A problem concerning the Measurable function Suppose that $f\in \mathcal{L}(\mathbb{R})$ and $g\in \mathcal{L}(\mathbb{R})$, $\phi(x,y)=f(y-x)g(x)$, prove that $\phi$ is measurable in $\mathbb{R}^2$.
I try to prove this problem by the definition of measurable function:
For any $\alpha\in\mathbb{R}$, $E=\{(x,y)\in\mathbb{R}^2|\phi(x,y)<\alpha\}$ is measurable in $\mathbb{R^2}$.But I failed.
Are there some better ideas to solve this problem about measurable function in $\mathbb{R^2}$ ?
 A: First of all, the $\sigma$-algebras $\mathcal{B}(\mathbb{R}^2)$ and $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$ are the same, since $\mathbb{R}$ is a separable metric space.
So, a function which takes values in $\mathbb{R}^2$ is measurable if and only if its two coordinates functions (which take values in $\mathbb{R}$) are measurable.
The map $(x,y) \in \mathbb{R}^2 \to x - y \in \mathbb{R}$ is continuous, and so is also measurable.
Hence, the map $(x,y) \to f(x-y)$ is measurable (by composition).
The map $(x,y) \to g(y)$ is measurable (composition of the second coordinate map with $g$).
We can deduce that the map $(x,y) \to (f(x-y),g(y))$ is measurable.
Since the map $(x,y) \to xy$ is measurable (because continuous), by composition we have that $(x,y) \to f(x-y)g(y)$ is measurable.
A: $\newcommand{\R}{\Bbb R}\newcommand{\Z}{\Bbb Z}\newcommand{\N}{\Bbb N}\newcommand{\Q}{\Bbb Q}$ 
I don't know how many tools you have available, but a solution can go like this.
If $F$ and $G$ are measurable functions defined in $\R^d$ then $FG$ is a measurable function. A proof of this fact can be found here.
Then it's enough to show that the maps
$$(x,y)\mapsto f(y-x),\qquad (x,y)\mapsto g(x)$$
are measurable functions.
First, we need the following Lemma.
Lemma. Let $E\subseteq \R$ a measurable set. If $[a,b[$ is a finite interval, then $E\times [a,b[$ and $[a,b[\times E$ are measurables in $\R^{2}$.
Notice that it's enough to prove the Lemma for the interval $[0,1[$. Then the proof of the Lemma is in stages:


*

*First consider $E$ with $m(E)=0$

*Then consider $E=(c,d)$ a finite open interval.

*$E$ open set with finite measure.

*$E$ a $G_\delta$ set with finite measure.

*$E$ a measurable set with finite measure.

*$E$ a measurable set.


Notice that from the Lemma follows:
Corollary. If $E\subseteq \R$ is a measurable set, then $E\times \R$ and $\R\times E$ are a measurable sets of $\R^2$. 
To prove the Corollary, note that $$E\times\R=\bigcup_{n\in\Z} E\times [n,n+1[.$$
Finally
Lemma. Let $f:\R\to\R$ a measurable function. Then the functions $F,G:\R^2\to\R$ defined by $$F(x,y)=f(y),\qquad G(x,y)=f(x)$$ are measurable.
Proof. Let $\alpha\in\R$. Notice that
$$\lbrace (x,y):F(x,y)\lt\alpha \}=\R\times \lbrace y:f(y)\lt\alpha\},$$
by the corollary above, we are done.
 To finish note that
$$\lbrace (x,y)\in\Q^2: f(y-x)\lt\alpha\}=\bigcup_{x\in\Q}\lbrace x\}\times(\lbrace y\in\Q : f(y)\lt \alpha\}+x),$$
is measurable, therefore $(x,y)\mapsto f(y-x)$ is measurable. 
