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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)

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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!

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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.

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    $\begingroup$ Thank you, Daniel for the elucidation. It is very much appreciated - spent 2 hours trying to make sense out of the proof and so I decided to come here. It's the first one I've encountered in this book, thus far, which took me longer than usual to figure out. I realized that it was partly due to how the sets were named in the proof! $\endgroup$
    – Tomas Jorovic
    Sep 28, 2013 at 23:51
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    $\begingroup$ It would've been better if the set $E$ was not used to describe the set of measure zero because the set $E$ that was defined initially is not necessarily of zero measure. This is what I was confused about. Everything else though I understood. But once again, many many thanks!!! $\endgroup$
    – Tomas Jorovic
    Sep 28, 2013 at 23:52
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    $\begingroup$ But why $\tilde{E}_k \subset \bigcap_n (\mu^{-1}(O_n)\cap \{(x,y) : \lVert y\rVert < k\})$ is null? It seems that it is equivalent to say $\bigcap_n\mu^{-1}(O_n)$ is null if $O_n$ tends to a null set and this is what we want to prove. $\endgroup$
    – mnmn1993
    Feb 9, 2018 at 16:37
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    $\begingroup$ @mnmn1993 Yes, $\lambda^r$ is the Lebesgue measure on $\mathbb{R}^r$. We typically don't have $\lambda^{2d}\bigl(\mu^{-1}(O)\bigr) = \lambda^d(O)$. What we do have is $$\lambda^{2d}\bigl(\mu^{-1}(O)\bigr) = \int_y \lambda^d\bigl(\{ (x \in \mathbb{R}^d : x - y \in O\}\bigr)\,d \lambda^d(y).$$ By the translation-invariance of the Lebesgue measure, since for every fixed $y$ we have $\{x : x-y \in O\} = y + O$, we have $\lambda^d\{ x : x - y \in O\} = \lambda^{d}(O)$. And thus $\lambda^{2d}(\mu^{-1}(O)) = \lambda^d(O) \cdot \lambda^d(\mathbb{R}^d)$. Which of course is $+\infty$, unless (since $O$ $\endgroup$
    – Daniel Fischer
    Mar 3, 2018 at 18:49
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    $\begingroup$ is open) $O = \varnothing$. But that's why we intersect with $A_k := \{(x,y) \in \mathbb{R}^{2d} : \lVert y\rVert < k\}$. For then we have $$\lambda^{2d}\bigl(\mu^{-1}(O) \cap A_k\bigr) = \int_y \lambda^d\bigl(\{ x : x - y \in O, (x,y) \in A_k\}\bigr)\,d\lambda^d(y) = \int_{\lVert y\rVert < k} \lambda^d\bigl(\{x : x - y \in O\}\bigr)\,d\lambda^d(y) = \lambda^d(O)\cdot \lambda^d\bigl(\{y : \lVert y\rVert < k\}\bigr).$$ And since the ball has finite measure, we see that (still for fixed $k$) $\lambda^{2d}\bigl(\mu^{-1}(O_n) \cap A_k\bigr) = C_k \cdot \lambda^d(O_n)$, $\endgroup$
    – Daniel Fischer
    Mar 3, 2018 at 18:49

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