$f(x-y)$ considered as a function of $(x,y)\in \mathbb{R^{2n}}$ is measurable if $f$ is measurable

Question

I know there are similar questions up proving this, but I had a question specific to the following proof (specifically in bold):

Let $f$ be a Lebesgue measurable function on $\mathbb{R^n}$. Then the function $f(x)$ considered as a function of $(x,x)\in \mathbb{R^n}$ is Lebesgue measurable.

The linear transformation given by $T:(x,y)\rightarrow (x−y, y)$ is invertible, and so $f(x−y)$ is a Lebesgue measurable function of $(x,y)\in \mathbb{R^n}$. Thus, we see that $f(y)g(x − y)$ is measurable on $\mathbb{R^2n}$.

I know it has something to do with the fact that $T$ maps measurable sets to measurable sets, meaning that $F\circ T = f(x-y)$ is measurable if $F(x,y) = f(x)$, and I'm guessing it has something to do with proving that $T^{-1}$ is Lipschitz, but I just can't make the connection.

Answer 1

For every open $U$, $\{(x, y) : f(x - y) \in U\} = \{(x, y) : x - y \in f^{-1}[U]\}$. So it is enough to show that the function $(x, y) \mapsto x - y$ pull back Leb. measurable sets to Leb. measurable sets. But this follows from the fact that it is continuous and pulls back null sets to null sets. For the latter fact, you can use Fubini's theorem.