$f: \mathbb R^n\to \mathbb R^m$ Lebesgue measurable, is it true that $(x,y)\mapsto f(x+y)$ is Lebesgue measurable on $\mathbb R^{2n}$?
I know in general $f \circ g$ is not Lebesgue measuable provided $f$ Lebesgue measurable and $g$ continuous, but I don't know whether it is true when $g$ is addition.
I'm interested in this question because in the book Measure Theory and Fine Properties of Functions, in Step 7 of the proof of the coarea formula, it is assumed that $(y,w) \mapsto \mathcal H^{n-m}(A\cap f^{-1}(y-\epsilon w))$ is $\mathcal L^{2m}$- measurable, provided $A$ $\mathcal L^n$- measurable and $f:\mathbb R^n\to \mathbb R^m $ Lipschitz. However, in the previous section of the book, it is only proved that $y \mapsto \mathcal H^{n-m}(A\cap f^{-1}(y))$ is $\mathcal L^m$- measurable.
Let $T:\Bbb{R}^{2n}\to\Bbb{R}^{2n}$ be $T(x,y)=(x+y,y)$. This is a linear isomorphism, hence Lipschitz with Lipschitz inverse (all linear transformations on finite-dimensional normed spaces are Lipschitz continuous), hence it and its inverse send Lebesgue-measurable subsets of $\Bbb{R}^{2n}$ to Lebesgue measurable sets of $\Bbb{R}^{2n}$. Also, let $\pi_1:\Bbb{R}^{2n}\to\Bbb{R}^n$, $\pi_1(x,y)=x$ be the projection to the first $n$-coordinates. Let us define $g:\Bbb{R}^{2n}\to\Bbb{R}^m$ as $g(x,y)=f(x+y)$. Note that then we have \begin{align} g&=f\circ \pi_1\circ T \end{align} So, for any Borel set $B\subset\Bbb{R}^m$, we have \begin{align} g^{-1}(B)&=(f\circ \pi_1\circ T)^{-1}(B)\\ &=T^{-1}\left(\pi_1^{-1}\left(f^{-1}(B)\right)\right)\\ &=T^{-1}\left(f^{-1}(B)\times \Bbb{R}^n\right) \end{align} Now, by assumption, $f$ being Lebesgue measurable means preimage of Borel sets in the target $\Bbb{R}^m$ are Lebesgue-measurable in the domain $\Bbb{R}^n$, so $f^{-1}(B)\subset\Bbb{R}^n$ is Lebesgue-measurable. Its product with $\Bbb{R}^n$ is thus Lebesgue measurable in $\Bbb{R}^{2n}$, and finally by my remark above about $T$ having Lipschitz inverse, it follows that $g^{-1}(B)$ is indeed a Lebesgue-measurable set in $\Bbb{R}^{2n}$.