What is the definition of a measurable set?

Question

I have seen multiple definitions for what a measurable set is (all of which come together to form a sigma algebra). I was wondering if they are all equivalent and if not what situation would one be used over another?

• Definition 1

Let $(X, \Sigma)$ be a measurable space, then any set $S \in \Sigma$ is a measurable set.

Measurable Space: The pair $(X, \Sigma)$ where $X$ is a set and $\Sigma$ is a $\sigma$-algebra on $X$

• Definition 2

Given a space $X$ let there exist an outer measure $\mu : 2^{X} \to [0, \infty]$ (where $2^{X} = \mathcal{P} \left( X \right) = $ all the subsets of $X$) then a set $S$ is measurable iff for every $A \in 2^{X}$ $$ \mu(A) = \mu(A \cap S) + \mu(A \cap S^{c}) = \mu(A \cap S) + \mu(A \setminus S) $$

• Definition 3 (I'm drawing this one from memory from baby rudin, and it's defined on $\mathbb{R}$)

Begin by defining a (outer$_1$) measure $\mu$. Next put $\mathcal{M}_f$ to be the set of countable unions of intervals. Then a set $S$ is measurable iff $$ \exists \{ S_n \}_{n=0}^{\infty} \, s.t. \mu(S_n) \to \mu(S) \text{ as } n \to \infty $$

Now I know that the sets described in definition 2 form a sigma algebra on $X$, and likewise I know the sets $S$ in definition 3 form a sigma algebra on $\mathbb{R}$, but definition 1 seems to imply that any possible sigma algebra can be used.

In definition 3 I cannot remember if he defines this using the outer measure (I believe he does).

The only conclusion I have been able to draw from this is definition 2 and 3 must be equivalent, but we call these sets $\boldsymbol\mu$-measurable (a specific measure is defined). However in definition 1 we call any set like that just measurable, and that given any sigma algebra $\Sigma \, \exists \mu$ an outer measure s.t. the sigma algebra formed by definition 2 with this outer measure is $\Sigma$. Is this correctly put?

Answer 1

There is no universal and objective definition of what is a measurable subset of a general space $X$. The general concept of a measurable subset has its origins in the problem of measure in Euclidean space:

Problem of measure: Given an object $A\subset\mathbb R^n$, how does one assign a measure $m(A)\in[0,\infty]$ to $A$? (In the case $n=1,2$ and $3$, the measure $m(A)$ is traditionally referred to as the length, the area, and the volume of $A$, respectively).

When the objects considered are very simple, this question is very easy to answer. For example, given a line segment $A=[a,b]\subset\mathbb R$, the measure of $A$ should obviously be of $m(A)=b-a$. Given a $n$-dimensional rectangle $$A=[a_1,b_1]\times\cdots\times[a_n,b_n]\subset\mathbb R^n,$$ the answer is equally obvious: $$m(A)=\prod_{i=1}^n(b_i-a_i).$$ Then, one can easily extend the measure of rectangle to slightly more general sets, such as disjoint unions of rectangles $$A=R_1\dot\cup\cdots\dot\cup R_k$$ by assigning $$m(A)=\sum_{i=1}^km(R_i)$$ (Indeed, for a theory of measure to make any kind of geometrical sense, the measure of a union of disjoint parts should be the sum of the measures of the constituent parts.) The real problem comes when trying to measure more complicated subsets of $\mathbb R^n$.

---

A classical solution to the measure problem consists in attempting to approximate the measure of a complicated set using simple sets. More precisely, suppose we have a class of simple sets $S$ which we know how to measure (these would contain rectangles and finite unions of rectangle for example). Then, given some arbitrary set $A$, we can define an inner measure $m_I(A)$ and an outer measure $m_O(A)$ of $A$ by letting $$m_I(A)=\sup\{m(E):E\subset A,~E\in S\}\text{ and }m_O(A)=\inf\{m(E):E\supset A,~E\in S\}.$$ (Note that the inner and outer measures of sets in $S$ are clearly the same as the measure we have already assigned to them.) In this framework, one calls a set $A\subset\mathbb R^n$ measurable if $m_O(A)=m_I(A)$, in which case we assign $m(A)=m_O(A)=m_I(A)$. In other words, we call a set measurable if our theory of measure is capable of giving a sensible answer to "what is the measure of $A$?"

---

The solution of the measure problem I discussed in the previous paragraph gave rise to the Jordan theory of measure, as well as the more modern Lebesgue measure. The concepts of outer measure and general measures you have written down in your questions are answers to generalizations of the problem of measure to arbitrary spaces $X$, with the intent of extending the theory of Lebesgue integration to those spaces. The exact axioms (i.e., the definitions of a $\sigma$-algebra and measure) are in place in order to ensure that we obtain a theory of integration that is similar to the theory of Lebesgue measure/integration, that is, with similar theorems such as countable subadditivity, countinuity from above/below, monotone/dominated convergence, etc.

---

Going into more detail would require a lot of explanation. If you'd like to know more, I personally recommend reading sections 1.1 - 1.4 of An introduction to measure theory by Terence Tao. I only really started to understand measure theory when I read it.