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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?

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  • $\begingroup$ I'm not completely convinced that this is true, as per this. There seems to be some subtlety involved. Can anyone help me out? $\endgroup$ – ArchieR577 Mar 29 at 21:18
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    $\begingroup$ 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. $\endgroup$ – Yu Ding Mar 29 at 21:21
  • $\begingroup$ 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 $\mathcal{F}$? $\endgroup$ – ArchieR577 Mar 29 at 21:39

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