Some confusion regrading $\sigma$-algebras I have the following proposition:

Let $X$ be a non-empty set and $\mathcal{B}$ a collection of subsets
  of $X$. Then there exists a smallest $\sigma$-algebra $\sigma(\mathcal{B})$ containing $\mathcal{B}$.

What I have trouble reconciling, is that I understand a $\sigma$-algebra to be a collection of subsets of $X$ satisfying certain properties (i.e. it is a collection of measurable sets). So, what if for example $\mathcal{B}$ is a collection of non-measurable sets? Then, I assume, there is no $\sigma$-algebra containing the sets from $\mathcal{B}?$
 A: To be measurable is a relative notion : you are measurable for a specific $\sigma$-algebra. You seem to be confused and to think that being measurable is an internal property...measurability is defined after having defined a $\sigma$-algebra
So, if you have a $\sigma$-algebra $\mathcal{B_1}$, you can take a collection of non-measurable set (for $\mathcal{B_1}$) and consider the $\sigma$-algebra $\mathcal{B_2}$ generated by this collection. Then your sets will be measurable for the new $\sigma$-algebra $\mathcal{B_2}$ BY DEFINITION of measurability. This is just that $\mathcal{B_2}$ has no reason to be included in $\mathcal{B_1}$
On the other hand, if you take a collection of measurable (still for $\mathcal{B_1}$) sets as a generator, then you will indeed create a sub $\sigma$-algebra of $\mathcal{B_1}$. But this is not the case in general
A: The notion 'measurable' only has a meaning with respect to a $\sigma$-algebra or a measure (which has a $\sigma$-algebra as its domain). 
(If no $\sigma$-algebra is mentioned, then 'measurable' usually means that 'measurable with respect to the Lebesgue measure' on $\Bbb R^n$, if I remember correctly.)
And, a subset $S$ is measurable w.r.t. the $\sigma$-algebra $\mathcal A$, by definition, iff $S\in\mathcal A$.
For the proof, consider the intersection of all $\sigma$-algebras that contain $\mathcal B$. (Note that the whole powerset, $P(X)$ is such.).
