Usbrth

Let $f:mathbb Rto mathbb R,$ and suppose $f$ is differentiable at every point of a measurable $Esubset mathbb R,$ 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.

I will prove two points : first, in response to a comment of zhw, we check that $f(E)$ is measurable, and then that its measure is $>0$.

We recall that $m^*$ denotes here Lebesgue outer measure, defined for every subset $S subset mathbbR$. For $I$ an open interval of $mathbbR$, we note $l(I) = sup I - inf I$. Then we define $$m^*(S) = inf left sum limits_k=1^+infty l(I_k), (I_k)_k ge 1 textrm is a sequence of intervals with S subset bigcup limits_k in mathbbN^* I_k right $$

For both results, the following lemma will be needed.

Lemma : Given $f : mathbbR rightarrow mathbbR$ differentiable at any point of a set $S$, assuming that there exists $C ge 0$ such that $forall x in S, |f'(x)| le C$, then $m^* big( f(S) big) le C cdot m^*(S)$.

Proof : Let $varepsilon > 0$. We define an increasing sequence of sets $(S_n)_n ge 1$ by : $$S_n = left le (C+varepsilon)$$

Using the hypothesis $|f'|le C$, we get that $S = bigcup limits_n=1^infty S_n$. For every $n ge 1$, we can take a sequence of open intervals $(I_n,k)_k ge 1$ covering $S_n$ and such that
$sum limits_k=1^+infty m^*(I_n,k) le m^*(S_n)+varepsilon.$

Without loss of generality, we can assume that for every $n$ and $k$, $m^*(I_n,k) le frac1n$.

Then, for $n in mathbbN^*$, for all $k in mathbbN^*$, for $x,y in S_n cap I_n,k$, we have $|y-x|le frac1n$ and $x,y in S_n$, so we can write $|f(y)-f(x)| le (C+varepsilon)|y-x|le (C+varepsilon)cdot m^*(I_n,k)$.
Thus, for $n ge 1$, $$m^*(f(S_n)) le sum limits_k=1^+infty m^*big(f(S_n cap I_n,k) big) le sum limits_k=1^+infty (C+varepsilon)cdot m^*(I_n,k) le (C+varepsilon)cdot (m^*(S_n)+varepsilon)$$

Letting $n to +infty$ and then $varepsilon to 0^+$, we get $m^* big( f(S) big) le C cdot m^* (S)$.

Now we prove that $f(E)$ is measurable. Classically (see Problem about $G_delta$ and $F_delta$ sets),

Claim 1 : There exists a subset $H subset E$ which is $F_delta$ (i.e. a countable union of closed sets) such that $N=E backslash H$ is a null set.

Write $H = bigcup limits_k=1^+infty F_k$ where the $F_k$ are closed. For $k ge 1$, for all $M>0$, $[-M,M] cap F_k$ is compact and $f$ is continuous (because it is differentiable) on this set, so $f([-M,M]cap F_k)$ is closed (it is a compact set).
So for all $k$, $f(F_k) = bigcup limits_M in mathbbN^* fbig([-M,M]cap F_kbig)$ is a Borel set, so $f(H)$ is measurable.

Now we prove that $f(N)$ is a null set. For $k in mathbbN^*$, we denote $N_k = x in N, f'(x)<k $. $N_k subset N$ so $N_k$ is a null set, so $m^*(N_k)=0$ for $k ge 1$. Plus, we can use the previous lemma on $N_k$, because $0 le f' le k$ on $N_k$, so $m^* big(f(N_k) big)le 0$.
Thus $f(N_k)$ is a null set. As $f(E) = f(H) cup f(N)$, we can conclude that $$f(E) textrm is measurable.$$

Now back to the original problem : we have some measurable set $E$ with positive measure, $f$ differentiable on $E$, $f'>0$ on E. We suppose that $m big( f(E) big)=0$.

For $x in E$, $f'(x)>0$ so $fracf(x)-f(y)x-y>0$ for all $y in E backslash x $ in some neighborhood of $x$. Thus
beginalign*
E & = bigcup limits_q in mathbbQ left x > q textrm and forall y in ]q,x[, fracf(x)-f(y)x-y>0 right \
& = bigcup limits_q in mathbbQ left x in E
endalign*

because $mathbbQ$ is dense. Moreover, $E$ has positive measure and $mathbbQ$ is countable. Hence, there exists $q_0 in mathbbQ$ such that $B = left x > q_0 textrm and forall y in ]q_0,x[, f(x)>f(y) right $ has positive measure.

Plus, for $(x,y) in B^2$ with $x<y$, we have $q_0 < x <y$, so $f(y)>f(x)$. Hence $f_$ is increasing.

Finally it is a well-known fact (see Can we have an uncountable number of isolated points) that $B$ has countably many isolated points, and thus we have a measurable subset $A subset B$ such that $m(A)=m(B)>0$ and $A$ has no isolated points. Note that we also have $f_$ increasing, $f'>0$ on A, and $m big( f(A) big)=0$.

Now we just need a stronger version of our lemma :

Lemma (bis) : Given $A subset mathbbR$ with no isolated points, and $f : A rightarrow mathbbR$, we say that $f$ is differentiable at $x in A$ whenever $lim limits_t to x^neq fracf(x)-f(t)x-t$ exists, and we note $f'(x)$ the limit. Assuming that $f$ is differentiable over $A$, and that there exists $C ge 0$ such that $|f'| le C$, we have $$m^* big( f(A) big) le C cdot m^*(A)$$

Proof : the proof is exactly the same as the one we gave for the first lemma.

Finally, we denote $g = f_$. As $g$ is stricly increasing, $g^-1$ is well defined. Moreover, as $f'>0$ on $A$, it is classical (see Inverse functions and differentiation) to show that $g^-1$ is differentiable in the sense of the Lemma bis on $g(A)$. As $g(A)$ is a null set, we can use our lemma (as we did with the set $N$ - see above) to conclude that $g^-1big( g(A) big)$ is a null set, so $A$ is a null set, which is absurd.

$$textrmHence we have m big (f(E) big) > 0.$$

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 $mathbbR$: just write $mathbbR$ 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.

From $f'>0$, there exists $c>0$ such that $f'>c$. Then $$m(f(E))=int_E f' dm>int_E c dm=cint_E 1 dm=cm(E)>0.$$