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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?

Ice Tea
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    Recall that usual arithmetic preserves measurability. (Of course division needs a bit extra treatment as you need to argue that the set of points where the denominator is zero is measure-zero.) – Sangchul Lee Jan 13 '18 at 18:20
  • you mean $1-x_1x_2$ measureable implies my f is measureable?? –  Jan 13 '18 at 18:22
  • The precise statement depends on how measurability is defined on your textbook, mainly because $f$ is not defined everywhere on $[0,1)^2$. What you can claim is that, if we set $$E = { (x_1, x_2) \in [0, 1)^2 : \text{$f$ is defined at $(x_1,x_2)$}}$$ then (1) $E$ is Lebesgue measurable with $\operatorname{Leb}([0,1)^2\setminus E) = 0$, and (2) $f$ is measurable on $E$. But in application to integration theory this will never bother you as two functions which agree almost everywhere will be considered the same. – Sangchul Lee Jan 13 '18 at 18:29
  • @SangchulLee why is $f$ not defined everywhere on $[0,1)^2$? – la flaca Jan 13 '18 at 18:33
  • I think it is only not defined when $x_1 x_2=1$? –  Jan 13 '18 at 18:34
  • @laflaca, Oops, you are right. I did not notice that indeed the denominator never vanishes on $[0, 1)^2$. :s – Sangchul Lee Jan 13 '18 at 18:36

2 Answers2

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Continuous functions are Borel measurables, hence Lebesgue measurable. Can you use that fact, or do you see how to prove it?

la flaca
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  • I know i can use borelsets $\subset$ lebesguesets. I will try to find in my book about that continuous functons are borel measureables now, i am not sure if i can use it. Thank you. –  Jan 13 '18 at 18:32
  • @wheretostart It is very easy to prove that continuous functions are Borel measurables, it just basically uses mesurability definition and the fact that Borel sigma algebra is generated by open sets. – la flaca Jan 13 '18 at 18:35
  • Oooh of course! –  Jan 13 '18 at 18:36
  • @wheretostart you will probably need to use this fact to prove continuous functions are Bprel measurable: if you have two measurable spaces $(X,\mathcal{M})$ and $(Y,\mathcal{N})$, and $\mathcal{N}$ is generated by a family $G$ of subsets of $Y$, then a function $f:X\to Y$ is measurable iff $f^{-1}(A)\in\mathcal{M}$ for every $A \in G$ – la flaca Jan 13 '18 at 18:46
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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.

user284331
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