As in the title, suppose $f:[a,b]\to \mathbb R$ is continuous and non-decreasing, and maps sets of measure zero to sets of mesure zero. Then, $g(x)=x+f(x)$ defines a function that also maps sets of measure zero to sets of measure zero. This appears in Rudin's Real and Complex Analysis, without proof. My proof goes as follows, and I'd like feedback, improvements or corrections. It seems a bit fiddly.
Suppose $E\subset [a,b]$ and $m(E)=0.$ Then, there are countably many intervals $(a_j,b_j)$ that cover $E$ and satisfy $\sum |b_j-a_j|<\epsilon.$ Then, the intervals $(a_j+f(a_j),b_j+f(b_j))$ cover $g(E)$.
Now, the intervals $(f(a_j),f(b_j))$ cover $f(E)$ so there are $a_j\le a'_j\le b'_j\le b_j$ such that the intervals $(a'_j,b'_j)$ cover $E$ and satisfy $\sum |f(b'_j)-f(a'_j)|<\epsilon.$ Of course, $\sum b'_j-a'_j|\le \sum |b_j-a_j|<\epsilon.$
But then the $(a'_j+f(a'_j),b'_j+f(b'_j))$ cover $g(E)$ so $m(g(E))\le \sum |f(b'_j)-f(a'_j)|+\sum |b'_j-a'_j|<2\epsilon.$