I want to show if $f(x_1,x_2)=\dfrac1{1-x_1x_2}$ is Lebesgue-measurable or not on $[0,1)^2$. How do I start in this case, because the function is 2 variables?
Normally, I would look if the set $\{f>a\}$ is in the $\sigma$-algebra $\forall$ $a$, as I have practised it before with Borel sets. I do not think it is useful here. i read on Internet that a.e.-continuous implies measurable?
As this is homework, could please someone provide a hint?
On
For $(x_{1},x_{2})\in[0,1)^{2}$, $x_{1}x_{2}\ne 1$ (if it were, then $x_{1}=x_{2}^{-1}>1$ since $0<x_{2}<1$, a contradiction).
The map $\varphi:(x_{1},x_{2})\rightarrow 1-x_{1}x_{2}$ is clearly continuous, now on $[0,1)^{2}$, $\varphi(x_{1},x_{2})\ne 0$. Now the map $\xi:u\rightarrow 1/u$ is continuous except at $u=0$, so $\xi$ is continuous at $u=\varphi(x_{1},x_{2})$, so $f=\xi\circ\varphi$ is continuous at $(x_{1},x_{2})$.
Continuous maps are Borel measurable and hence Lebesgue measurable.
Continuous functions are Borel measurables, hence Lebesgue measurable. Can you use that fact, or do you see how to prove it?