I have that $u:\mathbb{R} \rightarrow \mathbb{R}$ given as
$$ u(t) = \begin{cases} 3 & t<0\\ t & t\geq 0 \end{cases} $$ and I want to find $\{u\geq a\}$ for all $a \in \mathbb{R}$ and I am given this hint: look at the situations $a\leq0$ , $0<a\leq3$ and $a>3$
I know I am supposed to generate some open (or semi-open) intervals cause I think that u(t) should be Borel-measurable on $\mathbb{R}$, but I don't know how to set it up?
Can anyone point me in the right direction?
To prove that the function $u$ is Borel measurable, it suffices to show that each preimage $u^{-1}([a, \infty))$ is a Borel set for all $a \in \mathbb{R}$ since the intervals $\{[a, \infty)\}_{a \in \mathbb{R}}$ generate the Borel $\sigma$-algebra.
If $a > 3$, then $u^{-1}([a, \infty)) = (3, \infty)$.
If $0 < a \leq 3$, then $u^{-1}([a, \infty)) = [a, \infty) \cup (-\infty, 0)$.
If $a \leq 0$, then $u^{-1}([a, \infty)) = (-\infty, \infty)$.
In each case, the preimage is either an open interval or a pairwise union of intervals and thus, a Borel set.
For part (c), if $G(x, y) = x$ and $H(x, y) = y$ are Borel measurable, then so is
$$[G(x, y)]^{2} - [H(x, y)]^{2} = x^{2} - y^{2} = F(x, y)$$
If $ a\in R$, then $G^{-1}([a, \infty)) = [a, \infty) \times (-\infty, \infty)$ and $H^{-1}([a, \infty)) = (-\infty, \infty) \times [a, \infty)$. The preimages of both functions are measurable sets in $\mathbb{R}^{2}$ so $G$ and $H$ are Borel measurable.
Finally, since $F$ and $u$ are both Borel measurable, so is their composition $u(x^{2} - y^{2})$.