Can someone explain what is the smallest sigma field? I need to know this


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    $\begingroup$ It seems to me that you're omitting quite a lot of context. $\endgroup$ Sep 12, 2014 at 3:15
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    $\begingroup$ Field vs Sigma Field. $\endgroup$
    – vadim123
    Sep 12, 2014 at 3:24
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    $\begingroup$ Is your question what your subject line says, or is it what you wrote in the body of your posting? $\endgroup$ Sep 12, 2014 at 3:40

1 Answer 1


Well, it is easy. In fact field and sigma-field are algebra and sigma-algebra of Real Analysis in probability. The difference is in one condition. In Sigma-field you need being closed in respect of countable(finite and infinite countable) union but in field (without sigma) you only need being closed in respect of finite union.Here there is an example which is field but not sigma-field.

Let $X=\mathbb{N}$, define $\mathfrak{m}=\{\mathbb{N}\}\cup\{A\subseteq\mathbb{N} : |A|<\infty\text{ or }|\mathbb{N}-A|<\infty\}$. $\mathfrak{m}$ is closed under completion and finite union and contains $\emptyset$ and $X$ but it's not closed under infinite countable union. Because for every odd number $2n-1$ we have $|\{2n-1\}|=1<\infty$ so $\{2n-1\}\in\mathfrak{m}$ but $\bigcup_{n=1}^\infty\{2n-1\}$ is the whole odd numbers which is not finite and its completion is the whole even numbers which again is not finite.

Now for smallest sigma-field. For this matter you are better to determine smallest sigma-field on what space or smallest sigma-field containing what, something like this. But let see what can I say for a general answer.

In the above example you can easily check that the smallest sigma-field containing that field, $\mathfrak{m}$, is the whole power set of $\mathbb{N}$. Denote the smallest sigma-field containing $\mathfrak{m}$ (which we haven't compute it yet) by $M$. Because every subset of $\mathbb{N}$ say $A$ can be written as $A=\cup_{a\in A}\{a\}$ And $|\{a\}|=1<\infty$ so $\forall a\in A\;:\;\{a\}\in\mathfrak{m}\subseteq M$ plus $A\subseteq\mathbb{N}\Longrightarrow |A|\leq|\mathbb{N}|$. $A$ is a countable union of elements of $M$ and thus it should be at $M$. But we showed this for an arbitrary subset of $\mathbb{N}$ so $M=P(\mathbb{N})$. And there is a simple famous exercise that infinite sigma-fields are uncountable.

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    $\begingroup$ useful and nice $\endgroup$
    – Olórin
    Sep 12, 2014 at 3:28
  • $\begingroup$ what does $|\{2n-1\}|=1$ denote here? $\endgroup$
    – vbm
    Mar 26, 2020 at 16:27
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    $\begingroup$ @vbm It means the cardinal of the set containing only one element which is the odd number $2n-1$ is equal to 1. $\endgroup$ Mar 27, 2020 at 9:34

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