I am trying to prove the following statement:
For $h:\mathbb{R}^2\longrightarrow \mathbb{R}$, given by $h(x,y)=x-y$, if $L\in\mathcal{M}(\mathbb{R})$, where $\mathcal{M}(\mathbb{R})$ denotes the set of Lebesgue measurable sets in $\mathbb{R}$, then $h^{-1}(L)\in \mathcal{M}(\mathbb{R}^2)$.
Here's the proof I came up with:
Take $L\in\mathcal{M}(\mathbb{R})$, we have that: \begin{align*} h^{-1}(L)& =\{(x,y)\in\mathbb{R}^2: (x-y)\in L\}\\ & =\{(x,y)\in\mathbb{R}^2: x=z+y, \forall z\in L\}\\ & =\{(x,x-z)\in\mathbb{R}^2: z\in L\}\\ \end{align*}
Now, let $F:L\times\mathbb{R}\longrightarrow \mathbb{R}^2$ be the linear transformation (shearing map in this case) associated with the matrix $A=\begin{bmatrix}1 &1\\0 & 1\end{bmatrix} $ in the standard basis for $\mathbb{R}^2$. Clearly, $F$ is injective. Notice, since $L\in\mathcal{M}(\mathbb{R})$, then $L\times\mathbb{R}\in\mathcal{M}(\mathbb{R^2})$, and since $F$ is injective then: $$\left(h^{-1}(L)=F(L\times \mathbb{R})\right)\in \mathcal{M}(\mathbb{R}^2)$$
Is this proof correct? Are there faster ways of going about this by wrapping $h^{-1}(L)$ in $\mathcal{G}_{\delta}$ or some other method?