# Sigma Algebra

## Homework Statement

Prove that the σ-algebra generated by the collection of all intervals in Rn coincides with the σ-algebra generated by the collection of all open subsets of Rn.

## Homework Equations

A σ-algebra is a nonempty collection Σ of subsets of X (including X itself) that is closed under the complement and countable unions of its members.

## The Attempt at a Solution

1. The σ-algebra generated by the collection of all open subsets of Rn would also contain all closed subsets by complementation. It would also contain their unions to generate all partly open subsets. Therefore the σ-algebra generated by the collection of all open subsets of Rn contains all subsets of Rn including ∅.

2. The σ-algebra generated by the collection of all intervals in Rn contains all intervals of Rn including ∅. Since an interval is a subset, every element of the σ-algebra generated by the collection of all intervals in Rn would coincide with the σ-algebra generated by the collection of all open subsets in Rn.

How far off am I? Is there a better way to make this argument?

Bacle2
You're on the right track, but careful on #2 : there are subsets of R^n that are not measurable.

It is enough to show that you can generate _all intervals_ by using the open intervals

alone.

jbunniii
Homework Helper
Gold Member
1. The σ-algebra generated by the collection of all open subsets of Rn would also contain all closed subsets by complementation. It would also contain their unions to generate all partly open subsets. Therefore the σ-algebra generated by the collection of all open subsets of Rn contains all subsets of Rn including ∅.

No, this isn't correct. The σ-algebra generated by the open subsets certainly does not contain ALL subsets of Rn.

Let's introduce some notation to make things easier to express.

Let I denote the set of all open intervals contained in Rn. Let O denote the set of all open subsets of Rn.

Let $\Sigma(I)$ be the sigma algebra generated by I, and $\Sigma(O)$ be the sigma algebra generated by O. You need to show that $\Sigma(I) = \Sigma(O)$. The standard way to prove set inequality is to prove inclusion in both directions:
$$\Sigma(I) \subset \Sigma(O)$$
and
$$\Sigma(O) \subset \Sigma(I)$$
Clearly $I \subset O$. I suggest starting with this to make the argument that $\Sigma(I) \subset \Sigma(O)$. This is the easier part of the proof.

Bacle2
Just to add that there are infinitely many sets that are not generated by the

open intervals. These last are "nice-enough" sets, in that they are measurable

sets, but not "Borel Sets" , i.e., sets in the sigma-algebra generated by open

intervals. You can show this using a cardinality argument to show that the

cardinality of the collection of measurable sets is strictly larger than the cardinality

of the sigma algebra generated by the open set.

Thanks for the help Bacle2. Here is what I am turning in.

Proof: Let I be the collection of all intervals of Rn, and Ʃ(I) be the σ-algebra generated by I. Let O be the set of all open subsets of Rn, and Ʃ(O) be the σ-algebra generated by O.

To show: Ʃ(I) is a subset of Ʃ(O).

Pick an arbitrary interval i from Ʃ(I). i is either open, closed, or partly open. If i is open, then i is a member of O → i is a member of Ʃ(O). If i is closed, then i is a member of Oc → i is a member of Ʃ(O). If i is partly open, then i can be constructed using the intersections of finite number of elements of O and Oc → i is a member of Ʃ(O).

Therefore Ʃ(I) is a subset of Ʃ(O).

jbunniii
Homework Helper
Gold Member
To show: Ʃ(I) is a subset of Ʃ(O).

Pick an arbitrary interval i from Ʃ(I). i is either open, closed, or partly open. If i is open, then i is a member of O → i is a member of Ʃ(O). If i is closed, then i is a member of Oc → i is a member of Ʃ(O). If i is partly open, then i can be constructed using the intersections of finite number of elements of O and Oc → i is a member of Ʃ(O).

Therefore Ʃ(I) is a subset of Ʃ(O).

 - Sorry, my original post assumed that $I$ contained only OPEN intervals, because I misread something. Here is a modified comment:

In fact, your argument only proves that $I \subset \Sigma(O)$. This is because $\Sigma(I)$ contains sets that are not intervals. For example, the union of two intervals is not generally an interval.

You still need an additional step to conclude that $\Sigma(I) \subset \Sigma(O)$. Hint: what does it mean that $\Sigma(I)$ is the sigma algebra generated by $I$?

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Bacle2
Just to make sure, ChemEng1: when you talk about open intervals in R^n, are

you defining these as open "n-cubes" , i.e., sets of the form:

(x1,x2,...,xn):

a1< x1< b1 or a1 ≤ x1 <b1, or a

a2<x2<b2 ,

..........

.........

an <xn <bn , ....., etc. ?

i.e., all half-open, half closed combinations of these ?

EDIT: these should be the components of the open intervals.

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I got confirmation from the professor that he uses the word "coincide" to mean "equal". So both directions of containment need to be proved. I am also changing the notation from Ʃ(x) to Ʃx to avoid confusion that Ʃ is a function.

This was my first attempt to show $\Sigma_{G}\subset\Sigma_{I}$, but I don't think that it's right.

Pick an arbitrary open $G\in\Sigma_{G}$. Notice $G=\bigcup_{k}G_{k}$ where $G_{k}=G\bigcap I_{k}$. $I_{k}=[-k,k]\times...\times[-k,k]$ for k=1,2,... Thus $G\in\Sigma_{I}$. Since $\Sigma_{I}$ is a σ-algebra, it contains all intervals. Then $\Sigma_{I}\supset\left\langle\left\{all open sets \right\}\right\rangle=\Sigma_{G}$. Therefore $\Sigma_{G}\subset\Sigma_{I}$

jbunniii
Homework Helper
Gold Member
Pick an arbitrary open $G\in\Sigma_{G}$.

This is already problematic, because an arbitrary set in $\Sigma_{G}$ is not necessarily open.

Notice $G=\bigcup_{k}G_{k}$ where $G_{k}=G\bigcap I_{k}$. $I_{k}=[-k,k]\times...\times[-k,k]$ for k=1,2,... Thus $G\in\Sigma_{I}$.

I assume you are claiming that this follows because $G\bigcap I_{k} \in \Sigma_{I}$. But you haven't demonstrated why this is true.

Since $\Sigma_{I}$ is a σ-algebra, it contains all intervals.
This is true because $\Sigma_{I}$ contains $I$, the set of all intervals, not because $\Sigma_{I}$ is a σ-algebra. There are σ-algebras that don't contain any intervals, for example, the σ-algebra generated by all of the singletons in R^n.

Then $\Sigma_{I}\supset\left\langle\left\{all open sets \right\}\right\rangle=\Sigma_{G}$.

This equality is false. It is true that

$\left\langle\left\{all open sets \right\}\right\rangle \subset \Sigma_{G}$

but the containment is proper/strict.

 Actually that depends on the meaning of the notation $\left\langle\left\{all open sets \right\}\right\rangle$. What does the angle bracket notation mean here? Are you using it as a shorthand to mean "the sigma algebra generated by..."?

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Yeah. I noticed those problems when I went back to the problem (again). Here's where I am with the problem:

Let I be the collection of all intervals of Rn and ƩI be the σ-algebra generated by I. Let G be the set of all open subsets of Rn and ƩG be the σ-algebra generated by G.

Pick an arbitrary interval, i, from Rn. By definition, i is closed. → i is a member of CG. → <{I}> is contained in ƩG. → ƩI is contained in ƩG.

Pick an artibrary open set, g, from Rn. g is a member of CI. → <{G}> is contained in ƩI. → ƩG is contained in ƩI.

ƩI is contained in ƩG. ƩG is contained in ƩI. → ƩG equals ƩI.

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Bacle2