If we define

$$\mathcal{J} = \{\text{ all intervals contained in [0,1]}\} $$ Then $$ B_0 = \{ \text{ all finite unions of elements of }\mathcal{J}\} $$ Is an algebra (so it is closed under the formation of complements (but it is not a $\sigma$ algebra) . Now consider $$ B_1 = \{\text{ all finite or countable unions of elements of }\mathcal{J}\} $$ and let $K$ be the Cantor set (everything that is left over after removing all the middle thirds over the interval $[0,1]$. Then the complement of the Cantor set is in $B_1$. That is, $$ K^c \in B_1 $$ which I believe is because the cantor set $K$ can be written as the countable intersection of closed intervals, and therefore its complement $K^c$ can be written as a countable union of open intervals. By definition, then, $$ K^c \in B_1 $$ However, $$ K \not\in B_1 $$ Since the complement of something in $B_1$ is not in $B_1$, then $B_1$ is not an algebra, even though $B_0$ is? I don't see how allowing the countable union of intervals would ruin closure under complements? Can someone explain this to me or point me in the right direction? Or let me know if there is something else going on that I am missing?


  • $\begingroup$ No source? :|.. $\endgroup$ – BCLC Jan 5 '16 at 4:15

You're absolutely right (assuming that, in the definition of $B_0$, "interval" means "open or closed interval") - $B_0$ is an algebra, but $B_1$ is not, even though $B_1$ is gotten from $B_0$ by adding a closure property. What's going on, basically, is that the complement of an infinite union is more complicated to describe than you might suspect based on just looking at finite unions. In particular, the complement of a finite union of intervals is again a finite union of intervals, so closure under complementation is in $B_0$ "for free," but this fails badly for countable unions.

For a simpler example of how adding closure in one direction can weaken closure in another, consider the following: let $C_0$ be the set of all subsets of $[0, 1]$ which are either finite or cofinite (=have finite complement). Then clearly $C_0$ is an algebra.

Now let $C_1$ be $C_0$, closed off under countable unions. Now $C_1$ is not an algebra, since e.g. the set of irrationals is not in $C_1$ but the set of rationals is.

  • $\begingroup$ Thank you. Then, since the borel sigma algebra of subsets of $[0,1]$ contains all finite and countable unions of open intervals on $[0,1]$, and is closed under complements (since it is a sigma field), I'm guessing that whatever other stuff the borel sigma field has (sorry, I am not able to really imagine this "other stuff" or explain it), allows it to contain the complement of an infinite union? $\endgroup$ – majmun Dec 15 '15 at 3:22
  • 1
    $\begingroup$ @majmun The Borel sets have a nice hierarchical description, as follows. We say a set is $\Sigma^0_1$ if it is open; a set is $\Pi^0_n$ if its complement is $\Sigma^0_n$; and a set is $\Sigma^0_{n+1}$ if it is the countable union of $\Pi^0_n$ sets. Then we have obvious inclusions $\Sigma^0_n, \Pi^0_n\subseteq \Sigma^0_{n+1}, \Pi^0_{n+1}$, but we can prove that these are the only inclusions: for instance, the set of rationals is $\Sigma^0_2$ but not $\Pi^0_2$. This can be proved using the Baire category theorem. (cont'd) $\endgroup$ – Noah Schweber Dec 15 '15 at 3:25
  • $\begingroup$ The point is: every set which is $\Sigma^0_n$ for some $n$ is Borel. (Actually, this turns out to be just the beginning - in order to get all the Borel sets, we have to define $\Sigma^0_\alpha$ and $\Pi^0_\alpha$ for arbitrary countable ordinals $\alpha$! But looking at the finite levels should give you a good idea of what the "extra stuff" is.) $\endgroup$ – Noah Schweber Dec 15 '15 at 3:26

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