Baire sets and Borel sets

Question

I need to show that a subset of $\mathbb{R}^{n}$ is a Baire set if and only if it is a Borel set.

The Baire set is $\sigma(\kappa)$ where $\kappa$ is the set of all compact subsets of $\mathbb{R}^{n}$.

Let $K$ be a compact set. Then it is closed and hence it is Borel.

Conversely, let U be an open set.

Consider $K_{m}=\{x\in \mathbb{R}^{n}:\|x\|\leq m,d(x,U^{c})\ge\frac{1}{m}\}$

Then $K_{m}$ is compact and $U=\cup K_{m}$ and hence $U\in \sigma(\kappa)$.

My question is what is this set $K_{m}$ as I have no idea how it looks like? Why is the open set $U=\cup K_{m}$?

Answer 1

The set $K_m$ is bounded (first condition) and closed (by the two conditions). It is a subset of $U$ (second condition). Now show: If $x\in U$ then because $U$ is open, $x$ must belong to some $K_m$ for $m$ large enough and you are done.