Using Lebesgue measure complete to prove: $B$ is measurable then $f(B)$ is measurable provided "$m(C)=0$ then $m(f(C))=0$" I'm struggling to understand the solution to the following problem:

$f: \mathbb{R} \to \mathbb{R}$ is continuous. Prove that the following are equivalent:

*

*$m(C)=0$ then $m(f(C))=0$

*$B$ is measurable then $f(B)$ is measurable.


In the solution for $1 \Rightarrow 2$, they used the completion of Lebesgue measure to write $B$ as the disjoint union of a null set $N$ and a Borel set $A$, so that: $B=A\cup N$. So far so good, but then they wrote: $f(B)$ is measurable since $m(f(A))=m(f(A)\cup f(N))$.
Why does the latter equality mean $f(B)$ is measurable? From my understanding this would only be useful if we knew that $f(A)$ is Borel, and then, from no. 1, we have that $f(N)$ is null so then we would conclude that $f(B)$ is the union of a Borel set and a null set. But my problem is that we don't know that $f(A)$ is Borel. What am I missing?
 A: I am always a bit irritated when a textbook or instructor assigns some technical looking exercise without any historical context or meaning.  So is this just some exercise of little importance other than measure theory calisthenics?
Problem:  Suppose that $f:\mathbb R\to\mathbb R$ is a continuous function.  Determine necessary and sufficient conditions so that $f$ maps measurable sets to measurable sets.
The condition you are asked to consider is a famous one, useful  in many other situations.  It is due to the Russian  mathematician Lusin (or Luzin as it is more often now transcribed).
Definition:  Suppose that $f:\mathbb R\to\mathbb R$.  Then $f$ is said to fullfil Lusin's condition (N) on a set $E$ provided $f$
maps measure zero subsets of $E$ to measure zero sets.
This definition is from a 1915 paper of Lusin.
About our situation here Saks, Theory of the Integral,  has this to say:

"The condition (N) was introduced by N. Lusin who was the first to
recognize the importance of this condition in the theory of the
integral.  It is easy to see that in the domain of continuous
functions the condition (N) is necessary and sufficient in order that
the function should transform every measurable set into a measurable
set (cf. H. Rademach [1916] and H. Hahn, [1921])."

So you have it on good authority that your problem is "easy to see."
Remarkably enough, being assured that something is easy to see is helpful.  Nothing too sophisticated is involved.  Do  remember, however, that continuous functions map compact sets to compact sets.
There are other questions on StackExchange concerning Lusin's condition (N) that are worth consulting.
