TITLE: If $f$ maps sets of measure zero to sets of measure zero, then so does $g(x)=x+f(x)$.
QUESTION [9 upvotes]: I want to prove the following.

Let $f:[a,b]\to\mathbb{R}$ be continuous and non-decreasing, and suppose that $f$ maps sets of (Lebesgue) measure zero to sets of measure zero. Then, so does
  $$g(x)=x+f(x).$$

This is used in a proof in Rudin Real and Complex Analysis, but I can't understand the argument. He simply say that this "follows easily" from the fact that:

"If the $f$-image of some segment of length $\eta$ has length $\eta'$, then the $g$-image of that same segment has length $\eta+\eta'$."

I am able to prove that this statement is true, but how does that imply that $g$ maps null sets to null sets?

REPLY [2 votes]: If $A$ is measure zero, then it can be covered by an open set $G$ of a small measure. The same way we construct an open set $H\supset f(A)$. Take $U=f^{-1}(H)\cap G$ - it is open and thus is a countable union of disjoint intervals. $f(U)\subset H$, so the sum of the lengths of the $f$-images of these intervals is small (they can intersect in no more than one point). Since $U$ itself has small measure as well, the initial lengths of these intervals sum to something small. Using the Rudin's $\eta+\eta'$, $(x+f)(U)$ is small, yet still contains $(x+f)(A)$. 
P.S. The word "small" can mean "$<C\varepsilon$".
P.P.S. I wonder if something along these lines holds true in higher dimensions.