Let $R$ be the smallest $\sigma$-algebra containing all compact sets in $\mathbb R^n$.
I know that based on definition the minimal $\sigma$-algebra containing the closed (or open) sets is the Borel $\sigma$-algebra. But how can I prove that $R$ is actually the Borel $\sigma$-algebra?
[Math] is the smallest $\sigma$-algebra containing all compact sets the Borel $\sigma$-algebra
compactnessmeasure-theoryreal-analysis
Best Answer
Well the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the open (or closed) sets of $\mathbb{R}^n$. I believe (I don't want to put words in your mouth) you are asking whether the sigma-algebra generated by the compact sets is equivalent to the sigma algebra generated by the open sets.
Since $R$ is not the greatest choice when referring to a $\sigma$-algebra over the reals, let us denote the $\sigma$-algebra generated by the compact sets by $\mathfrak{C}$ and the Borel $\sigma$-algebra by $\mathfrak{B}$.
Now every compact set is closed so it's the complement of an open set; hence $\mathfrak{C} \subset \mathfrak{B}$. Now, we want to show $\mathfrak{B} \subset \mathfrak{C}$. Let $F \subset \mathbb{R^n}$ be a closed set. Consider $F_n = F \cap \overline{B(0, n)}$ where $B(0,n)$ denotes the open ball of radius $n$ centered at the origin. Now $F_n$ is a sequence of compact sets whose union equals $F$. This means $F \in \mathfrak{C}$. Hence, all closed and open sets are in $\mathfrak{C}$. By countable union and intersection we see that $\mathfrak{B} \subset \mathfrak{C}$. Thus, the two $\sigma$-algebras are equal.