0

I have this function and I want to prove it is Borel measurable. $$\begin{equation} f:\mathbb{R}^2\rightarrow\mathbb{R}:f(x,y)=\begin{cases} \sin(\frac{1}{x-y}) &\text{ for } x>y \\\ x^2+y^2 &\text{ for } x\leq y \end{cases}\end{equation}$$ I thought you could start with only looking at the first part (so for $x>y$) and then the other part. Because if they were both Borel measurable then the function is too. But now I don't know how to do them apart.

NDewolf
  • 1,663
wisk
  • 57
  • How about using the characteristic functions of the sets $A= {(x, y) \in \mathbb{R}^2,, x > y}$ and $B = {(x, y) \in \mathbb{R}^2,, x \leq y}$ –  Jan 03 '20 at 10:01
  • what do you mean? – wisk Jan 03 '20 at 10:27
  • For $A$ and $B$ as avove, let $\mathbf{1}_A$ be the function that is $1$ on $A$ and zero on its complement. Same for $\mathbf{1}_B$. Then, your function can be written using these functions, as sum, product and compositions of measurable functions. This is essentially Kavi Rama Murthy's answer. –  Jan 03 '20 at 11:41

1 Answers1

1

Let $h(x,y)=\frac 1 {x-y}$ if $x >y$ and $0$ otherwise. Then $f(x,y)=\sin (h(x,y))I_A(x,y)+(x^{2}+y^{2})I_B$ where $A=\{(x,y): x>y\}$ and $B=\{(x,y): x\leq y\}$. Since products of Borel measurable functions, compositions of Borel measurable functions and sums of Borel measurable are Borel measurable it is enough it is enough to check that $h$ is Borel measurable. For this it is enough to check that $\{(x,y): h(x,y) <a\}$ is measurable for any $a \geq 0$ I will leave this last part to you.

Kavi Rama Murthy
  • 311,013
  • I understand that you can split that function into two different functions. Then you need to prove that they are Borel measurabel. How do you prove that for g(x,y)=$sin(1/(x-y))$ for x>y and h(x,y)=$x^{2}+y^{2}$ for x<y? – wisk Jan 28 '20 at 14:03
  • @wisk First check that $h$ is measurable. Then conclude that $\sin (h(x,y)$ is measurable. It then follows that $\sin (h(x,y)I_B(x,y)$ is measurable. – Kavi Rama Murthy Jan 28 '20 at 23:18