# “measurable sets are an algebra/a sigma-algebra.” ?<=>? “unions or intersections of finite/countable collections of measurable sets are measurable.”

Measurable sets are an algebra/sigma-algebra

In Chapter 3 of Real Analysis by Royden and Fitzpatrick, they sometimes say that measurable sets are an algebra/a sigma-algebra while sometimes they say that unions or intersections of finite/countable collections of measurable sets are measurable.

Do they not imply each other?

"measurable sets are an algebra/a sigma-algebra." ?<=>? "unions or intersections of finite/countable collections of measurable sets are measurable."

Edit: To clarify, I mean to ask 2 things:

1 Is it correct to say that measurable sets are an algebra iff complements, unions or intersections of finite collections of measurable sets are measurable?

2 ///ly, is it correct to say that measurable sets are a sigma-algebra iff complements, unions or intersections of countable collections of measurable sets are measurable?

Edited to include complements

Edit: Here are samples from the text:

• @rschwieb Sorry for the confusion. Algebra is for finite and sigma-algebra is for countable right? Anyway, yes. Two equivalences. Is it correct to say that measurable sets are an algebra iff unions or intersections of finite collections of measurable sets are measurable? ///ly, is it correct to say that measurable sets are a sigma-algebra iff unions or intersections of countable collections of measurable sets are measurable? – BCLC May 29 '14 at 16:50

Is the fact that the measurable sets are a sigma-algebra equivalent to saying that unions or intersections of countable collections of measurable sets are measurable?

To be precise, the collection of measurable sets forms a $\sigma$-algebra, and measurable sets are the elements inside this $\sigma$-algebra.

If you want to prove that the collection of measurable sets forms a $\sigma$-algebra, you will show the collection contains the empty set, closed under complement and countable union (or intersections) as you said.

Edit: Given an abstract space $X$, you do not need a $\sigma$-algebra and a measure for it to become a measure space, you only need to have an algebra and a premeasure, $(X,\mathcal{A}_0, \mu_0)$. You can then extend this into a measure space $(X,\mathcal{A}, \mu)$, but the extension might not be unique if the measure is not $\sigma$-finite.

I will give a concrete example: take the set of real numbers $\mathbb{R}$.

1. the collection of finite union of intervals is an algebra $\mathcal{A}_0$.

2. a premeasure on this algebra $\mu_0$, using the fact that any finite union of intervals $\cup I_k$ can be rewritten as a finite union of disjoint intervals $\cup I'_{k'}$, define $$\mu_0 (\cup I_k) = \sum |I'_{k'}|.$$

3. Define an outer measure $\mu^*: 2^\mathbb{R} \rightarrow [0,\infty]$ $$\mu^*(E) = \inf \{\sum_{k=1}^\infty \mu_0(E_k) : E\subset \cup E_k \text{ and } E_k\in \mathcal{A}_0\}.$$

4. Using the definition of Caratheodory measurability, a set $E$ is called $\mu^*$-measurable if $$\mu^* (A) = \mu^*(A\cap E) +\mu^* (A\cap E^c)$$ for every $A\subset \mathbb{R}$.

5. Now we have the collection of $\mu^*$-measurable sets, one can show that this collection has the structure of a $\sigma$-algebra. When we restrict $\mu^*$ to this $\sigma$-algebra $\mathcal{A}$, we have the measure $\mu$ and the measure space $(\mathbb{R}, \mathcal{A},\mu)$.

• @rschwieb I edited my post as a reply. – Xiao May 29 '14 at 18:14
• Ugh...thanks, I'll assume that's a yes. I added pictures. Are R&F then being inconsistent, or am I interpreting R&F wrong? – BCLC Jun 3 '14 at 14:19
• @BCLC The important idea is that the collection of (Caratheodory type) measurable sets (w.r.t a measure ) forms a $\sigma$-algebra, thus any countable union or countable intersection of measurable sets is also measurable. In the picture that you posted, this argument is used in the opposite way, that is, if you want to show a set $A$ is measurable ($A\in \mathcal{B}$), it is enough to show $A = \cup_n A_n$ (or $A = \cap_n A_n$) where each $A_n$ is measurable ($A_n\in \mathcal{B}$). – Xiao Jun 3 '14 at 15:19
• And by "measurable sets are an algebra", this is a simple remark of the trick that I mentioned above. If you need to show $A\in \mathcal{A}$ where $\mathcal{A}$ is an algebra, all you need to show is that $A = A_1 \cup A_2$ (or $A = A_1 \cap A_2$) for some $A_1, A_2 \in \mathcal{A}$. – Xiao Jun 3 '14 at 15:26
• "In the picture that you posted, this argument is used in the opposite way" --> Okay I think this explains a lot assuming I understand you correctly. So they do imply each other but one is used instead of the other depending on what is being assumed and what is being asked? And thanks! – BCLC Jun 24 '14 at 4:08

The definition of the σ-algebra from Wikipedia: σ-algebra on a set X is a collection of subsets of X that is closed under countably many set operations (complement, union and intersection). You statement in Question 2 is close, but you forgot complements.

• ...and similarly for (non-$\sigma$) algebras. – Eric Stucky May 29 '14 at 17:17
• Edited. So is it a yes now? – BCLC May 30 '14 at 11:21
• @PA6OTA Included pictures. – BCLC Jun 3 '14 at 14:12
• @EricStucky Included pictures. – BCLC Jun 3 '14 at 14:16
• Ah right so are R&F then being inconsistent, or am I interpreting R&F wrong? – BCLC Jun 3 '14 at 14:27

To the best of my knowledge, a measure space is defined by three pieces of data $(X,\Sigma,\mu)$ where $\Sigma$ is a $\sigma$-algebra on a set $X$, and $\mu$ a function from $\Sigma$ into the extended real line.

"Measurable sets" just means "the elements of $\Sigma$," so the measurable sets are a $\sigma$-algebra by definition.

If you relax the definition so that it only requires finitely many intersections and unions, then you are talking about a premeasure, and the collection of sets that are "premeasurable" only form a ring of sets.

First red box: $f^{-1}(\infty)$ is measurable since it is the intersection of a countable collection of measurable sets. Since that set is an intersection of measurable sets, and the measurable sets are closed under countable intersections, the set is measurable.

Second red box: By definition of a measurable function, those two inverse images are measurable sets. Since $f^{-1}(\mathcal O)$ can be written in terms of countably many set operations of these mesurable sets, it is also measurable.

Third red box: the intersection of two measurable sets is a measurable set.

Fourth an fifth red boxes: The measurable sets form an algebra. Of course they do, in fact they form a $\sigma$ algebra, so they form an algebra too.

All of these things follow from the definition of "measure" and "measurable function." There is no equivalence being implied anywhere here, they are just all straightforward applications of the definition of "measurable sets."

• @BCLC What is your definition of $\sigma$ algebra, exactly? If we're not working from the same definition, I'll never be able to answer :) – rschwieb May 29 '14 at 17:31
• @BCLC In that case it seems weird to ask if being closed under countable unions and countable intersections "is equivalent" to being a $\sigma$ algebra since you already know being a $\sigma$ algebra means being closed under countable intersections, countable unions and complements. – rschwieb May 30 '14 at 12:40
• @BCLC It's OK :) Most of us remember what it's like to be in your position. I hope my comments aren't coming off in a bad way: I'm just trying to nudge you in the right direction. – rschwieb May 30 '14 at 13:06
• They aren't. I know stackexchange has higher good faith standards than Wikipedia meta.stackexchange.com/a/55844 I just found it weird that the book alternates their language. I'm probably missing something. More of this later – BCLC May 30 '14 at 13:18
• @BCLC When learning mathematics, much time is spent (invested?) getting used to unfamiliar language and concepts :) – rschwieb May 30 '14 at 13:29