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Another practice preliminary question here. Similar to this one, but the statement is different and I would prefer a non-topological hint or solution since my knowledge of topology is very limited. i.e. IF your solution refers to the topology of the spaces, please elaborate.

Problem Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable function such that there exists $m,M > 0$ for which $0 < m \leq f'(x) \leq M < \infty$ for all $x \in \mathbb{R}$. If $A \subseteq \mathbb{R}$ is Lebesgue measurable, then prove that $f^{-1}(A)$ is also Lebesgue measurable.

I've already deduced the following:

My attempt so far ...

Of these three things I am sure...

  1. $f$ is uniformly continuous, but I'm not sure if that comes into play here.

  2. $f$ is bijective from $\mathbb{R}$ to $\mathbb{R}$ because of its continuity and strict monotonicity.

  3. For any open set $O_\epsilon$, since bjiective functions map open sets to open sets, $f(O_\epsilon)$ is open.

After that my line of reasoning is less certain...

Using the above open set we can approximate $A$ with $f(O_\epsilon)$; i.e. $A \subseteq f(O_\epsilon)$ and $m^*(f(O_\epsilon)\setminus A) < \epsilon$. Thus, by the equivalence theorem we have that the measurability of $f$ means that $f^{-1}(f(O_\epsilon))$ is measurable (I'm not sure what the theorem's name is. It states for any open set $E$, $f^{-1}(E)$ is measurable if and only if $f$ is measurable). Since we can make $f(O_\epsilon)$ as close to $A$ as we want, ... I feel like this is close to the correct direction for solution, but I am probably way off base. Letting $\epsilon = 1/n$ and taking the limit we have $m^*(f(O_{1/n}) \setminus A) < 1/n \rightarrow 0$, but does this imply that $A$ is measurable since it is arbitrarily close to $\lim_{n \rightarrow \infty} f(O_{1/n})$, each of which is measurable, in outer measure? I feel rather lost in the woods.

Thanks in advance for any help!

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Antoine Love
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  • The downvote is a mystery to me. – Randall Jun 03 '20 at 19:53
  • @Randall My first downvote. Oh well. Perhaps the reader thought it was too similar to the other question. I'm happy to remove the question if it truly is equivalent to another which I was unable to find, but in my non-expert opinion it appears unique from the linked problem. – Antoine Love Jun 03 '20 at 19:55
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    @Randall The downvote is unfair, indeed, just because the question is a bit... strange. Continuity of $f$ is sufficient (actually, equivalent) so that $f^{-1}(A)$ is open for an open set $A$, measurability for measurable $A$ follows by a standard argument. –  Jun 03 '20 at 20:01
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    Have you already used the fact that the sigma algebra of closed intervals is equal to the sigma algebra of measurable sets in $\mathbb{R}$? Because the inverse of a closed interval will be a closed interval. Thus, you can generate a countable cover by closed intervals and use $m^(\bigcup_{I \in \mathcal{C}} I) - m^(f(A)) < \dfrac{\epsilon}{2}$ and $m^(f^{-1}(I_n))-m^(f^{-1}(A)\cap f^{-1}(I_n)) < \dfrac{\epsilon}{2^{n+1}}$. I am hand-waving a bit, but with continuity, it should be fairly easy to choose the endpoints of each interval. – SlipEternal Jun 03 '20 at 20:04
  • @ProfessorVector Do we know that $A$ is an open set? – Antoine Love Jun 03 '20 at 20:05
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    @Antoine Love The set of all $A$ with $f^{-1}(A)$ measurable is a $\sigma$-algebra, and contains all open sets. Thus, it contains all measurable sets. –  Jun 03 '20 at 20:10
  • @InterstellarProbe You mean, let $A \subseteq \cup_{I \in C} I$ where $C$ is any countable collection of closed intervals with the given $\epsilon$ properties you described? No, but I can work on that approximation of $A$ as well. – Antoine Love Jun 03 '20 at 20:10
  • @AntoineLove Yes, that was my suggestion. It can probably also work with open intervals, since the inverse of an open interval will be an open interval. – SlipEternal Jun 03 '20 at 20:13
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    Or, you might use the fact that any measurable set $K$ can be represented as $K = N\cup E$ where $N$ is closed (or open) and $m^(E) = 0$. Then, it is a simple matter to show that $$m^(f^{-1}(f(A))) = m^(f^{-1}(f(N)\cup f(E))) = m^(N) + m^(f^{-1}(f(E)))$$ and since $N$ is closed (or open, as you desire), it is measurable, so you just need to show that $m^(f^{-1}(f(E))) = 0$. – SlipEternal Jun 03 '20 at 20:23
  • @ProfessorVector: That is incorrect. There are continuous functions that are not Lebesgue measurable. The are, of course, Borel measurable but that is not what was asked. – copper.hat Jun 03 '20 at 20:57
  • Note that $f$ has an inverse that is Lipschitz continuous and hence maps null sets into null sets. – copper.hat Jun 03 '20 at 21:21

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Note the distinction between Lebesgue & Borel measurablilty.

If $f$ is continuous then it is automatically Borel measurable, however there are continuous functions (see here) that are not Lebesgue measurable. This is because there are many more Lebesgue measurable sets.

It is not too hard (see here) to show that any Lebesgue measurable set can be written as the union of a Borel set and a null set (a set of Lebesgue measure zero).

It is also straightforward (see here) to show that if a function is Lipschitz then it maps null sets to null sets.

It is straightforward to show that the $f$ in the question is a homeomorphism and the inverse function theorem shows that $g$ has differentiable inverse $g$ and ${1 \over M } \le g'(y) \le {1 \over m}$ for all $x$. In particular, $g$ is Lipschitz.

Finally, suppose $A$ is Lebesgue measurable, then we can write $A = B \cup N$ where $B$ is Borel and $N$ is a Lebesgue null set. Then $f^{-1}(A) = f^{-1}(B) \cup f^{-1}(N) = f^{-1}(B) \cup g(N)$, and since $f^{-1}(B)$ is Borel and $g(N)$ is null we see that $f^{-1}(A)$ is Lebesgue measurable.

copper.hat
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