Purpose of Sigma Algebras for Measures In Folland chapter 1 he constructs a pathological subset of $\mathbb{R}$ to show that a function satisfying the properties of a measure and that allows this set in its domain does not exist.  He then introduces sigma-algebras as the "correct" domain for measures.
However, since $\mathcal{P}(\mathbb{R})$ is a sigma-algebra, it seems like we haven't really gotten anywhere since this pathological set is now in this sigma-algebra, yet it is a valid domain for a measure.
What am I overlooking?
 A: We can only do everything we want to do with measures if the domain is a $\sigma$-algebra. Specifically, if the domain isn't a $\sigma$-algebra, then all the nice existence of limits that we want for the Lebesgue theory will not hold. For example, if $\{ A_n \}_{n=1}^\infty$ are measurable sets and $\bigcup_{n=1}^\infty A_n$ is not measurable, then the monotone convergence theorem fails for the sequence $f_N(x)=\sum_{n=1}^N 1_{A_n}(x)$.
Of course we could do all of the set theory stuff we want with the Lebesgue measure if the domain was $\mathcal{P}(\mathbb{R})$, but various examples (requiring various set theory axioms) show that if the domain is $\mathcal{P}(\mathbb{R})$ then the Lebesgue measure cannot agree with length and be translation-invariant. (Actually, it's worse: a measure on $\mathcal{P}(\mathbb{R})$ cannot be translation-invariant as long as the measure of an interval is finite and positive.) So we choose a $\sigma$-algebra other than $\mathcal{P}(\mathbb{R})$ for the domain of the Lebesgue measure.
Nevertheless you can put some other measure on $\mathcal{P}(\mathbb{R})$; for example the counting measure can be put on $\mathcal{P}(A)$ for any set $A$. But the counting measure on $\mathbb{R}$ doesn't assign finite measure to an interval. Similarly you can put the zero measure on $\mathcal{P}(A)$ for any set $A$, but the zero measure on $\mathbb{R}$ doesn't assign positive measure to an interval.
A: You CAN define a measure on $P(\Bbb{R})$, but you can't define Lebesgue measure on $P(\Bbb{R})$ (for a rather boring example, define $\mu(S)=0$ for all $S\in P(\Bbb{R})$). 
Recall the definition of Lebesgue measurable: 
A set, $A$, is Lebesgue measurable iff:
$$\mu^{\ast}(E)=\mu^{\ast}(E \cap A)+\mu^{\ast}(E \cap A^c)$$
for all $E \subseteq \mathbb{R}$
Vitali sets (or other terrible sets) are NOT Lebesgue measurable, but that does NOT mean that we can't assign ANY measure to them
A: From the book:

The moral of these examples is that $\mathbb R^n$ contains subsets
  which are so strangely put together that it is impossible to define a
  geometrically reasonable notion of measure for them

In other words, while we can define a measure on $\mathcal P(\mathbb R^n)$, it could not satisfy the usual notion of area or volume. 
