When determining the smallest sigma-algebra generated by a finite collection of sets (and hence the smallest algebra containing that collection), is there any faster way to do this than by direct computation? On a related note, is there a criterion for selecting a set which generates the finest sigma-algebra (that is, the power set)?


Consider set $E_1,\dots,E_N$. Take $S_I:=\bigcap_{i\in I}E_i\cap\bigcap_{i\in [N]\setminus I}E_k^c$ for $I\subset [N]$. These sets are pairwise disjoint and generate the same $\sigma$-algebra as $E_1,\dots,E_N$. For each $I\subset [2^N]$, define $F_I:=\bigcup_{i\in I}S_i$. Then the $\sigma$-algebra generated by the $E_j$ is $\{F_I,I\subset [N+1]\}$.

Note that the $\sigma$-algebra generated by a finite collection is finite.

  • 1
    $\begingroup$ The sets $E_j\setminus\bigcap_{k\neq j}E_k$ will not in general be pairwise disjoint; I conjecture you meant $\bigcup$ rather than $\bigcap$. But even then the $\sigma$-algebra generated by $N$ sets is in general bigger than what you described. It can have up to $2^{2^N}$ elements, not $2^{N+1}$. $\endgroup$ – Andreas Blass Mar 3 '13 at 22:14

Given finitely many subsets $E_i$ of some set $X$, you can define an equivalence relation on $X$ by declaring two elements to be equivalent iff they belong to the same $E_i$'s. Then the sets in the $\sigma$-algebra (of subsets of $X$) generated by the $E_i$'s are exactly the unions of equivalence classes.

Incidentally, since only finitely many sets are involved here, the "$\sigma$" is irrelevant; the same sets constitute the (Boolean) algebra of sets generated by the $E_i$'s.


1) take intersections and complementations until you get the atoms. Then these atoms will generate the sigma algebra. Moreover, the sigma algebra is the power set of the atoms.
2) if in the process of taking intersections, you get to sets which form a partition of the "sample set" consisting of atoms, then you are done. So take intersections until you obtain such a partition. That's fun. Such a partition determines uniquely the subalgebra.
3) it results form 1) that the sigma algebra has cardinal $2^n$, where n is the number of atoms; so the cardinal should be of the form 2,4,8,16 and so on. That's a nice why to check if you got it right.

  • $\begingroup$ I think it should be $2^{2^n}$ i.e. $2^1, 2^2, 2^4, 2^8, 2^{16}, ...$ . Because the number of atoms grows as $2^n$ with the number of sets $n$. $\endgroup$ – UnadulteratedImagination Apr 24 '13 at 8:11
  • $\begingroup$ @UnadulteratedImagination No, you can have any number of atoms. $\endgroup$ – Theta33 May 28 '13 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.