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

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    $\begingroup$ A set $E$ has measure $0$ iff given $\epsilon >0$ there exist open intervals (not necessarily disjoint) $(a_i,b_i)$ such that $ E \subset \cup (a_i,b_i)$ and $\sum (b_i-a_i) <\infty$. If you drop the word 'disjoint' from both places your argument is fine. $\endgroup$ – Kavi Rama Murthy Oct 11 '18 at 5:47
  • $\begingroup$ @KaviRamaMurthy you only mention $\epsilon$ once $\endgroup$ – mathworker21 Oct 11 '18 at 9:00
  • $\begingroup$ @mathworker21 There was a typo. I meant $\sum (b_i-a_i) <\epsilon$. Thanks for pointing out. $\endgroup$ – Kavi Rama Murthy Oct 11 '18 at 9:02
  • $\begingroup$ The answer to the question in your TITLE is NO. An example is given in Exercise 21.G (p. 155) of Rooij/Schikhof's A Second Course on Real Functions. See also An example that shows the space of functions with N Luzin property is not closed under addition. Incidentally, the Mazurkiewicz paper mentioned in a comment there shows there exists a continuous $N$-function $f(x)$ such that $f(x) + cx$ is NOT an $N$-function for each nonzero value of $c.$ $\endgroup$ – Dave L. Renfro Oct 11 '18 at 10:24
  • $\begingroup$ @DaveL.Renfro You are right. The title does not include all that the OP wanted to assume. Rudin's proof is correct and the result becomes true with the added hypothesis in the body of the question. $\endgroup$ – Kavi Rama Murthy Oct 11 '18 at 11:53

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'| $


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