# Constructing the generated $\sigma$-algebra

Consider a collection $$\mathcal{A} = \{A_i\}_{i \in I}$$ where $$I$$ is some indexing set. I want to construct $$\sigma(\mathcal{A})$$.

Let $$\mathcal{F}_0 = \mathcal{A}$$. We now construct inductively: let $$\mathcal{F}_{i+1}$$ be the collection of sets which are formed by countable union, intersection, complementation of sets in $$\mathcal{F}_i$$. This is an increasing sequence of collections.

Let $$\mathcal{F} = \bigcup_{i \geq 0} {\mathcal{F}_i}$$. Is $$\mathcal{F} = \sigma(\mathcal{A})$$?

Clearly $$\mathcal{F}$$ contains $$\mathcal{A}$$ and the aim is to show that $$\mathcal{F}$$ is a $$\sigma$$-algebra. If we show this, then we should be done as $$\sigma(\mathcal{A})$$ must contain all the sets in our construction.

It is easy to show that $$\mathcal{F}$$ is closed under complementation and finite intersection, but how can one show it is closed under countable union?

Let $$B_1, B_2, \dots \in \mathcal{F}$$. Then $$B_i \in \mathcal{F}_{n_i}$$ for some $$n_i$$, Let $$B^{'}_i = \bigcup_{j=1}^i {B_j}$$, $$m_j = \operatorname{max} \{n_1,\dots,n_j\}$$, we see that $$B^{'}_i \in \mathcal{F}_{m_i}$$, and we would like to show that $$B = \bigcup_{i\geq 0} {B^{'}_i} \in \mathcal{F}$$.

Can we proceed further? If not, what is missing?

• I'm not completely convinced that this is true, as per this. There seems to be some subtlety involved. Can anyone help me out? – ArchieR577 Mar 29 at 21:18
• Seems to be not so simple. There are many books concerning this issue, see e.g. Hewitt-Stromberg "Real and abstract analysis", page 133, the PROOF OF Theorem (10.23), there a construction of $\sigma({\mathcal A})$ is given using kind of transfinite induction. – Yu Ding Mar 29 at 21:21
• Yeah, looks like there are many more operations that we need to do and this is not sufficient. Is there an easy example of a set in $\sigma(\mathcal{A})$ that is not in $\mathcal{F}$? – ArchieR577 Mar 29 at 21:39