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On the page 25 of Prof. Folland's Real analysis:

If $X$ is a set and $M\subset P(X)$ is a $\sigma$-algebra, $(X,M)$ is called a measurable space and the sets in $M$ are called measurable sets. ($P(X)$ is the family of all subsets of a set $X$)

My question is we know $P(X)$, power set, is also a $\sigma$-algebra:

https://proofwiki.org/wiki/Power_Set_is_Sigma-Algebra

  1. What about the set of subsets of $X$: $\{E:E\in P(X), E\notin M\}$? Are they also measurable sets in the above definition? Also, why don't we care about these subsets, are there any specific example to say more on this?

  2. Moreover, can we choose $M = P(X)$ in the above definition?

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    $\begingroup$ 1. By definition, no. 2. $P(X)$ is a sigma-algebra. $\endgroup$ – Lord Shark the Unknown Jun 3 '17 at 3:58
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    $\begingroup$ Why are you wondering about this obscure subset (All the sets that aren't in a given sigma algebra)? The question isn't why don't we care about them, but why should we? $\endgroup$ – mathematician Jun 3 '17 at 4:11
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    $\begingroup$ Only the sets in $M$ are measurable. $\endgroup$ – md2perpe Jun 3 '17 at 4:11
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  1. $P(X)$ \ $M$ can fail to be a $\sigma$-algebra. For example, when $X=\mathbb R$ and when $M$ is the family of Lebesgue-measurable subsets of $\mathbb R$, there exists a countable $F\subset P(X)$ \ $M$ with $\cup F=X\in M....$ So $P(X)$ \ $M $ can fail to be closed under countable unions.

2.$P(X)$ is indeed a $\sigma$-algebra. For example, for the counting measure $\mu$ on $X$,with $M=P(X)$, where $\mu(Y)=\infty$ if $Y$ is an infinite subset of $X,$ and $\mu(Y)$ is the number of members of $Y$ if $Y$ is a finite subset of $X.$

  1. A great deal has been said about $P(X)$ \ $M$ in various contexts. It is not without interest. But $\sigma$-algebras are interesting too.
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  • $\begingroup$ How to say the existing of $\cup F = X\in M$? Thanks! $\endgroup$ – sleeve chen Jun 3 '17 at 4:42
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    $\begingroup$ @sleevechen. There is a standard textbook example of such F. If you post this as a Q you will surely get an answer. An answer might be too long for a comment, depending on writing style. $\endgroup$ – DanielWainfleet Jun 3 '17 at 4:54

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