Measurable function of a transformation still measurable I'm looking for a maybe simpler or more elemental proof of the following statement:
Let $f:\mathbb{R^n}\to \mathbb{R}$ be (Lebesgue-)measurable.
Then $F:\mathbb{R^{2n}}\to \mathbb{R}$ such that $F(x,y)=f(x-y)$ is (Lebesgue-)measurable.
The proof I know works in the following way:
First show that $f$ is measurable on $\mathbb{R^{2n}}$.
Then $T(x,y)=(x-y)$ is Lipschitz and hence $f\circ T$ is measurable (by a theorem ).
Can one somehow argue without the Lipschitz continuity in the last step or with a different ansatz? Several books I've seen introduce the convolution using that $f(x-y)$ is measurable but proof that Lipschitz $T$ maps measurable sets onto measurable sets later in the text.
 A: Alright, so this is as elementary and intuitive as I could think of. We assume only that if $U \subset R^n$ is measurable and $T$ is a rotation on $R^n$, then $T(U)$ is measurable. This is a standard fact about the Lebesgue measure and should be covered quite early in most books. For simplicity, consider $n = 1$.
Define $G:R^2 \to R$ by $G(x,y) = y-x$. Since $F^{-1}(U) = G^{-1}(f^{-1}(U))$ for each $U \in R$ and $f$ is assumed measurable, we only have to show that $G^{-1}(U) \subset R^2$ is measurable for each measurable $U \subset R$ in order to prove that $F$ is measurable.
If $U = \{c\}$ consists of a single point, then
$$G^{-1}(\{c\}) = \{(x,y) \in \mathbb{R}^2: y-x = c\} = \{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}$$
which is the graph of the function $y(x) = x + c$. The slope of this function is $1$ which corresponds to an angle with the $x$-axis of $45$ degrees.
If we let $T: R^2 \to R^2$ be the linear transformation that rotates clockwise by $45$ degrees, then
$$T(G^{-1}(\{c\})) = T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}) = \{(x,c) \in R^2: x \in R\}$$
Now if $U \subset R$ is measurable, then
$$U = \bigcup_{c \in U} \{c\}$$
which means that
$$T(G^{-1}(U)) = \bigcup_{c \in U} T(G^{-1}(\{c\})) = \bigcup_{c \in U}T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\})$$
$$ = \bigcup_{c \in U}\{(x,c) \in R^2: x \in R\} = R \times U$$
which is the product of two measurable sets and hence measurable (by definition).
Thus
$$G^{-1}(U) = T^{-1}T(G^{-1}(U))$$
is measurable since $T^{-1}$ is also a rotation.
A: The rotation technique needs to be preceded by a translation to the origin, and then shifted back. If you rotate the line y=x+1 by 45 degrees, it doesn't become parallel to the x axis unless you do this shifting stuff...
