Are all sets in sigma-algebra measurable? In the Wikipedia article it says:

the collection of those subsets for which a given measure is defined
  is necessarily a $\sigma$-algebra.

Fine, but is the opposite true? Do we know for sure that all sets of sigma algebra are measurable? If the answer is no, then is it the reason why Borel sigma algebra is so widely used in probability theory? 
 A: Here is one answer, which may answer the question as I understand it (but perhaps I do not understand the question correctly).
Start with a measure $\mu$ defined on a $\sigma$-algebra $\mathcal A$. Then define the outer measure $\mu^*$ associated with $\mu$, in the sense of Caratheodory. Once you have this outer measure, you can define the class $\mathcal A(\mu^*)$ of all $\mu^*$-measurable sets (still in the sense of Caratheodory). 
Then, it is part of the Caratheodory extension theorem that $\mathcal A\subseteq \mathcal A(\mu^*)$. So, in this sense, the answer to the question is "Yes".
On the other hand, it is not necessarily tru that all measurable sets belong to the original $\sigma$-algebra $\mathcal A$...
A: I think you possibly miss an concept about measurable. A set $a$ in an sigama-algebra $\mathcal{A}$ is called a $\mathcal{A}$-measurable set.
So, measurable is not an general idea. It must correspond to a sigama-algebra. Measurable is given with respect to a given sigama-algebra. Therefore, you cannot generally say that a set is measurable without saying which sigma -algebra it belongs to.
Back to your question, the correct question should be "Are all sets in sigma-algebra $\mathcal{A}_1$ $\mathcal{A}_1$-measurable?". The answer is Yes. The problem can also be "Are all sets in sigma-algebra $\mathcal{A}_1$ $\mathcal{A}_2$-measurable?". The answer is Yes if the set is in $\mathcal{A}_2$ too. The answer is No if the set is not in $\mathcal{A}_2$.
A: The term “measurable” is defined only in relation to a given measure. For example, in the counting measure, all sets are measurable (the finite sets have as measure the number of elements, the infinite sets have measure $\infty$). In the Lebesgue measure, there are unmeasurable sets (assuming the axiom of choice).
If you mean “there exists a measure in which every set is measurable” then yes, every member of a given sigma algebra is measurable for some measure. In particular, since all sets are measurable in the counting measure, this in particular is true for all the sets in the given sigma algebra.
Note also that all subsets of a set form a sigma algebra. Therefore for any measure that has unmeasurable sets, there exists a sigma algebra in which not all sets are measurable, namely the sigma algebra of all subsets.
Therefore if you mean “for a given arbitrary measure, are the sets in every sigma algebra measurable” the answer is no: Since every set is member of the sigma algebra of all subsets, any measure with unmeasurable sets is a counterexample.
A: Given a set $X$ and any $\sigma$-algebra (denoted $\mathcal{P}_\sigma\left(X\right)$) of $X$, is there always an outer measure $\mu^*$ on $X$ (which, by definition, applies to all subsets of $X$ including $\mathcal{P}_\sigma\left(X\right)$), so that $\mathcal{P}_\sigma\left(X\right)$ is the set of all $\mu^*$-measurable subsets of $X$?
At this point, if the answer is yes, we don't assume that this outer measure is unique. In other words, if the answer is yes, we should presume that there are many such outer measures. Therefore, to find the condition of the positive answer to this question means to at least raise one example.
Since the answer asks for something not necessarily exist for a given $\sigma$-algebra, we must start with something always exist for any $\sigma$-algebra. What we do is direct define such a thing on $\sigma$-algebra, and better it can be related in some way to outer measure.
Since we know that 1) Given an outer measure $\mu^*$ on $X$, the set of all $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra of $X$; and 2) Any outer measure $\mu^*$ is countably additive on pair-wise disjoint $\mu^*$-measurable sets. So, what we try to define for a $\sigma$-algebra $\mathcal{P}_\sigma\left(X\right)$ of a set $X$ is an outer measure $\mu^*$ that is countably additive on pair-wise disjoint elements of $\mathcal{P}_\sigma\left(X\right)$. We call the latter a measure.
The next several steps to reach the answer to the original question is similarly written in most textbooks. Namely, check if the measure space is both $\sigma$-finite and complete. If so, extent the measure $\mu$ to an outer measure $\mu^*$ in the sense of Caratheodory and we get one what the question asked for. If not so, try another measure. There are many examples of general measure spaces that are both $\sigma$-finite and complete.
Therefore, the answer to the question is conditionally yes: if on this $\sigma$-algebra there exists a measure by which the measure space is $\sigma$-finite and complete then yes.
Of course, this only converts the question into a next one: is there always a measure on any $\sigma$-algebra of a set that is $\sigma$-finite and complete? Fact: the Borel $\sigma$-algebra on $\mathbb{R}$ is not complete with respect to any $\sigma$-finite measure.
