Prove that $F(x,y)=f(x-y)$ is Borel measurable

Question

Suppose $A$ is a subset of $\Bbb R$, let $s(A)=\{ (x,y)\in \Bbb R \times \Bbb R :x-y\in A\}$. I already showed: If $A\in \Bbb B$ (Borel measurable set), then $s(A)\in \Bbb B \times \Bbb B$.

I want to use this to prove that if $f$ is a Borel measurable function on $\Bbb R$ to $\Bbb R$, then the function $F$ defined by $F(x,y)=f(x-y)$ is measurable with respect to $\Bbb B \times \Bbb B$. Could someone help to provide a proof please? Thanks.

Answer 1

For any Borel set $B$ $$ F^{-1}(B)=\{(x,y):F(x,y)\in B\}=\{(x,y):f(x-y)\in B\}=\{(x,y):x-y\in f^{-1}(B)\}, $$ and $f^{-1}(B)$ is Borel, the statement that you already showed implies that the above set is Borel (in $\mathbb{R}\times\mathbb{R}$).

Answer 2

Compositions of Borel measurable functions are Borel measurable. In this case you have a Borel measurable function composed with a continuous function, and a continuous function is of course Borel measurable.