Why is this function measurable? If $f$ is Borel measurable on $\mathbb{R}$, why is then $f(x-y)$ Bore-measurable on $\mathbb{R}^2$?
I tried showing that the set $\{(x,y): x-y \in f^{-1}(a,\infty)\}$, is measurable, we know that $f^{-1}(a,\infty)$ is in $\mathcal{B}$, but I am not quite sure how to finish the argument. Any tips?
 A: The function $g(x,y) = (x - y)$ is continuous.  Why?  Let $\epsilon > 0$.  Choose $\delta = \frac{\epsilon}{2}$.  Then if $||(x,y) - (x_{0}, y_{0})||:=\max \{|x - x_{0}|, |y - y_{0}| \} < \delta$, we have $|x - y - (x_{0} - y_{0})| = |x - x_{0} + y_{0} - y| \leq |x - x_{0}| + |y - y_{0}| \leq 2 \max \{|x - x_{0}|, |y - y_{0}| \} < 2 \frac{\epsilon}{2} = \epsilon$.
Since $g(x,y) : \Bbb R^{2} \to \Bbb R$ defined by $g(x,y) = x - y$ is continuous, it is Borel measurable.  Since $f$ is Borel measurable, $f \circ g = f(x - y)$ is Borel measurable.  (Since the composition of Borel measurable functions is always Borel measurable.)
A: Let $I=\{(x,y):f(x-y)<a,\:a\in\Bbb{R}\}$. Then
\begin{align}
I&=\bigcup_{r\in\Bbb{Q}}\{x:f(x-r)<a,\:a\in\Bbb{R}\}\times r
\\
&=\bigcup_{r\in\Bbb{Q}}\{(x+r):f(x)<a,\:a\in\Bbb{R}\}\times r
\\
&=\bigcup_{r\in\Bbb{Q}}(\{x:f(x)<a,\:a\in\Bbb{R}\}+r)\times r
\end{align}
Clearly for any $r$, $(\{x:f(x)<a,\:a\in\Bbb{R}\}+r)$ is Borel measurable. So $f(x-y)$ is Borel measurable.
A: The problem is essentially to show that 
$$
A=\left\{ (x, y)\in \mathbb{R}^2\ :\ x-y\in B\right\}$$
is Borel measurable if $B\subset \mathbb{R}$ is. The linear change of coordinates 
$$
\begin{cases}
\xi=x+y \\
\eta =x-y
\end{cases}$$
preserves Borel sets and maps $A$ onto 
$$
\left\{(\xi, \eta)\in\mathbb{R}^2\ : \eta \in B\right\}=\mathbb{R}\times B.$$
The fact that $\mathbb{R}\times B$ is Borel follows from the fact that the Borel sigma algebra on $\mathbb{R}^2$ is the product sigma algebra of two copies of the Borel sigma algebra on $\mathbb{R}$. (See for example the book "Analysis" by Lieb and Loss, second edition, page 8 - although there is no proof). This is actually less trivial than it seems. 
