Prove that if $f$ is Lebesgue measurable on $\mathbb{R^n}$, then $f(x-y)$ is Lebesgue measurable on $\mathbb{R^n}\times \mathbb{R^n}$. 
Prove that if $f$ is Lebesgue measurable on $\mathbb{R^n}$, then $f(x-y)$ is Lebesgue measurable on $\mathbb{R^n}\times \mathbb{R^n}$.

My Work:
Define $S:\mathbb{R^n}\times \mathbb{R^n}\rightarrow \mathbb{R^n}$ by $S(x,y)=x-y$. Since $S$ is continuous it is Borel measurable. Hence, if $f$ is Borel measurable on $\mathbb{R^n}$ then $f\circ S=f(x-y)$ is Borel measurable on $\mathbb{R^n}\times \mathbb{R^n}$. But how can I prove that Lebesgue measurability of $f$ implies Lebesgue measurability of $f(x-y)$? Any help would be appreciated.
 A: Let $h:\mathbb{R}^{2d}\to\mathbb{R}^d$ be such that $h(x,y)=x-y$ and $f:\mathbb{R}^d\to\mathbb{R}$ a Lebesgue measurable function. We want to show that $K:\mathbb{R}^{2d}\to\mathbb{R}$ with
$
K(x,y):=f\circ h(x,y)=f(x-y)
$
is Lebesgue measurable. 
Now let $E$ be a Borel set of $\mathbb{R}$. It suffices to 
show that $K^{-1}(E)=h^{-1}\circ f^{-1}(E)$ is Lebesgue measurable in $\mathbb{R}^{2d}$. Observe that $f^{-1}(E)$ is Lebesgue measurable. It thus suffices to show that $h^{-1}(F)$ is Lebesgue measurable whenever $F\subset\mathbb{R}^d$ is Lebesgue measurable. 
This is true when $F$ is Borel measurable since $h$ is continuous. On the other hand, one can check that the preimage of a set of zero outer measure is again of zero outer measure, so null sets pull back to null sets. Since Lebesgue measurable sets differ by a null set from a Borel measurable set, the claim for Lebesgue measurable sets then follows from the corresponding claim for Borel sets.

[Added for the "one can check" part] Note that $h(x,y)=\psi\circ T(x,y)$ with
$$
\psi(x,y)=x,\quad T(x,y)=(x-y,x).
$$
It is also known that $m(T(E))=|\det T|m(E)$. To deal with $\psi$, a warming-up question would be to show that if $E\subset\mathbb{R}$ has zero outer measure, then the set  $E\times[0,1]\subset \mathbb{R}^2$ has zero outer measure; using the countable subadditivity of outer measure, this implies that $E\times\mathbb{R}$ also has zero outer measure.
