$\sigma$ algebra question So I'm learning a little bit of measure theory currently, and I am a little in the dark as to the motivation for the definition of the $\sigma$ algebra. My biggest question is why the $\sigma$ algebra must be used rather than the power set (since a $\sigma$ algebra is a proper subset of the power set, right?). I had read somewhere that it has something to do with the Banach-Tarski paradox, but I am unsure why if that is the case. Can someone help shed some light on this for me? Thanks!
 A: The relevant theorem about using the whole power set rather than some smaller $\sigma$-algebra is the following. There is no function $\mu$ assigning to every subset $A$ of $\mathbb R$ an ext4ended-real number $\mu(A)\in[0,\infty]$  such that: (1) For any interval $[a,b]$ (for $a<b$ in $\mathbb R$), we have $\mu([a,b])=b-a$ (so $\mu$ agrees with our usual notion of length when the latter is available), (2) For any $A\subseteq\mathbb R$ and any $t\in\mathbb R$, we have $\mu(\{a+t:a\in A\})=\mu(A)$ (i.e., $\mu$ is invariant under translations), and (3) If $A_n$ (for $n\in\mathbb N$) are pairwise disjoint, then $\mu(\bigcup_nA_n)=\sum_n\mu(A_n)$. 
In other words, there is no "reasonable" measure defined on the whole power set of $\mathbb R$.  There is, however, a reasonable measure defined on the "nice" subsets of $\mathbb R$, and the meaning of "nice" here is broad enough to encompass the sets people (or at least analysts and probabilists) normally need to work with. The idea is to use (1) above to define $\mu$ on intervals and then use (3) (and its consequence that, if $A\subseteq B$ then $\mu(A)\leq\mu(B)$) to extend $\mu$ to lots of other sets. Roughly speaking, the sets that get assigned a measure this way are the Lebesgue measurable sets, and they form a $\sigma$-algebra.
It turns out that lots of other situations lead to measure functions defined on $\sigma$-algebras. For example, if you had a material object with density $\rho$ varying in space, then there would be a measure, as described above, except it's defined on "nice" subsets of space $\mathbb R^3$ and, in place of (1) you'd have $\mu$ agreeing with ordinary mass on rectangular boxes. Similarly for probability densities. So the $\sigma$-algebra context is broad enough to cover lots of important situations.
On the other hand, the $\sigma$-algebra context is narrow enough to allow proofs of useful theorems. Specifically, this context allows us to impose requirement (3) above, countable additivity, on measures, and that requirement plays a role in numerous proofs. 
A: $\sigma$ algebras are defined this way so that a suitable measure can be defined on them. It’s impossible to define a reasonable measure on the power set of $\mathbb R$ or many other sets, because you can always find sets which can’t be assigned a proper value (in a similar way to the Banach-Tarski paradox, where the sum of values of the pieces should simultaneously be “twice” the sphere and “all” of the sphere).
