2
$\begingroup$

I am teaching myself measure theory over the summer, and would appreciate some feedback on this basic result, since this type of reasoning is quite new to me.

Solution:

By Heine-Borel, the compact sets of $\mathbb{R}$ are exactly the closed bounded subsets. Let $\mathcal{K}(\mathbb{R})$ denote the $\sigma$ algebra generated by the compact sets. The Borel $\sigma$ algebra on $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, is generated by the open subsets of $\mathbb{R}$, and therefore the closed subsets of $\mathbb{R}$. Additionally, $\mathcal{B}(\mathbb{R})$ is generated by all the sets of the form $(-\infty,t]$ for real $t$. Clearly, the compact sets are contained in the collection of closed sets, so $\mathcal{K}(\mathbb{R}) \subset \mathcal{B}(\mathbb{R})$. Now, each set of the form $(-\infty,t]$ is the union of countably many closed bounded sets, so the $\sigma$ algebra generated by these sets will be contained in $\mathcal{K}(\mathbb{R})$. In other words $\mathcal{B}(\mathbb{R}) \subset \mathcal{K}(\mathbb{R})$, and the result follows.

$\endgroup$
4
  • $\begingroup$ You only did the $(-\infty,t]$ intervals. Still missing all other closed sets. But an advice, the open sets are easier to do. It is easier to fit "large" compacts in them. Divide $\mathbb{R}$ in little intervals of size $1/n$ for $n$ very large. Take those close intervals completely inside the open and inside $[-n,n]$. Let $n\to\infty$. $\endgroup$
    – Peyton
    Jul 30 '17 at 13:08
  • $\begingroup$ @Peyton I am not sure I understand your objection. It is known that the $\sigma$ algebra $\mathcal{B}(\mathbb{R})$ is generated by the sets of the form $(-\infty,t]$ for real t. I then showed that this algebra is contained within the algebra generated by the compact sets. This together with the fact that the algebra generated by the compact sets is contained within $\mathcal{B}(\mathbb{R})$ gives $\mathcal{B}(\mathbb{R}) \subset \mathcal{K}(\mathbb{R}) \subset \mathcal{B}(\mathbb{R})$, so $\mathcal{K}(\mathbb{R}) = \mathcal{B}(\mathbb{R})$, no? $\endgroup$ Jul 30 '17 at 13:18
  • $\begingroup$ The sentence "and therefore the closed subsets of $\mathbb{R}$, as well as all the sets of the form $(−\infty,t]$" sounded like you were basing your proof on knowing that it was generated by the collection of closed sets and $(-\infty,t]$, all together. From what you are saying now I see that the last part of the sentence was enumerating yet another separate set of generators. $\endgroup$
    – Peyton
    Jul 30 '17 at 13:22
  • $\begingroup$ @Peyton I realise that now, and I apologize for the misunderstanding. I am very interested in your input and your idea of a solution though, and I shall look into it! Thank you. $\endgroup$ Jul 30 '17 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.