proof that $\hat{f}(x,y)=f(x-y)$ is measurable if $f$ is measurable, Stein & Shakarchi Prop 3.9 I am following Stein and Shakarchi's book on analysis (Real Analysis: Measure Theory, Integration, and Hilbert Spaces) and in Proposition 3.9 on p.86 they present a proof that if $f$ is a measurable function on $\mathbb{R}^d$ then $\hat{f}(x,y)=f(x-y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d$.
The strange thing is, I can follow all the arguments in the proof but I cannot make sense of all of it, I can't string the facts together.
A snipped of the proof is as follows: (from Google books)

The book concludes the proof by saying that any measurable set $E$ can be written as a difference of a $G_\delta$ and a set of measure 0.
Alright so if I proceed with this then because $E=\{z \in \mathbb{R}^d:f(z)<a \} $ as defined in the book is measurable, then it can be written as $A-B$ where $A$ is a $G_\delta$ set while $m(B)=0.$ Now how do I relate this to $\tilde{E}$ as defined in the proof?
help very much appreciated!
 A: Let's introduce a bit of further notation, and define
$$\mu \colon \mathbb{R}^d\times\mathbb{R}^d \to \mathbb{R}^d;\quad \mu(x,y) = x-y.$$
Then the notation in the proof is that $\tilde{M} = \mu^{-1}(M)$ for all (perhaps only all measurable) $M \subset \mathbb{R}^d$.
To show that $\hat{f} = f\circ \mu$ is measurable, it is shown that $\hat{f}^{-1}\bigl((-\infty,a)\bigr)$ is measurable for all $a$. Since $f$ is assumed measurable, $E_a = f^{-1}\bigl((-\infty,a)\bigr) \subset \mathbb{R}^d$ is known to be measurable for all $a$.
The proof now proceeds to show that for all (Lebesgue) measurable $E\subset \mathbb{R}^d$, the preimage $\mu^{-1}(E) \subset \mathbb{R}^d\times\mathbb{R}^d$ is also (Lebesgue) measurable. (Note that it would be trivial for Borel measurable sets, since $\mu$ is continuous.)
The first part of the proof treats a subset of the Borel sets, open sets and $G_\delta$ sets. The preimage of an open resp. $G_\delta$ set is open resp. a $G_\delta$ set, since $\mu$ is continuous.
The remainder of the proof treats the nontrivial case, that the preimage of a null set is also measurable. That finishes the proof because any measurable set is the difference of a $G_\delta$ set and a null set, so its preimage
$$\mu^{-1}(G_\delta \setminus N) = \mu^{-1}(G_\delta)\setminus \mu^{-1}(N)$$
is the difference of a $G_\delta$ set and a measurable set, hence measurable.
To show that the preimage of a null set $E\subset\mathbb{R}^d$ is measurable, the preimage is approximated by constraining the $y$ component,
$$\tilde{E}_k = \mu^{-1}(E)\cap \{(x,y) : \lVert y\rVert < k\}.$$
Then furthermore $E$ is approximated by open sets $O_n$ whose measure tends to $0$. Since the measure of $\mu^{-1}(O) \cap  \{(x,y) : \lVert y\rVert < k\}$ is bounded by the measure of $O$ times the measure of the ball of radius $k$, it follows that $\tilde{E}_k \subset \bigcap (\mu^{-1}(O_n)\cap \{(x,y) : \lVert y\rVert < k\})$ is in fact a null set, hence measurable. Now $\mu^{-1}(E) = \tilde{E} = \bigcup \tilde{E}_k$ is seen to be the countable union of null sets, hence a null set.
