# Is the product operator lebesgue measurable?

Define $$T(x,y):=xy$$ for $$(x,y)\in \mathbb{R}^2$$.

Since $$T$$ is continuous(moreover is of $$C^\infty$$), it is Borel measurable. However, is $$T:\mathbb{R}^2\rightarrow \mathbb{R}$$ Lebesgue measurable?

That is, if $$A$$ is an 1-dim Lebesgue measurable set, then is $$T^{-1}(A)$$ 2-dim Lebesgue measurable set?

EDIT

I wrote my proof for this as an answer below. I hope someone verifies this.. thank you in advance!

• Are you sure that what you've said is what you want "Lebesgue measurable function" to mean? – paul garrett Nov 28 '18 at 2:28
• Yes, I am sure. Would you verify my answer below? – Rubertos Nov 28 '18 at 2:38

Let $$E$$ be an 1-dim Lebesgue measurable set such that $$m(E)=0$$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $$T^{-1}(E)$$ is measurable. Let us first fix $$k\in\mathbb{Z}^+$$. Define $$B_k:=\{(x,y)\in \mathbb{R}^2|1/k \leq |y| \leq k\}$$.
Since $$m(E)=0$$, we can find a sequence $$O_n$$ of open sets such that $$E\subset O_n$$ and $$\lim_{n\to \infty} m(O_n)=0$$.
Since $$T$$ is continuous, $$T^{-1}(O_n)$$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)\cap B_k)=\int \int \mathbb{1}_{O_n}(xy) \mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= \int \frac{1}{|y|} \mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.
Taking $$n\to \infty$$, the right hand side tends to $$0$$. Hence, $$m(T^{-1}(O_n)\cap B_k) \to 0$$ as $$n\to \infty$$. Thus, $$m^*(T^{-1}(E)\cap B_k)=0$$ for all $$k$$. (Here, $$m^*$$ denotes the 2-dim Lebesgue outer measure)
By taking $$k\to \infty$$, we have $$m^*(T^{-1}(E)\cap \{(x,y):|y|>0\})=0$$. Since $$m^*(\{(x,y): |y|=0\})=0$$, we have $$m^*(T^{-1}(E))=0$$. Q.E.D.