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

My proof:
Since $f$ is differentiable then it's continuous and hence it preserves both compact sets and intervals.
Now since $m(E)&gt;0$ 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 $&gt;0$.

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

For both results, the following lemma will be needed. 


  Lemma : Given $f : \mathbb{R} \rightarrow \mathbb{R}$ 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 $\varepsilon &gt; 0$. We define an increasing sequence of sets $(S_n)_{n \ge 1}$ by : $$S_n = \left \{ x \in S,\ \forall t \in S,\ |t-x| \le \frac{1}{n} \Rightarrow |f(t)-f(x)| \le (C+\varepsilon)|t-x| \right \}$$

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 $k$, $m^*(I_{n,k}) \le \frac{1}{n}$.

Then, for $n \in \mathbb{N}^*$, for all $k \in \mathbb{N}^*$, for $x,y \in S_n \cap I_{n,k}$, we have $|y-x|\le \frac{1}{n}$ 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 $M&gt;0$, $[-M,M] \cap F_k$ 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 $k$, $f(F_k) = \bigcup \limits_{M \in \mathbb{N}^*} f\big([-M,M]\cap F_k\big)$ is a Borel set, so $f(H)$ is measurable.

Now we prove that $f(N)$ is a null set. For $k \in \mathbb{N}^*$, we denote $N_k = \{ x \in N,\ f'(x)&lt;k \}$. $N_k \subset N$ 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 $E$, $f'&gt;0$ on E. We suppose that $m \big( f(E) \big)=0$.

For $x \in E$, $f'(x)&gt;0$ so $\frac{f(x)-f(y)}{x-y}&gt;0$ for all $y \in E \backslash \{x \}$ in some neighborhood of $x$. Thus 
\begin{align*}
E &amp; = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x &gt; q \ \textrm{ and }\ \forall y \in ]q,x[,\ \frac{f(x)-f(y)}{x-y}&gt;0 \right \}\\ 
&amp; = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x &gt; q \ \textrm{ and }\ \forall y \in ]q,x[,\ f(x)&gt;f(y) \right \}
\end{align*}

because $\mathbb{Q}$ is dense. Moreover, $E$ has positive measure and $\mathbb{Q}$ is countable. Hence, there exists $q_0 \in \mathbb{Q}$ such that $B = \left \{ x \in E\ | \ \ x &gt; q_0 \ \textrm{ and }\ \forall y \in ]q_0,x[,\ f(x)&gt;f(y) \right \}$ has positive measure. 

Plus, for $(x,y) \in B^2$ with $x&lt;y$, we have $q_0 &lt; x &lt;y$, so $f(y)&gt;f(x)$. Hence $f_{|B}$ 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 $m(A)=m(B)&gt;0$ and $A$ has no isolated points. Note that we also have $f_{|A}$ increasing, $f'&gt;0$ on A, and $m \big( f(A) \big)=0$.

Now we just need a stronger version of our lemma : 


  Lemma (bis) : Given $A \subset \mathbb{R}$ with no isolated points, and $f : A \rightarrow \mathbb{R}$, we say that $f$ is differentiable at $x \in A$ whenever $\lim \limits_{t \to x^{\neq}} \frac{f(x)-f(t)}{x-t}$ 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 $g = f_{|A}$. As $g$ is stricly increasing, $g^{-1}$ is well defined. Moreover, as $f'&gt;0$ 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 $g^{-1}\big( g(A) \big)$ is a null set, so $A$ is a null set, which is absurd.

$$\textrm{Hence we have }\ m \big (f(E) \big) &gt; 0.$$
From $f'&gt;0$, there exists $c&gt;0$ such that $f'&gt;c$. Then $$m(f(E))=\int_{E} f' dm&gt;\int_{E} c dm=c\int_{E} 1 dm=cm(E)&gt;0.$$