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Define $A:=\{(x,y,z) \in \mathbb R^{3}: \frac{y^2}{4}\leq 1, x^2+z^2\leq 1\}$

Background: I want to calculate the integral $\int_{A}d\lambda^{3}$ which is $\int_{\mathbb R^{3}}1_{A}d\lambda^{3}$

In order to use Tonelli I need to show that $1_{A}$ is measurable and I can show that by showing that $A$ is indeed Borel measurable.

Is my proof below sound?

Define: $g(x,y,z)=\frac{y^2}{4}-1$ and $h(x,y,z)= x^2+z^2 -1$. Note that $g$ and $h$ are measurable. It is clear that for $A_{1}:=\{(x,y,z) \in \mathbb R^{3}:\frac{y^2}{4}\leq 1\}$ and $A_{2}:=\{(x,y,z) \in \mathbb R^{3}: x^2+z^2\leq 1\}$

$A_{1}=g^{-1}((-\infty,0])$ which by definition is $\in \mathcal{B}^{3}$

And $A_{2}:=h^{-1}((-\infty,0])$ which also by definition of measurability is $\in \mathcal{B}^3$

Note $A=A_{1}\cap A_{2}\in \mathcal{B}$ which is what was required to prove.

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    $\begingroup$ Any closed set is Borel. Why don't you just verify that $A$ is closed? This is easy. $\endgroup$ – Kavi Rama Murthy Jan 18 at 23:58
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    $\begingroup$ Define $f(x,y,z) =(x^2+y^2,y^2)$ then $A = f^{-1}((-\infty, 1] \times (-\infty, 4]).$ $\endgroup$ – Will M. Jan 19 at 0:09
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To show that the set $A$ is closed we can reason as follows. Suppose that $(x_n,y_n,z_n)\in A$ is such that $(x_n,y_n,z_n)\to (x,y,z)$ i.e. $x_n\to x$, $y_n\to y$ and $z_n\to z$. Then we note that $$ \frac{y_n^2}{4}\leq 1\stackrel{n\to\infty}{\implies}\frac{y^2}{4}\leq 1 $$ and $$ x_n^2+z_n^2\leq 1\stackrel{n\to\infty}{\implies}x^2+z^2\leq 1 $$ by properties of limits. In particular, $(x,y,z)\in A$. So the set $A$ is closed and hence a Borel set (as closed sets are the complement of open sets).

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