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.

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

3 Answers 3


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 $N_k = \{ x \in N,\ f'(x)<k \}$. $N_k \subset N$ 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 $\frac{f(x)-f(y)}{x-y}>0$ 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 $x<y$, we have $q_0 < x <y$, 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.

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

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    $\begingroup$ 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 ? $\endgroup$ Jan 12, 2017 at 14:10
  • $\begingroup$ Also you mean at the end the set A is it self the set E :)? $\endgroup$ Jan 12, 2017 at 14:11
  • $\begingroup$ 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) $\endgroup$
    – charmd
    Jan 12, 2017 at 14:26
  • $\begingroup$ In the end, I managed to use only classic tools, but still, this is not an easy question for an exam $\endgroup$
    – charmd
    Jan 12, 2017 at 16:34
  • $\begingroup$ @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. $\endgroup$
    – 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.

  • $\begingroup$ Have you read Bogachev in its entirety ? :D $\endgroup$ Jan 13, 2017 at 12:25
  • $\begingroup$ 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? $\endgroup$ Jan 13, 2017 at 16:27

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

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    $\begingroup$ How does m(f(E))=∫f'. ?can you give me a reference? $\endgroup$ Jan 8, 2017 at 16:30
  • $\begingroup$ @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$ $\endgroup$
    – kayak
    Jan 15, 2017 at 13:08
  • $\begingroup$ Isn't the monotonicity of $f$ usually required for such a substitution ? $\endgroup$
    – charmd
    Mar 12, 2017 at 12:12

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