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Suppose $F$ is a sigma-algebra, $A\subset B$, $B\in F$. Is it the case that $A\in F$?

I'm familiar with the definition of a sigma-algebra (closed under complements and countable unions and intersections). The intuition of a sigma-algebra as information suggests that if we have some information that we know ($B\in F$), we should also have the subset ($A\in F$). But I'm not seeing it via the definition.

Thanks in advance!

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2 Answers 2

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HINT: the whole space $X$ is an element of any $\sigma$-algebra, and every subset of $X$ is, well, a subset of $X$. So if this were true, every $\sigma$-algebra would have to consist of **all* sets. Can you think of a counterexample?

Also, can you clarify "the intuition of a $\sigma$-algebra as information"? I think I know what that refers to, but before I respond I want to make sure I won't just be making things more confusing. :P

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  • $\begingroup$ Ah, good point! A non-measurable subset of $[0,1]$ would not be contained in the Borel subsets of $[0,1]$. I may have been imprecise in thinking of $\sigma$-algebras as information sets. I've recently been reading about filtrations, increasing $\sigma$-algebras that represent information in time. Maybe the analogous interpretation doesn't exist for $\sigma$-algebras generally. Is there one? $\endgroup$
    – manofbear
    Commented Jun 15, 2016 at 0:54
  • $\begingroup$ @manofbear Yes. Now: how do we know there's a non-Borel set of reals? (This is actually a little bit subtle . . .) $\endgroup$
    – Noah Schweber
    Commented Jun 15, 2016 at 0:56
  • $\begingroup$ Ok, maybe I'm being loose again, but I remember seeing a proof (which used the axiom of choice) of a "non-measurable" set in $\mathbb{R^d}$ in Stein and Shakarchi. I take non-measurable to mean not measurable with respect to any $\sigma$-algebra? $\endgroup$
    – manofbear
    Commented Jun 15, 2016 at 1:02
  • $\begingroup$ @manofbear No, non-measurable just means not Lebesgue measurable (but this also implies non-Borel); every set is an element ofsome $\sigma$-algebra, namely the $\sigma$-algebra it generates. You can also just prove it by counting the Borel sets - you can show that there are only continuum-many Borel sets, whereas there are $2^{2^{\aleph_0}}$-many sets of reals. $\endgroup$
    – Noah Schweber
    Commented Jun 15, 2016 at 1:29
  • $\begingroup$ Oh, right (regarding every set being an element of some $\sigma$-algebra). Thanks for these clarifications! They've been quite helpful $\endgroup$
    – manofbear
    Commented Jun 15, 2016 at 3:41
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Let $B$ be the universal set. Is every sigma algebra the power set?

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