I am following a construction of a measure on subsets of $\mathbb{R}$. The idea is to first define the semi-algebra of intervals and a "measure" $\mu$ on this semi-algebra which acts as we expect a length function should. i.e. $\mu(I)=b-a$ for intervals $[a,b)$ (note: the intervals can be open or closed on either side), $\mu(I+x)=\mu(I)$ (length is translation invariant), and $\mu(E)=\sum\mu(E_i)$ where $E_i$ are the disjoint union of $E$. Given this, I can quite easily show that we can extend this to a $\sigma$-additive measure $\nu$, on the algebra generated by this semi-algebra of intervals.
My question is regarding the extension of this $\nu$ to $\pi$ on the $\sigma$-algebra containing this algebra.
The idea as I understand is to first define an outer measure $\pi^*$ on all subsets of $\mathbb{R}$. An outer measure satisfies the following properties:
- $\mu(\emptyset)=0$.
- For $E\subseteq F$, $\mu(E)\leq\mu(F)$.
- For $E\subseteq\bigcup E_i$, $\mu(E)\leq\sum\mu(E_i)$
We define $\pi^*(E)$ as $\inf\limits_{\{E_i\}}\nu(E_i)$, where the $E_i$ are elements of the algebra and cover $E$. It is easy to check this is an outer measure.
Here is where I am confused about intutition: The next step is to define a collection of measurable sets $\mathcal{U}$ using this outer measure by stipulating that a set $S\subseteq\mathbb{R}$ is measurable if for every $E\subseteq\mathbb{R}$ $\pi^*(E)=\pi^*(A\cap E)+\pi^*(A^c\cap E)$. I am struggling to understand intuitively why these sets should be the measurable ones? Can anyone elucidate this condition for me?
I understand that from here we can show that this measurable space is a $\sigma$-algebra containing the algebra and that $\pi^*$ on this space extends $\nu$. But I'm struggling with the definition of $\mathcal{U}$.
Here's how I like to think of the definition of measurability; it may or may not match the historical development of the definition.
Instead of working with the whole real line, let's first deal with a space $Y$ of finite measure, like an interval $[b,c]$. Every subset $A$ of our interval has, as you said, an outer measure $\nu(A)$ obtained by approximating $A$ by supersets whose measure was already known (the sets $E_i$ in your definition of $\nu$). Symmetrically, one could define the inner measure of $A$ by approximating $A$ by subsets whose measure was already known. The inner and outer measures of $A$ are reasonable lower and upper bounds (respectively) for any sensible notion of "measure" of $A$. In particular, if the inner and outer measure of $A$ are the same number, then that number has a good claim to being called the measure of $A$. So one defines the measure of $A$ to be the outer measure of $A$ when that coincides with the inner measure of $A$. (When they don't coincide, we leave the measure of $A$ undefined.)
Using the fact that subsets of $A$ are the same thing as complements of supersets of the complement of $A$, one easily finds that the inner measure of $A$ is given by $\nu(Y)-\nu(A^c)$. So to say that the inner and outer measures of $A$ agree is just to say that $\nu(A)+\nu(A^c)=\nu(Y)$.
That looks somewhat like the definition of measurability that you quoted, but it's not exactly the same, and it's only for the situation of a space $Y$ of finite measure, not something like the whole real line.
So what about the whole real line? When should a set $A\subseteq\mathbb R$ be measurable? The first idea that occurs to me is to require $A\cap[n,n+1]$ to be measurable (as a subset of the finite measure space $[n,n+1]$) for all integers $n$. Then we can define (and we want to define, for the sake of $\sigma$-additivity) $\pi(A)$ to be $\sum_{n\in\mathbb Z}\nu(A\cap[n,n+1])$.
The second idea that occurs to me is to require $A\cap[-n,n]$ to be measurable (as a subset of the finite measure space $[-n,n]$) for all natural numbers (or even all positive reals) $n$. Then define $\pi(A)$ to be $\lim_{n\to\infty}\nu(A\cap[-n,n])$.
Fortunately, these two ideas are equivalent to each other, i.e., they produce the same $\sigma$-algebra of measurable sets and the same measure function. The definition you quoted broadens the two ideas by using arbitrary sets $E$ in place of intervals like $[n,n+1]$ or $[-n,n]$, but it too is equivalent to my two ideas. I personally prefer to think of just the simpler case where $E$ is an interval, but using unrestricted $E$'s makes the definition shorter (and maybe easier to use).