I understood the definition of a $\sigma$-algebra, that its elements are closed under complementation and countable union, but as I am not very good at maths, I could not visualize or understand the intuition behind the meaning of "closed under complementation and countable union".

If we consider the set X to be a finite set, then what would be a good real life example of a $\sigma$-algebra, for a noob to understand.

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    $\begingroup$ A trivial example is the empty set and the whole set that form a sigma algebra. $\endgroup$ – Spock Feb 1 '14 at 21:41
  • $\begingroup$ The basic two trivial $\sigma$-algebra definition I got was, (empty set and the whole set) due to closed under complementation, and all possible subsets due to closed under union. But what would be a bit non trivial example which would explain the interplay between the closure under complementation and union. $\endgroup$ – Shambo Feb 1 '14 at 21:51
  • $\begingroup$ It probably should be mentioned that any finite example is not going to be able to distinguish between "countable union" and "union", because in a finite set you've only got a countable number of subsets in the first place. The distinction there is between a union indexed on $\Bbb N$ and a union indexed on $\Bbb R$ (for example). $\endgroup$ – tabstop Feb 1 '14 at 21:58

Let $X = \{a, b, c, d\}$, a possible sigma algebra on $X$ is $Σ = \{∅, \{a, b\}, \{c, d\}, \{a, b, c, d\}\}$.
I think this is a good example.

  • $\begingroup$ suppose we have $X = \{ a \ldots j\}$, so $\Sigma$ must have the empty set, $X$ , and suppose it also has $\{a,b,c,d\}$, $\{e,f,g\}$, $\{h,i,j\}$, so it must also contain $\{e,f,g,h,i,j\}$ and similarly the other combinations also. And if it also contains $\{a,b\}$, $\{c,d\}$, $\{e,f\}$ and $\{g\}$, then it must also contain $\{c,d,e,f\}$, $\{e,f,g,h,i,j\}$... and so on. So how is it different from having all possible subsets of $X$ $\endgroup$ – Shambo Feb 1 '14 at 22:00
  • $\begingroup$ I don't understand your question very well. If it doesn't have all possible subsets of the set, well, then it's different. $\endgroup$ – Spock Feb 1 '14 at 22:08
  • $\begingroup$ Sorry, I asked the question wrong. They are different. But how can we intuitively view such a partitioning. Can we say, it exactly consists of all possible combinations of the smallest(or unpartitioned subsets) disjoint subsets. $\endgroup$ – Shambo Feb 1 '14 at 22:16
  • $\begingroup$ Not really. The example I gave was super nice in its behavior but if we have a larger set then it becomes tricky since you need to check the union/intersection thing. It's easier to prove that something is sigma algebra than to construct one. If you try partitioning the set you gave in your example then you'll get what I mean. $\endgroup$ – Spock Feb 1 '14 at 22:21

Its power set, i.e. the set $2^X$ (or $\mathcal{P}(X)$ depending on the notations) of all subsets of $X$.


steps to form the smallest sigma field step 1: break the whole space into sets that are mutually exclusive and exhaustive . step 2: then form a new set using these sets(sets obtained in step 1) in this way{ take single sets ,take sum of two at a time ,take sum of three at a time ,..........,take sum of all } . the set so formed is sigma field you need

  • $\begingroup$ for example read modern probability theory by B R bhat $\endgroup$ – Shivam Singh Oct 11 '16 at 19:45

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