Prove $f^{-1}(A)$ is measurable if A is measurable and $0<m \leq f'(x)\leq M <\infty$ on $\mathbb{R}$.

Question

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!

Answer 1

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.