[Math] topology and Borel – $\sigma-$algebra

lebesgue-measuremeasure-theoryreal-analysis

(Topology): Let $\Omega\neq \emptyset $ be an arbitrary set. A class of sets $\tau \subset 2^{\Omega} $ is
called a topology on $\Omega$ if it has the following three properties:

$(i)\,\, \emptyset,\Omega \in \tau $

$(ii)\,\, A\cap B \in \tau $ for any $A,B\in\tau$

$(iii)\,\, (\bigcup_ {A\in\mathscr {F} }A)\in \tau $ for any $\mathscr {F}\subset \tau.$

In contrast with $\sigma-$algebras, topologies are closed under finite intersections only, but
they are also closed under arbitrary unions.

(Borel $\sigma-$algebra): Let $(\Omega,\tau)$ be a topological space . The $\sigma-$algebra

$$\mathscr {B}(\Omega):= \sigma(\tau)$$

that is generated by the open sets is called the Borel $\sigma-$algebra on $\Omega$.

Consider $\Omega= (0 ,1]$. Let $\mathscr {F_{0}}$ consist of the empty
set and all sets that are finite unions of the intervals of the form
$(a,b]$. A typical element of this set is of the form

$$F=(a_{1},b_{1}] \cup (a_{2},b_{2}] \cup… \cup (a_{n},b_{n}] $$

where ,$0\le a_{1}\lt b_{1}\le a_{2} \lt b_{2} \le \,… \le a_{n}
\lt b_{n}$ and $n\in \mathbb N.$

Lemma:

a) $\mathscr {F_{0}}$ is an algebra

b) $\mathscr {F_{0}}$ is not a $\sigma-$algebra

c) $\sigma(\mathscr {F_{0}})=\mathscr {B}(\Omega)$

Proof (b) : "to see this ,note that $(0,\frac{n}{n+1}]\in\mathscr
{F_{0}}$ for every n , but
$\bigcup_{n=1}^{\infty}(0,\frac{n}{n+1}]=(0,1)\notin \mathscr {F_{0}} $"

Problem:
I have seen this lemma here Lecture pdf . my problem is in part (b) and (c) , if $\mathscr {F_{0}}$ is not closed under countable union , it's not closed under arbitrary unions (from $(iii)$) , and so $\mathscr {F_{0}}$ is not topology on $\Omega$ and from (Borel $\sigma-$algebra) definition I can't understand how $\sigma(\mathscr {F_{0}})=\mathscr {B}(\Omega)$ (lemma (c)). is $\mathscr {F_{0}}$ topology ? or it's not important in Borel $\sigma-$algebra? I can't fully understand what is Borel $\sigma-$algebra.Could someone help me, where I have made a mistake. Any help is greatly appreciated

Best Answer

Sometimes we call $\mathscr {F_{0}}$ as a ring. Do not try to compare it with the topology, they are different.

Actually topology is an abstraction/generalization of open sets, so for the usual topology on $\mathbb R$ they are just open sets in the normal sense, and the Borel sets are sigma algebra generated by open set - if this could help you to gain some insights.

I think the last sentence in your question actually well explained why $\mathscr {F_{0}}$ is not a sigma- algebra.

To see why $\sigma(\mathscr {F_{0}}) = \mathscr B(\Omega)$, just try to show that Borel sigma algebra is generated by each of the following collection of sets: $\{(a,b)\}$, $\{[a,b]\}$, $\{(a,b]\}$. And the other direction is obvious as every open set is in $\sigma(\mathscr {F_{0}})$.

EDIT to answer three questions in the comment

  1. You are right that it is the smallest $\sigma$-algebra. Notice that Borel sigma algebra is defined as $\sigma(\tau(\Omega))$, and this by itself meaning the sigma-algebra generated by the topology of $\Omega$, i.e the smallest sigma-algebra containing all open subsets of $\Omega$.
  2. Such definition of Borel $\sigma$-algebra is good for any topological space (i.e. a set with a topology defined, and you could think that a topology defined on an arbitrary set is to define a bunch of its subsets as "open")
  3. Borel $\sigma$-algebra is defined by the sigma algebra generated by collection of open sets, but that does not mean the collection of open sets is the only way to generate such sigma algebra. The generated sigma algebra is usually much larger than the collection it is generated from, containing lots of sets that is not in original collection. And for our case, it's easy to see that we could use the sets in $\mathscr F_0$ to get open sets by intersection and countable union of them.
Related Question