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Given a set $\Omega$ with its powerset denoted by $\mathscr P(\Omega)$ and a class of its subsets $\mathscr C\subset \mathscr P(\Omega)$:

the generated $\sigma$-algebra $\sigma(\mathscr C)$ is the smallest $\sigma$-algebra which contains $\mathscr C$

Due to the definition of $\sigma$-algebra, $\sigma(\mathscr C)$ always exists (I guess, provided AC) and can be given by the intersection of all $\sigma$-algebras containing $\mathscr C$. However, such implicit construction gives me a hard time when I am trying to understand which properties does $\sigma(\mathscr C)$ have for some particular choices of $\mathscr C$.

Proving the claim in this answer I realized that the following method can be applied. Say, we would like to show that any $A\in \sigma(\mathscr C)$ satisfies a particular property $S$, equivalently it can be stated as $$ \sigma(\mathscr C)\subset \mathrm S \tag{1} $$ where $\mathrm S\subset \mathscr P(\Omega)$ is the class of all subsets which satisfy $S$. Then we can do the following:

  1. verify that $\mathscr C\subset \mathrm S$ and $\{\emptyset\}\in \mathrm S$ which is a necessary condition;

  2. prove that $\sigma(\mathscr C)\cap \mathrm S$ is a $\sigma$-algebra.

By the minimality of $\sigma(\mathscr C)$ it will imply $(1)$. For example: let $(\Omega,\mathscr F,\mathsf P)$ be a probability space. Define two classes: $$ \mathscr F_B = \{A\in \mathscr F:A\subset B\}\text{ for }B\in\mathscr F $$ $$ \mathscr N = \{A\in \mathscr F:\mathsf P(A) = 0\}. $$ For a $\sigma$-algebra $\mathscr F^\prime_B = \sigma(\mathscr F_B\cup\mathscr N)$ I want to show that $$ \mathsf P(A\setminus B) \in \{0,1-\mathsf P(B)\}\tag{2} $$ for any $A\in \mathscr F^\prime_B$ so, $\mathrm S$ is given by $(2)$. Clearly, $\mathscr F_B\cup \mathscr N\subset \mathrm S$ - and further I show that $\mathscr F^\prime_B\cap\mathrm S$ is closed under taking complements and countable unions.

My issue is the following: I don't know how to apply the same method to the following problem without using results of an example above: let $(\Omega,\mathscr F,\mathsf P) = ([0,1],\mathscr B([0,1]),\lambda)$ where $\mathscr B$ is the Borel $\sigma$-algebra and $\lambda$ is the Lebesgue measure. For the case $B = [0,\frac12]$ I want to show that $\mathscr F^\prime_B$ (defined as above) does not contain sets of the form $[a,1]$ for $a>\frac12$.

Say, I consider $A\in \mathscr F^\prime_B\cap \mathrm S$ so it does not have the form $[a,1]$; hence $A^c\in \mathrm F^\prime_B$ but I don't know how to show that $A^c$ is also not of the from $[a,1]$. With countable unions the problem is similar.

I would appreciate any help with the problem above - also I will be happy to learn about more neat methods of how to verify if the generated $\sigma$-algebra satisfy some property or not.

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For $a>1/2$, $\lambda[a,1]<1/2$ and hence $\lambda[a,1]\notin\{0,1/2\}=\{0,1-\lambda B\}$. If follows that $[a,1]$ is not in $S$ and therefore also not in $\mathcal{F}_B'$. – Michael Greinecker Mar 20 '12 at 16:54
@MichaelGreinecker: True, but I've mentioned in OP that I was trying to prove it directly, i.e. without using results of an example above. – Ilya Mar 20 '12 at 17:01
up vote 3 down vote accepted

I read your question. You wrote a lot of good things, but still it is hard to take a lot of them. First, about your specific question, you can use the fact that $\sigma(\mathcal C ) \cap A=\sigma\left (\mathcal C \cap A \right)$(the rest is obvious). Second, if you want to verify, whether some statement $S$ is true for $\sigma(\mathcal{C})$, you should do the following:

  1. Prove that the set $\mathcal{P}\subset \sigma(\mathcal C), \mathcal{P}=(A\in \sigma(\mathcal C)| S \mbox{ is true for } A)$ is sigma-algebra(verify all conditions on sigma-algebra)
  2. $\mathcal C\subset \mathcal P$, that is $S$ is true for $\mathcal C$, then $S$ is true for $\sigma(C)$.

It is obvious algorithm, however it is very helpful.

You can look at P.Billingsley ``Convergence of probability measures'' Theorem 1 is proved in the same fashion, that could be good example for you.

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$\sigma(\mathcal C)\cap A$ - I guess you mean here that $A$ is a set rather than a collection of sets, don't you? – Ilya Mar 21 '12 at 8:58

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