I have a problem with a statement in Rudin's book "Real and Complex Analysis" (3rd edition) - proof of Theorem 7.18
Let $f:[a,b] \mapsto \mathbb{R}$ continuous non decreasing. If $f$ maps sets of measure $0$ to sets of measure $0$, then the function $g(x) = x + f(x)$ satisfies the same property. Rudin just states that it is a trivial consequence of the equality $m(g(I))=m(I)+m(f(I))$ for any interval $I$, with $m$ the Lebesgue measure.
I can prove it if $f$ is increasing as follows: If $A$ is measurable s.t. $m(A)=0$ ($m$ is the Lebesgue measure), then we can find a decreasing sequence of open sets $(O_n)$ such that $A \subset O_n$ and $m(O_n)\downarrow_n 0$. If $O=\bigcap_{n=1}^\infty O_n$, then $m(g(A)) \leq m(g(O)) = \lim_n m(g(O_n))=\lim_n \left(m(O_n) + m(f(O_n))\right)=m(O) + m(f(O))=0$ since in this special case $\bigcap_{n=1}^\infty f(O_n)=f(O)$. Otherwise, if $f$ is only non-decreasing, I am not able to prove that $m(f(O))=\lim_nm(f(O_n))=0$.
Thank you for your help!