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So I am studying measure theory and I have found myself struggling to fully understand the concept of the Borel $\sigma$-algebra in depth. We know that the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing all open sets. The part that I cannot clearly grasp is the word smallest. The way we can generate a Borel $\sigma$-algebra is take all the open sets and take all possible set operations between them. Won't this always produce a unique $\sigma$-algebra of sets? How can a $\sigma$-algebra be larger? Can anyone provide an intuitive example? Thank you in advance!

SpawnKilleR
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    Sigma algebras are supposed to tell you 'what you're able to measure. So a sigma algebra thats ``larger'' than other sigma-algebras is one that can measure more stuff. It's just a superset of other sigma algebras. For instance, the set ${a,b,c}$ admits these sigma algebras: $${{},{a,b,c},{a},{b,c}}$$ and just the power set: $${{},{a,b,c}, {a,b}, {b,c}, {a,c}, {a}, {b}, {c}}.$$ You can easily verify they're both sigma algebras, but the powerset is larger than the other. – Christian Chapman Apr 24 '15 at 23:56
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    You may want to read the first few pages of Lee's Topological Manifolds, which gives a nice, light but precise idea of what we mean when we call a set `open.' – Christian Chapman Apr 25 '15 at 00:05
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    If you take all subsets of $\mathbb R$, that is a $\sigma$-algebra which is larger than the Borel $\sigma$-algebra; it contains all of the Borel sets and various other sets. – bof Apr 25 '15 at 00:11

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Lets consider $\Omega=\{1,2,3,4\}$

$\sigma(\{1\})$ is the smallest $\sigma$-algebra which contains $1$.

So we must take any other elements of $P(\Omega)$ such that the conditions for being a $\sigma$-algebra are fulfilled.

It does clearly contains $1$. Also $1^C=\{2,3,4\}$. And $\Omega,\emptyset$.

So we have $\sigma(\{1\})=\{\Omega,\emptyset,\{1\},\{2,3,4\} \}$. Indeed this is a $\sigma$-algebra. Also we got this set for taking as much elements of the power until our conditions are fulfilled for the "first time".

An example for a $\sigma$-algebra with is not the smallest one but is containing $\{1\}$ is:

B:=$\{\Omega,\emptyset,\{1\},\{2\},\{1,2\},\{3,4\},\{2,3,4\},\{1,3,4\} \}$.

Clearly $\sigma(\{1\})\subset B$

Marm
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Say we are doing this on the real line $\mathbb R$. Let $\mathcal G$ be the collection of all open sets. We are interested in $\sigma$-algebras $\mathcal F$ such that $\mathcal F \supseteq \mathcal G$. There may be many such $\sigma$-algebras. For example, the power set $\mathcal P$, consisting of all subsets of $\mathbb R$ is one. But that is the largest one, we want the smallest one.

As a comment noted, this largest and smallest are in the sense of set inclusion. The Borel $\sigma$-algebra $\mathcal B$ satisfies $\mathcal B \subseteq \mathcal F$ for any $\sigma$-algebra $\mathcal F$ such that $\mathcal F \supseteq \mathcal G$. That is the sense in which $\mathcal B$ is smallest.

GEdgar
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  • just need to clarify something: in the second paragraph, we should also have $\mathcal B \supseteq \mathcal G$, right? – rims Jan 20 '20 at 19:47
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if you agree that the borel $\sigma$-algebra isn't neccessarily all the subsets of the topological space, then you might also agree that $P(X)$ (or $2^X$ in a different notation, the power set) is a larger $\sigma$-algebra. Larger means more sets in the $\sigma$-algebra.

Also note that the generated $\sigma$-algebra by the open sets, which is the borel $\sigma$-algebra, is the intersection of all $\sigma$-algebras containing the open sets - which is smaller than any other $\sigma$-algebra by the sense of being a subset.