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?


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

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

  • $\begingroup$ Looks OK to me. $\endgroup$ – Kavi Rama Murthy Nov 28 '18 at 5:58

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