Does the sum of two continuous non-decreasing functions that map sets of measure zero to sets of measure zero retain this property?

Question

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 $m(g(E))\le \sum |f(b'_j)-f(a'_j)|+\sum |b'_j-a'_j|<2\epsilon.$

Answer 1

Yes, your proof is correct with the additional changes you have incorporated from the comments.

A minor typo: you have written

Of course, $\sum b_j' - a_j' | \leq \sum |b_j - a_j| < \epsilon$.

One of the bars "$|$" is missing. It should be: $ \sum |b_j' - a_j'| $