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

Question

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:

Answer 1

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 $(\mathbb{R}, \mathcal{A},\mu)$.