I’m reading up on $\sigma$-algebra generated by an arbitrary collection $\mathcal{C}$ of subsets of $\Omega$, which is defined as the smallest $\sigma$-algebra on $\Omega$ containing $\mathcal{C}$ and denoted $\sigma(\mathcal{C})$. The example I’m working with is $\Omega = \mathbb{R}$, and $\mathcal{C}$ is the collection of open intervals. I’m having trouble seeing what kinds of subset of $\mathbb{R}$ would be in $\sigma(\mathcal{C})$. Please let me know if the following makes sense:

  • We have $\mathbb{R}, \emptyset \in \sigma(\mathcal{C})$ by definition of $\sigma$-algebra.
  • Let $I_i = (a_i,b_i)$. Then $I_i \in \sigma(\mathcal{C})$, and also $I_i^c = (-\infty,a_i] \cup [b_i, +\infty) \in \sigma(\mathcal{C})$.
  • $\bigcup_{i = 1}^{\infty}I_i = (a_1,b_1) \cup (a_2,b_2) \cup \dots \in \sigma(\mathcal{C})$.
  • Taking the complement of the above also gives $(\bigcup_{i = 1}^{\infty}I_i)^c = (-\infty,a_1] \cup [b_1,a_1] \cup [b_2,a_3] \cup \dots \cup [b_\infty,\infty) \in \sigma(\mathcal{C})$ (I have doubts about the last semi-infinite interval, but I don’t know what else could happen here).

At this point it gets pretty complicated, and I suspect this might not be the way to go. Am I missing something here?

Also I need to show that $\sigma(\mathcal{C}) = \sigma(\mathcal{D})$, where $\mathcal{D}$ is the collection of semi-infinite intervals $\{(\infty,x]: x \in \mathbb{R}\}$, but I’m not sure I should analyze $\sigma(\mathcal{D})$ in the same manner as above.

  1. I don't think you'll be able to characterize all subsets of that sigma-algebra in a few bullet points. It will contain all nice sets (unions of closed intervals, open intervals, half open intervals, semi-infinite intervals, etc.), but there will be more that are hard to describe.

  2. In general, to show $\sigma(\mathcal{C}) = \sigma(\mathcal{D})$, it suffices to show $\mathcal{D} \subseteq \sigma(\mathcal{C})$ and $\mathcal{C} \subseteq \sigma(\mathcal{D})$. (Do you see why?) Proving this amounts to constructing intervals of the form $(-\infty, x]$ by using operations on intervals of the form $(a_i, b_i)$, and vice versa.

  3. A way to get a simple closed interval from your second bullet point: note that $\bigcup_{n=1}^\infty ((-\infty, a-1/n] \cup [b+1/n, \infty)) = (-\infty, a) \cup (b, \infty)$ and take the complement to get $[a,b]$.


You can prove that pretty much every set you can imagine is in $\sigma(\mathcal{C})$. To find sets that are not in $\sigma(\mathcal{C})$ is quite tricky, see Vitali set.

However, keep in mind that the number of elements in $\sigma(\mathcal{C})$ has the same cardinality as the set of real numbers $\mathbb{R}$, so the sets which are not in $\sigma(\mathcal{C})$

The construction of the Borel Set is in the first section of the Borel set page of Wikipedia, but it uses transfinite induction, so it may be too much advanced for your current level.

Going back to your examples, you can prove that:

  • finite unions and intersections are in $\sigma(\mathcal{C})$ (take $A_{n+1}=A_{n+2}=...=\emptyset$, respectively $\mathbb{R}$)
  • singletons are in $\sigma(\mathcal{C})$ by writing $\{x\}=\bigcap_n(x-1/n,x+1/n)$
  • infinite open intervals are in $\sigma(\mathcal{C})$ (for example $(a,\infty)=\bigcup_n(a,n)$
  • closed and half-closed intervals are in $\sigma(\mathcal{C})$ (by attaching singletons at the ends of open intervals)
  • open sets in standard topology on $\mathbb{R}$ are in $\sigma(\mathcal{C})$ as countable unions of open intervals
  • closed sets in standard topology on $\mathbb{R}$ are in $\sigma(\mathcal{C})$ as complements of open sets

and so on.


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.