# Prove that $A \in \mathcal{B}(\mathbb R^{3})$

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.

• Any closed set is Borel. Why don't you just verify that $A$ is closed? This is easy. – Kavi Rama Murthy Jan 18 at 23:58
• Define $f(x,y,z) =(x^2+y^2,y^2)$ then $A = f^{-1}((-\infty, 1] \times (-\infty, 4]).$ – Will M. Jan 19 at 0:09

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).