i)) \leq |f(b_i)-f(a_i)|$. To get over this problem note that $\lambda (f((a_i,b_i)) \leq \lambda (f([a_i,b_i]) \leq \lambda [f(x_i),f(y_i)])=|f(y_i)-f (x_i)|$ where $x_i$ and $y_i$ are the points in $[a_i,b_i]$ at which $f$ attains its minimum and maximum values. If $I_j$ is the interval $(x_j,y_j)$ if $x_j<y_j$ and   $(y_j,x_j)$ if $y_j<x_j$ then  the intervals $I_j,1\leq j \leq  n$ form a finite disjoint collection of intervals with total length $<\delta$ and hence $\lambda (\cup _i f((a_i,b_i))\leq \sum_i |f(y_i)-f (x_i)|\leq  \epsilon$.