# 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 $$f'>0$$ 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.

• No, the set of irrational numbers has positive measure but doesn't contain any interval.
– user384138
Jan 7, 2017 at 16:28
• It's not even clear that $f(E)$ is measurable from the given hypotheses.
– zhw.
Jan 10, 2017 at 19:29
• I want to ask the OP: Is this really the problem? is this the exact wording?where does it come from?
– zhw.
Jan 11, 2017 at 20:23
• @charMD can you put your solution ,thanks in advance Jan 11, 2017 at 20:55
• @zhw. Yes I am almost sure,It's a final exam question , I will ask my prof at the beginning of next semester Jan 12, 2017 at 14:20

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 $$S \subset \mathbb{R}$$. 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 $$\varepsilon > 0$$. 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 $$M>0$$, $$[-M,M] \cap F_k$$ 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 $$N_k = \{ x \in N,\ f'(x). $$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 $$\frac{f(x)-f(y)}{x-y}>0$$ for all $$y \in E \backslash \{x \}$$ in some neighborhood of $$x$$. Thus \begin{align*} E & = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x > q \ \textrm{ and }\ \forall y \in ]q,x[,\ \frac{f(x)-f(y)}{x-y}>0 \right \}\\ & = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x > q \ \textrm{ and }\ \forall y \in ]q,x[,\ f(x)>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 $$B = \left \{ x \in E\ | \ \ x > q_0 \ \textrm{ and }\ \forall y \in ]q_0,x[,\ f(x)>f(y) \right \}$$ has positive measure.

Plus, for $$(x,y) \in B^2$$ with $$x, we have $$q_0 < x , so $$f(y)>f(x)$$. 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 $$m(A)=m(B)>0$$ and $$A$$ has no isolated points. Note that we also have $$f_{|A}$$ 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 $$A \subset \mathbb{R}$$ 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 $$f'>0$$ 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.

$$\textrm{Hence we have }\ m \big (f(E) \big) > 0.$$

• Thank you CharMD for your valuable answer .I follow up with your answer until the last step where you said that f^-1 is differentiable (where the result follow immediately as you proved above) so how can we guarantee differentiability of f^-1 ? Jan 12, 2017 at 14:10
• Also you mean at the end the set A is it self the set E :)? Jan 12, 2017 at 14:11
• If $f$ is stricly monotone and that $f'(x)$ exists, with $0<f'(x)<+\infty$, then $f^{-1}$ is differentiable at $f(x)$ with derivative $\frac{1}{f'(x)}$. This can be found in various course of calculus (see en.wikipedia.org/wiki/Inverse_functions_and_differentiation) Jan 12, 2017 at 14:26
• In the end, I managed to use only classic tools, but still, this is not an easy question for an exam Jan 12, 2017 at 16:34
• @charMD It is not necessarily the case that $f|_A$ is increasing. Think of $f(x)=x-\lfloor x\rfloor$ and $A=(0,1)\cup (1,2)$. However, as can be inferred following the proof in the link to the Bogachev book, there exists a subset of $A$ with positive measure on which $f$ is increasing, and from here you can apply the rest of the proof.
– Del
Jan 13, 2017 at 23:54

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.

• Have you read Bogachev in its entirety ? :D Jan 13, 2017 at 12:25
• I haven't; just some sections here and there. I mostly use it as a reference and for the abundant exercises (about a hundred per chapter!). Why do you ask? Jan 13, 2017 at 16:27

From $f'>0$, there exists $c>0$ such that $f'>c$. Then $$m(f(E))=\int_{E} f' dm>\int_{E} c dm=c\int_{E} 1 dm=cm(E)>0.$$

• How does m(f(E))=∫f'. ?can you give me a reference? Jan 8, 2017 at 16:30
• @user283366 $m(f(E))=\int_{y\in f(E)}1\ dy.$Substitute $y=f(x).$Then, $m(f(E))=\int_{x\in E}f'(x)\ dx$ Jan 15, 2017 at 13:08
• Isn't the monotonicity of $f$ usually required for such a substitution ? Mar 12, 2017 at 12:12