# Proving that $f(x)=\frac{1}{x^2 \ln x}$ is Lebesgue measurable on $(2, + \infty)$

I have that a set $E$ is Lebesgue measurable if the outer measure: $$\mu^*(E)=\inf_{I_1,...,I_n} \mu (I), E \subseteq I_1 \cup I_2 ,...\cup I_n , I_i-\text{intervals}$$

satisfy the three properties of measure. Proving that $$f(x)=\frac{1}{x^2 \ln x}$$ is Lebesgue measurable would suggest that I must prove that the set $f((2,+\infty))$ satisfies these properties. Is this correct, and how do I go about proving this? Maybe using the fact that the function is a continuous bijection and a map of an open set would also be open and this would be a subset of Borel- sigma algebra. Help ?

• No. You have to show that $f^{-1}(A)$ is a Borel in $(2,\infty)$ for all Borel set $A$ in $\mathbb{R}$ – sinbadh Jan 24 '16 at 22:13
• I do not know how to do this. Thank you for that insight. – Jerry West Jan 24 '16 at 22:14
• @JerryWest Your function is continuous, so $f^{-1}(A)$ is open when $A$ is open. Depending on your exact definitions, this might give measurability by itself. With sinbadh's definition you need to do a little extra work proving the equivalence. – Ian Jan 24 '16 at 22:22
• I see that $f^{-1}(A)$ is open when $A$ is open, I just dont know how to prove that $f^{-1}(A) \in (2,+ \infty ) \ \ \forall A \in \mathcal B( \mathbb R)$ – Jerry West Jan 24 '16 at 22:38
• @JerryWest If that is your definition of measurability (in many books it isn't), the important thing is to show that it is Borel, not that it is in $(2,+\infty)$ (which comes for free anyway). To do this you need a lemma: let $\mathcal{A},\mathcal{B}$ be $\sigma$-algebras on $X,Y$ respectively, and $f : X \to Y$. Let $\mathcal{C}$ be a base of $\mathcal{B}$ (for instance, if $\mathcal{B}$ is a Borel $\sigma$-algebra then $\mathcal{C}$ could be the open sets). If $f^{-1}(B) \in \mathcal{A}$ whenever $B \in \mathcal{C}$, then $f^{-1}(B) \in \mathcal{A}$ whenever $B \in \mathcal{B}$ as well. – Ian Jan 25 '16 at 3:22

The function $f$ is strictly decreasing and continuous hence injective.
Then $f^{-1}((a,b]) = [f^{-1}(b), f^{-1}(a))$ hence measurable.