# Borel sigma-algebra over [0,1]

I just started studying this, so forgive me if I get something wrong.

I have been given the following definition of a Borel $\sigma$-algebra over $\Omega=[0,1]$: It is the smallest $\sigma$-algebra that contains all intervals $(a,b)$ with $0\leq a<b\leq1$. Lets call this algebra $\mathcal{B}$.

Now apparently, every subinterval of $[0,1]$, including a half-open one like $[0,1)$, should be in $\mathcal{B}$. Even simpler, $\{0\}\in\mathcal{B}$ should be correct. Right?

I can not figure out how this would work - I know that complements, intersections and unions of any elements of $\mathcal{B}$ are again elements of $\mathcal{B}$.

With the given definition, I don't know how to obtain a subset that contains either $0$ or $1$ and not both: $0\leq a<b\leq 1, M = (a,b) => 0,1 \notin M => 0,1\in \overline{M}$.

I have searched using google and stackexchange, but I seem to have been given an uncommon definition. Is the definition wrong or am I missing something?

• Remember that the Borel $\sigma$-algebra is generated by open sets. So the first thing you have to ask yourself is, "What do open sets look like in $[0,1]$?" If you are familiar with topology and the notion of subspace topology, then in addition to all intervals of the form $(a,b)$, we also have all intervals of the form $[0,b)$, and $(a,1]$, and finally $[0,1]$ is also open in this subspace topology. Commented Apr 10, 2015 at 13:29

You're right; the definition you quoted is wrong. The family $\mathcal F$ of all subsets of $[0,1]$ that contain both or neither of $0$ and $1$ is a $\sigma$-algebra that contains the intervals $(a,b)$. So all Borel sets according to the quoted definition are in $\mathcal F$.

You should prove that if $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, then $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$.

Once you have the above fact, then since $[0, \frac{1}{n}) \in \mathcal{B}$ for each $n$, we have $\bigcap \limits_{n = 1}^{\infty} [0, \frac{1}{n}) = \{ 0 \} \in \mathcal{B}$.

Here is a proof for the first claim I made (try it yourself first, and when you want to see the proof, just move your mouse over the box below):

Claim: If $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, then $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$.

Proof:

Since $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, and $\mathcal{B}$ is a $\sigma$-algebra (which is closed under complements), then $A_{n}^{c} \in \mathcal{B}$ for each $n$. Then since by definition $\sigma$-algebras are closed under countable unions, we have $\bigcup \limits_{n = 1}^{\infty} A^{c}_{n} \in \mathcal{B}$. But then the complement of this set is in $\mathcal{B}$. But $(\bigcup \limits_{n = 1}^{\infty} A^{c}_{n})^{c} = \bigcap \limits_{n = 1}^{\infty} A_{n}$, which proves $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$, as desired.

UPDATE: As requested in the comments, if $\mathcal{B}$ represents the smallest $\sigma$-algebra containing sets of the form $(a,b)$ with $a, b \in [0,1]$, then there is no way to get $\{ 0 \}$. We can get $\{ 0, 1 \}$ by noticing this set is the complement of $(0,1)$. But there is no way to "separate" $\{0 \}$ from the other elements as a measurable set.

• What do you mean prove? The first fact comes from the definition of $\sigma$ - algebra
– Ant
Commented Apr 10, 2015 at 13:20
• @Ant Sorry, I should be more clear. By prove, what I meant was prove. One definition of a $\sigma$-algebra is a collection of subsets that contains the entire space, and is closed under complements, and also countable unions. If these are the only axioms you have, closure under countable intersections should be proven. Commented Apr 10, 2015 at 13:22
• Ah I see. I've used another definition for $\sigma$-algebra then :-)
– Ant
Commented Apr 10, 2015 at 13:25
• @Ant Can you share what it is? I'm relatively new to this material, too, and this definition is the only one I've seen so far. Commented Apr 10, 2015 at 13:26
• Thanks, I knew that this claim was true, but I hadn't looked at the proof yet. Useful! But I have an issue with the rest of your answer: You say "...since $[0,\frac{1}{n})\in B$ for each $n$, ..." which is pretty much exactly what I am doubting. $[0, ...)$ contains $0$ and not $1$, so I don't think it is in $\mathcal{B}$.
– Wutz
Commented Apr 10, 2015 at 13:37

Actually your definition is wrong: the Borel $\sigma$-algebra of $[0,1]$ is the smallest $\sigma$-algebra containing all open sets of $[0,1]$.

Clearly, every open set $U \subset (0,1)$ is a (at most) countable union of disjoint intervals of the form $(a_i, b_i)$. This means that $U \in \mathcal{B}$ (I denote by $\mathcal{B}$ your definition of Borel $\sigma$-algebra).

However, as you pointed in your question, there are some open subsets touching the boundary of $[0,1]$. As an example, $[0,1)$ is open in $[0,1]$, hence it is a Borel subset of $[0,1]$. Actually, $[0,1) \notin \mathcal{B}$: this shows that $\mathcal{B}$ is not the Borel algebra of $[0,1]$.