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My understanding at this point is that to assign a probability measure to a random variable defined on the real line, we need a Borel $\mathscr{B}$ sigma algebra, because otherwise we wouldn't be able to find a bijection from every subset of the power set ($2^\Omega$) to the Lebesgue measure, due to the fact that the cardinality of $2^\Omega$ is not just greater, but much greater than the subsets of the Borel algebra.

On the other hand, the Borel sigma algebra, even though it contains every number in the real line as a singleton set, has that same cardinality as $\mathbb{R}$.

How can this be proven? Is there a way to see this through the concept of standard topology?

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marked as duplicate by Nate Eldredge, Xander Henderson, Alonso Delfín, Claude Leibovici, JonMark Perry Oct 3 '17 at 10:19

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  • $\begingroup$ Do it step by step, try to prove the following collections have cardinality of $\mathbb{R}$. 1. Collection of open intervals. 2. Collection of open sets. 3. Collection of Borel sets. $\endgroup$ – gamma May 6 '16 at 4:12
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We want to see that $\mathscr{B}$ has the cardinality of $\mathbb{R}$, that is to say, the cardinality of continuum $c$.

Here: Cardinality of Borel sigma algebra , there is a proof of the fact that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$.

It is clear that it is the same than $\mathbb{R}$ or $\mathscr{P}(\mathbb{R})$, so it isn´t finite.

By definition $\mathscr{B}$ is the sigma algebra generated by the usual topology: $\mathscr{B}=\sigma(\tau_{us})$. Every open set $A$ in $\mathbb{R}$ can be seen as the union of the open (basic) neighbourhoods of the points in $A \cap \mathbb{Q}$ since $\mathbb{Q}$ is dense , that is,a numerable union of sets in $\tau_{us}$. So the collection

$$ \mathscr{A}=\{(q,p) \quad : q<p \quad ; \, q,p \in \mathbb{Q}\}$$

is countable and satisfy : $ \tau_{us} \subset \sigma(\mathscr{A}) $, so: $$\mathscr{B}=\sigma(\tau_{us})=\sigma(\mathscr{A})$$

However, usually $\mathscr{B}$ is presented as $\sigma(\{(-\infty,a] : a\in \mathbb{R}\})$

And it is easy to see that

$$\{(-\infty,a] : a\in \mathbb{R}\} \quad \subset \quad \sigma(\{(-\infty,q] : q\in \mathbb{Q}\})$$

since $(-\infty,a]=\bigcap_{q\geq a ,q\in \mathbb{Q}}(-\infty,q]$ ,

so the sigma-algebra which generates the first ($\mathscr{B}$) is the same that the one which generates the second (countable) set.

I don´t know what you mean by "a way to see this through the concept of standard topology". Actually, the fact that this collection generates $\mathscr{B}$ rely on the nature of the topology of $\mathbb{R}$ : it is second countable.

You can see many of the properties of Borel Algebras in these topology spaces in several posts, for example: cardinality of the Borel $\sigma$-algebra of a second countable space

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  • $\begingroup$ I have edited it to explain it a bit clearer. I hope now you can understand better what I mean $\endgroup$ – Acas Oct 2 '17 at 15:30

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