# Absolute continuity and sets of measure 0

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 $m(f(O))=\lim_nm(f(O_n))=0$.

Suppose $A$ is of measure zero. For any $\epsilon > 0$ you can cover $A$ by intervals $(c_i,d_i)$ such that $\sum (d_i - c_i) < {\epsilon \over 2}$, and because $f$ takes sets of measure zero to measure zero, you can similarly cover $f(A)$ by intervals $(g_i,h_i)$ such that $\sum (h_i - g_i) < {\epsilon \over 2}$.
Let $E = f^{-1}(\cup_i (g_i,h_i)) \cap (\cup_i (c_i,d_i))$ and write $E = \cup_i (l_i,m_i)$ as a union of open intervals covering $A$. Then
$$m(g(E)) \leq m(f(E)) + m(E)$$ $$\leq \sum_i(h_i - g_i) + \sum_i(d_i - c_i)$$ $$< {\epsilon \over 2} + {\epsilon \over 2}$$ $$= \epsilon$$ Since $A \subset E$ and $\epsilon$ can be made arbitrarily small, $g(A)$ is of measure zero.
Oh yes, I should have think of covering $f(A)$ as well... Thank you very much Zarrax, this definitely solves the question. – kkloo Oct 19 '11 at 16:17