A question about the definition of measurable sets.

Question

$\newcommand{\scrF}{\mathscr{F}}$ I've been thinking about some of the topics I've learned in basic measure theory and I'm having a bit of a conceptual issue. We introduced first the notion of a $\sigma$-algebra so that if $X$ is a non-empty set and $\scrF$ is a $\sigma$-algebra defined on $X$ then the pair $(X,\scrF)$ is a measurable space. Then we studied measures for a while and measurable functions etc; where the measurability of some $f:X \rightarrow \mathbb{R}$ is intimately related to the measurable space $(X,\scrF)$ since a function is measurable only if preimages of Borel sets in $\mathbb{R}$ happen to live in $\scrF$.

I suppose I never thought to question this critically but why do we say that $(X,\scrF)$ constitutes a measurable space? What about the structure of a $\sigma$-algebra makes the subsets it contains measurable? A reason I ask is because I was reading this answer about the construction of measures from outer measures and I realized that I didn't really know what was nice about the sets in a $\sigma$-algebra. Why do they play so well with integration and measure theory, and why couldn't we have a theory of integration/measure without $\sigma$-algebras? Thanks in advance for the clarification.

Answer 1

The motivation comes from trying to define a measure on the real numbers. To be useful, it should reflect our intuitive notions about length. We are trying to find a function $\mu : \mathcal{P}(\mathbb{R})\rightarrow \mathbb{R}$ that respects the following properties:

1. If $I$ is an interval $[a, b]$, then $\mu(I) = b - a$,
2. For any $a \in \mathbb{R}$, $\mu(A + a) = \mu(A)$ for all sets $A \subset \mathbb{R}$,
3. If $(A_i)_{i \in \mathbb{N}}$ is a countable collection of pairwise disjunct sets, then $\mu(\bigcup A_i) = \sum \mu(A_i)$.

It turns out that such a function cannot exist: An easy way to show this is by using a Vitali set, defined as follows. We define a relation $\sim$ on $[0,1]$ by $a \sim b$ if $a - b \in \mathbb{Q}$. It is easy to show this is an equivalence relation. Take the quotient $[0,1]/\sim$ and choose one representative for each equivalence class. Call the set of these representatives $V$.

It is now easy to show that any two copies of $V$ translated by rational numbers are disjunct (i.e. $V + p \cap V + q = \emptyset$ for any $p \neq q$, $p, q \in \mathbb{Q}$.

Thus, by property (3), $ \mu(\bigcup_{q \in \mathbb{Q}, |q| < 1} V + q \cap [0, 1]) = \sum_{q \in \mathbb{Q}, |q| < 1} V + q \cap [0, 1]$. However, it is easy to show that $\bigcup_{q \in \mathbb{Q}, |q| < 1} \mu(V + q \cap [0, 1]) = [0,1]$, thus, $$ 1 = \sum_{q \in \mathbb{Q}, |q| < 1} \mu(V + q \cap [0, 1])$$ by property (1). By property (2), $$ \sum_{q \in \mathbb{Q}, |q| < 1} \mu(V + q \cap [0, 1]) = \sum_{q \in \mathbb{Q}, |q| < 1} \mu(V \cap [0, 1]),$$ but it is clear that there is no real number $\mu(V \cap [0, 1])$ other than $0$ such that this series converges.

All this shows that a function defined on all of $\mathcal{P}(\mathbb{R})$ with the properties 1-3 cannot exist. Usually, one moves on from this by defining the outer Lebesgue measure and then finding ''good'' (measurable) sets for which these properties do hold. One then shows that these sets form a sigma algebra, which is useful for showing that certain sets are also measurable. When moving on to abstract measure spaces, one starts from the sigma algebra, having learned that one usually cannot define a good measure on the entire power set.