Let $\mathfrak{B}$ be a class of sets containing $\Omega$, such that:

1) $A, B \in \mathfrak{B}$, $A \subset B \Rightarrow B \setminus A \in \mathfrak{B}$ (closed under differences)

2) $A_1 \subset A_2 \subset \dots \subset A_n$, $A_i \in \mathfrak{B} \Rightarrow \bigcup_{n=1}^\infty A_n \in \mathfrak{B}$ (closed under increasing limits).

3) $\mathfrak{B}$ is closed under finite intersections.

Show $\mathfrak{B}$ is a $\sigma$-algebra.


It is easy to show that $\mathfrak{B}$ is closed under complementation. This, together with (3), implies that $\mathfrak{B}$ is closed under finite unions.

My problem is that I don't know how to get to countable unions (or intersection) from there; What I thought was something along these lines

Let $A_i \in \mathfrak{B}$ a family of sets. Let $B_i = \bigcup_{n=1}^i A_n$. Then, since $B_i \subset B_{i+1}$, we have $\bigcup B_n \in \mathfrak{B}$. Also $\bigcup B_n = \bigcup A_n$, , and so $\bigcup A_n \in \mathfrak{B}$.

The problem of course is that we have to show that $B_i \in \mathfrak{B}$; this is true for a finite index $i$, but does not suffice right? (If it would suffice, then the above proof would be useless because we would have already shown that countable unions of elements of $\mathfrak{B}$ are in $\mathfrak{B}$)

Can I have a little help?

Also, can some hypothesis be relaxed? (I don't think so though).

  • $\begingroup$ Why doesn't that suffice? Every $B_i$ is in $\mathfrak{B}$ and hence by 2) also $\bigcup B_i$ is in $\mathfrak{B}$ $\endgroup$ May 6, 2014 at 15:44
  • $\begingroup$ It seems to me like it's not very rigorous.. I mean if every $B_i$ is in $\mathfrak{B}$, this implies that $\bigcup_{n=1}^iA_n$ is in $\mathfrak{B}$ for every $i$.. So isn't that what we want to prove? To be honest I think I need a clear understanding of why closed under finite unions does not imply closed under countable unions. $\endgroup$
    – Ant
    May 6, 2014 at 15:52
  • $\begingroup$ There's a subtle (conceptual) difference between being closed under countable unions and being closed under increasing limits. Try to draw a picture of how they relate to each other, and you will have gotten the point of the exercise. $\endgroup$
    – nomen
    May 6, 2014 at 16:12
  • $\begingroup$ @Ant We want to prove that $\bigcup A_i \in \mathcal{B}$, which is not the same. For example, take $\mathcal{B} = \{ [-r,r] \mid r > 0 \}$, then any finite union of such interval is again in $\mathcal{B}$ (if $A_i=[-r_i,r_i]$ then $\bigcup_{i=1}^{n} A_i=[-r,r]$ where $r=\max_{1 \leq i \leq n}{r_i}$, but it is not closed under increasing limit, since with "countability" you can reach the whole set $\mathbb{R}=\bigcup_{i=1}^{\infty} [-i,i]$ $\endgroup$
    – yago
    May 6, 2014 at 16:21
  • 2
    $\begingroup$ The tag (abstract-algebra) is funny. $\endgroup$
    – Did
    May 6, 2014 at 16:29

1 Answer 1


The proof is actually sufficient. Since the OP seemd to have a confusion between finite and countable unions (or the difference between finite unions and increasing limit, but it is basically the same), then I'll post here my example :

Why can you reach potentially larger set with a countable union ? Well, let's take the set $\mathcal{B}=\{ [-r,r] \mid r > 0 \}$ that is the set of symetric intervals centered at zero. Then this set is closed by finite union : if we take a collection $A_i=[-r_i,r_i]$ for $i \in \{1,\ldots,n\}$, let $r=\max{r_i}$ then $\bigcup_{i=1}^nA_i=[-r,r] \in \mathcal{B}$

But, with countable union, you can reach the whole set $\mathbb{R}$, which is obviously not in $\mathcal{B}$ : for example, consider $\bigcup_{i=1}^{+\infty}[-i,i]$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .