TITLE: If $f'(x)>0$ on $E$ , where $m(E)>0,$ then $m(f(E))>0$
QUESTION [17 upvotes]: Let $f:\mathbb R\to \mathbb R,$ and suppose $f$ is differentiable at every point of a measurable $E\subset \mathbb R,$ with $f'>0$ on $E$.
Suppose also that $m(E)>0$ (where $m$ is Lebesgue measure).
Prove that $m(f(E))>0$.

My proof:
Since $f$ is differentiable then it's continuous and hence it preserves both compact sets and intervals.
Now since $m(E)>0$ we can find compact interval inside it (is this true or not?)
If this is true so the proof is completed.
I know that if $E$ is measurable then $E$ is either Borel set or a set of measure zero.
So here $E$ is Borel, but still not necessarily to be an interval.

REPLY [3 votes]: You can prove this by using the following result, which you can find in V. I. Bogachev's Measure Theory book (Springer, 2007). This is Lemma 5.8.13., which I quote almost verbatim but with adapted notation:

Proposition: Let $f$ be a function on $[a,b]$ and let $A$ be the set of all points at which $f$ has a nonzero derivative. Then, for every set $Z$ of measure zero, the set $f^{-1}(Z) \cap A$ has measure zero. In other words, $\lambda \circ f^{-1}|_A \ll \lambda|_A$, where $\lambda$ is Lebesgue measure.

Here is a link to the proof given by Bogachev. It is quite unwieldy (to me, at least) and I must say I haven't gone through the details. The proof relies on Vitali's covering theorem.
Note that the proposition remains true for a function $f$ defined on all of $\mathbb{R}$: just write $\mathbb{R}$ as a countable almost-disjoint union of intervals and apply the proposition to the restriction of $f$ to each interval.
To solve the problem at hand, we argue by contradiction and suppose that $m(f(E)) =0$. Since $E \subset f^{-1}(f(E))$ and $E \subset A$, we have $E \subset f^{-1}(f(E)) \cap A$. Since $f(E)$ has measure $0$, the proposition implies that $f^{-1}(f(E)) \cap A$ has measure $0$. This is a contradiction since we assumed that $m(E) >0$.
Remark: In the above we assumed that $f(E)$ is measurable. This does follow from the hypotheses. See CharMD's great answer for a proof, or alternatively Proposition 5.5.4. in Bogachev's book.