prove function in product space is measurable Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, define a function $F: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ by $(x,y) \rightarrow f(x-y)$, how do I show that this function is measurable? What I have tried is:
$F^{-1}(-\infty,\alpha) = \{(x,y) \in \mathbb{R} \times \mathbb{R}|  f(x-y) <  \alpha \} = \bigcup\limits_{x \in \mathbb{R}} \{x\} \times \{y \in \mathbb{R} | f(x-y) < \alpha\}$. Since Lebesgue measure is translation invariant, the function $f_{x}: y \rightarrow f(x-y)$ is also measurable, so the set $\{y \in \mathbb{R} | f(x-y) < \alpha\}$ is measurable. The problem is that the union is not over countable set, and I am not sure how to proceed.
 A: This is not specific to $\Bbb{R}$, so let me provide a proof in the general case. Suppose $f:\Bbb{R}^n\to\Bbb{R}$ is Lebesgue measurable, and $F:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}$ is defined as $F(x,y):=f(x-y)$. Now, define the function $T:\Bbb{R}^n\times\Bbb{R}^n\to \Bbb{R}^n\times \Bbb{R}^n$ as $T(x,y):= (x,x-y)$ and $\pi_2:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}^n$ as $\pi_2(x,y):=y$. Then, $F=f\circ\pi_2\circ T$, so for any $A\subset\Bbb{R}$, we have
\begin{align}
F^{-1}(A)&= T^{-1}\bigg[\pi_2^{-1}[f^{-1}(A)]\bigg]= T^{-1}\bigg[\Bbb{R}^n\times f^{-1}(A)\bigg].
\end{align}
Now, if $A$ is actually a Borel set, then by definition of $f$ being Lebesgue-measurable, we have that $f^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n$. Next, the product $\Bbb{R}^n\times f^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n\times\Bbb{R}^n$ (this should be one of the first few things you discuss when talking about Lebesgue measure on a product space). Finally, $T$ is an invertible linear transformation, which in particular means that the function $T^{-1}$ is actually Lipschitz continuous. As a result, it maps Lebesgue-measurable sets to Lebesgue-measurable sets. Thus, $F^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n\times\Bbb{R}^n$, hence $F$ is Lebesgue measurable.

Edit in response to comments regarding measurability
When talking about measurability of functions, one has to be very careful to specify the $\sigma$-algebras involved; not specifying them can easily cause a lot of confusions. If $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces and $f:X\to Y$ is a function, we say $f$ is $\mathcal{A}$-$\mathcal{B}$-measurable (or measurable with respect to $\mathcal{A}$ and $\mathcal{B}$... however you want to phrase things) if the preimage of every set in $\mathcal{B}$ lies in $\mathcal{A}$, i.e $f^{-1}(\mathcal{B})\subset \mathcal{A}$.
When dealing with a topological space $(Y,\tau)$, a natural choice for a $\sigma$-algebra is the one generated by $\tau$; this is called called the Borel $\sigma$-algebra of $Y$, and we shall denote this as $\mathcal{B}_Y$. So, if you have $(X,\mathcal{A})$ a measurable space, and $(Y,\tau)$ a topological space, you can consider the measurable space $(Y,\mathcal{B}_Y)$ so given a function $f:X\to Y$, we can talk about $\mathcal{A}$-$\mathcal{B}_Y$ measurability as I've defined above. By the way, it is a theorem (not too hard to prove) that to check a function is $\mathcal{A}$-$\mathcal{B}$ measurable, it suffices to check that for some (or equivalently, for any) generating set $\mathcal{S}$ of the $\sigma$-algebra $\mathcal{B}$, we have $f^{-1}(\mathcal{S})\subset \mathcal{A}$. So, if you apply this to the situation of the Borel $\sigma$-algebra of a topological space, we see that saying $f:X\to Y$ is $\mathcal{A}$-$\mathcal{B}_Y$ measurable is equivalent to saying preimage of every open set lies in $\mathcal{A}$ (i.e $f^{-1}(\tau)\subset \mathcal{A}$). This is precisely the definition you cite from Rudin.
So far we've talking quite generally. Let's specialize to $\Bbb{R}^n$. Well, this is a topological space, with the usual toplogy generated by (for example) the Euclidean norm. So we can consider the Borel $\sigma$-algebra $\mathcal{B}_{\Bbb{R}^n}$. This is certainly a natural choice for a $\sigma$-algebra. On the other hand, it is also quite natural to consider the Lebesgue $\sigma$-algebra $\mathcal{L}_{\Bbb{R}^n}$ (which you can for example define using the Caratheodory construction). Hence, we now have two fairly natural choices for $\sigma$-algebras on $\Bbb{R}^n$ (for the purposes we usually have in mind).
It is a fact that we have a strict inclusion: $\mathcal{B}_{\Bbb{R}^n}\subsetneq \mathcal{L}_{\Bbb{R}^n}$. So, if we now have a topological space $(Y,\tau)$, with the Borel $\sigma$-algebra $\mathcal{B}_Y$, and a function $f:\Bbb{R}^n\to Y$, then clearly, we have two different notions of measurability. We have the notion of $\mathcal{B}_{\Bbb{R}^n}$-$\mathcal{B}_{Y}$ measurability (for which we say $f$ is a Borel (measurable) function). We also have the notion of $\mathcal{L}_{\Bbb{R}^n}$-$\mathcal{B}_Y$ measurabiltiy (for which we say $f$ is a Lebesgue-measurable function). So as you can see the definition of Lebesgue-measurable is slightly nuanced.
You may now wonder the following question: for functions $f:\Bbb{R}^n\to\Bbb{R}^m$, why not define Lebesgue-measurability to mean $\mathcal{L}_{\Bbb{R}^n}$-$\mathcal{L}_{\Bbb{R}^m}$ measurability. Well, there aren't many such functions, because there exist even continuous functions which are not $\mathcal{L}_{\Bbb{R}}$-$\mathcal{L}_{\Bbb{R}}$ measurable, whereas every continuous function is $\mathcal{L}_{\Bbb{R}}$-$\mathcal{B}_{\Bbb{R}}$ measurable.
A: Since $f$ is measurable by Lusin's theorem, there is a sequence of continuous functions $f_n\to f$ a.e. Since $F_n:\mathbb R\times\mathbb R\to\mathbb R/F_n(x,y)=f_n(x-y)$ is continuous, $F_n$ is measurable and $F_n\to F$ a.e. so $F$ is measurable since the limit of measurable functions is measurable.
