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?
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In the second sentence, do you want to say open sets? – Dylan Moreland Sep 05 '12 at 00:16
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1I think the question should say "the Borel $\sigma$-algebra" – Trevor Wilson Sep 05 '12 at 00:19
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@Dylan Moreland that is another definition of it – Ana Sep 05 '12 at 00:21
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1@Ana Well, that's what the exercise shows :) But what is written seems like a tautology. – Dylan Moreland Sep 05 '12 at 00:26
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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.
Shankara Pailoor
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It is enough to show that every closed set is in the $\sigma$-algebra. So let $C$ be closed, $x$ be an arbitrary point and $K_n$ the closed ball with center $x$ and radius $n$. Then $K_n\cap C$ is compact for all $n$ and $\bigcup_n (K_n\cap C)=C$.
Michael Greinecker
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5The important fact that entered this argument (and the other answers) is that $\mathbb{R}^n$ is a countable union of compact sets. In a general topological space, the $\sigma$-algebra generated by the compact sets is strictly smaller than the Borel $\sigma$-algebra. The easiest example of this probably is the discrete topology on an uncountable set: the $\sigma$-algebra generated by the compact sets (=finite sets) is the countable-cocountable $\sigma$-algebra, while the Borel $\sigma$-algebra is equal to the $\sigma$-algebra of all sets. – t.b. Sep 05 '12 at 08:51
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Let $\mathcal{B}$ denote the $\sigma$-algebra of Borel sets, i.e. the smallest $\sigma$ algebra containing the closed sets. Let $\mathcal{C}$ is $\sigma$-algebra generated by all the compact subsets of $\mathbb{R}^n$.
As you mentioned in your previous question, Are all compact sets in $ \Bbb R^n$, $G_\delta$ sets? , all compacts sets are closed. Hence $\mathcal{C} \subset \mathcal{B}$.
My answer to your previous question showed that all closed sets are $G_\delta$. Hence all open sets are $F_\sigma$. Let $U$ be an arbitrary open set. Let $(F_n)$ be a sequence of closed sets such that $U = \bigcup_{n \in \mathbb{N}} F_n$. Let $\bar{B}_k$ be the closed ball of radius $k$ centered at the origin. Define $C_{n,k} = F_n \cap \bar{B}_k$ is a closed and bounded subset of $\mathbb{R}^n$. Hence $C_{n,k}$ is a compact set. So $U = \bigcup_{n,k} C_{n,k}$. Thus $U$ is a countable union of compact sets. So $\mathcal{C}$ contains all the open sets. $\mathcal{B} \subset \mathcal{C}$.
Finally, $\mathcal{B} = \mathcal{C}$. The two $\sigma$-algebras are identical.
