Image of set of measure zero has measure zero if the function is absolutely continuous

Solution 1:

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).

Solution 2:

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 $\lambda (\cup _i f((a_i,b_i))\leq \sum_i |f(y_i)-f (x_i)|\leq \epsilon$.