If $f, g: \mathbb{R} \to \mathbb{R}$ are measurable, is the product $F(x,y) = f(x)g(y)$ measurable? I need help with how to approach a homework problem. The question is:
Suppose $f, g: \mathbb{R} \to \mathbb{R}$ are measurable functions. Determine whether the following statement is true or false: "The function $F: \mathbb{R}^2 \to \mathbb{R}$ defined by $F(x,y) = f(x)g(y)$ is measurable." If true, provide a proof, if false provide a counterexample.
I haven't been able to construct a counterexample, so I think that it's true, but I have no idea how to show it. How could I start?
 A: I'ts easier to prove that if $f$ measurable then $f^2$ measurable. Use the fact that  $2fg = (f+g)^2 - f^2 - g^2$
A: If $\lambda=0$, then $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>0\}=(\{x\in \mathbb{R}|f(x)>0\}\times\{y\in \mathbb{R}|g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<0\}\times\{y\in \mathbb{R}|g(y)<0\})$. 
Since $f,g$ are measurable, then $\{x\in \mathbb{R}|f(x)>0\},\{y\in \mathbb{R}|g(y)>0\},\{x\in \mathbb{R}|f(x)<0\}\quad and\quad\{y\in \mathbb{R}|g(y)<0\}$ are measurable sets. Thus $\{x\in \mathbb{R}|f(x)>0\}\times\{y\in \mathbb{R}|g(y)>0\}$ and $\{x\in \mathbb{R}|f(x)<0\}\times\{y\in \mathbb{R}|g(y)<0\}$ are measurable. Therefore, $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>0\}$ is measurable. 
If $\lambda>0$, then 
$\{(x,y)\in\mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}
=\{(x,y)\in \mathbb{R}^{2}|f(x)>\lambda/g(y),g(y)>0\}\cup\{(x,y)\in\mathbb{R}^{2}|f(x)<\lambda/g(y),g(y)<0\}
=\bigcup_{i=1}^{\infty}(\{x\in\mathbb{R}|f(x)>r_{i}\}\times\{y\in\mathbb{R}|\lambda/g(y)\leq r_{i},g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<r_{i}\}\times\{y\in\mathbb{R}|r_{i}<\lambda/g(y),g(y)<0\}),r_{i}\in \mathbb{Q}$. 
Since $f,g$ are measurable, then 
$(\{x\in \mathbb{R}|f(x)>r_{i}\}\times\{y\in\mathbb{R}|\lambda/g(y)\leq r_{i},g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<r_{i}\}\times\{y\in\mathbb{R}|r_{i}<\lambda/g(y),g(y)<0\})$ is measurable. Thus when $\lambda>0$,$\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}$ is measurable.
Similarly, we can prove that $\lambda<0$, $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}$ is measurable. 
Thus, $F(x,y)=f(x)g(y)$ is measurable.
A: If $\lambda=0$, then $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>0\}=(\{x\in \mathbb{R}|f(x)>0\}\times\{y\in \mathbb{R}|g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<0\}\times\{y\in \mathbb{R}|g(y)<0\})$. Since $f,g$ are measurable, then $\{x\in \mathbb{R}|f(x)>0\},\{y\in \mathbb{R}|g(y)>0\},\{x\in \mathbb{R}|f(x)<0\}\quad and\quad\{y\in \mathbb{R}|g(y)<0\}$ are measurable sets. Thus $\{x\in \mathbb{R}|f(x)>0\}\times\{y\in \mathbb{R}|g(y)>0\}$ and $\{x\in \mathbb{R}|f(x)<0\}\times\{y\in \mathbb{R}|g(y)<0\}$ are measurable. Therefore, $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>0\}$ is measurable. \
If $\lambda>0$, then $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}=\{(x,y)\in \mathbb{R}^{2}|f(x)>\lambda/g(y),g(y)>0\}\cup\{(x,y)\in \mathbb{R}^{2}|f(x)<\lambda/g(y),g(y)<0\}=\bigcup_{i=1}^{\infty}(\{x\in \mathbb{R}|f(x)>r_{i}\}\times\{y\in\mathbb{R}|\lambda/g(y)\leq r_{i},g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<r_{i}\}\times\{y\in\mathbb{R}|r_{i}<\lambda/g(y),g(y)<0\}),r_{i}\in \mathbb{Q}$. Since $f,g$ are measurable, then $(\{x\in \mathbb{R}|f(x)>r_{i}\}\times\{y\in\mathbb{R}|\lambda/g(y)\leq r_{i},g(y)>0\})\cup(\{x\in \mathbb{R}|f(x)<r_{i}\}\times\{y\in\mathbb{R}|r_{i}<\lambda/g(y),g(y)<0\})$ is measurable. Thus when $\lambda>0$,$\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}$ is measurable. \
Similarly, we can prove that $\lambda<0$, $\{(x,y)\in \mathbb{R}^{2}|F(x,y)=f(x)g(y)>\lambda\}$ is measurable. \
Thus, $F(x,y)=f(x)g(y)$ is measurable.
