How is every subset of real numbers measurable despite the existence of a non-measurable set? We know the existence of nonmeasurable subsets of $\mathbb R$ by Vitali, but many books and lecture notes on real analysis still include the statement that every subset of real numbers is measurable. What am I missing?
Context
On page 2 under the title General Constructions in these notes, it is said that the collection of all subsets forms a sigma-algebra. On the other hand, every element of a sigma algebra is measurable by: Charalambos, Real Analysis (3rd edition), page 112, Theorem 15.3.
How to resolve this apparent contradiction?
 A: What the notes you linked to are saying is that it's always possible to form a $\sigma$-algebra consisting of all subsets of a set $S$ (which could be $\mathbf{R}$).
But when we talk about the theorem that there exist non-measurable subsets of reals, what we mean is that they are not Lebesgue-measurable. 
That is, a $\sigma$-algebra of subsets of $\mathbf{R}$ is defined in a certain way, and its members are called Lebesgue-measurable sets. The theorem then says that this particular $\sigma$-algebra does not consist of all subsets of $\mathbf{R}$.
EDIT: The definition of a "measurable" set in Aliprantis's book Principles of Real Analysis, 3rd edition (p. 112) that you mentioned is as follows. 
"A subset $E$ of a measure space $(X, S, \mu)$ will be called measurable (more precisely $\mu$-measurable) if $E$ is measurable with respect to the outer measure $\mu^{*}$ generated by $\mu$."
The book considers $S$ here to be a semiring rather than a $\sigma$-algebra.
This is indeed the definition of a Lebesgue-measurable set, when a certain measure space $(\mathbf{R}, S, \mu)$ is chosen. Here $S$ consists of all intervals of the form $(a,b]$, and $\mu$ is defined by $\mu((a,b]) = b-a$.
The book continues:
"Every member of $S$ is measurable."
So what the book is saying is that every interval of the form $(a,b]$ is Lebesgue-measurable, not that every subset of $\mathbf{R}$ is.
A: We usually construct Lebesgue measure $\mu$ on the real numbers $\mathbb{R}$ by first constructing an outer measure $\mu^*$ which is defined on all subsets of $\mathbb{R}$. We then restrict to subsets $E$ which satisfy, for any set $A \subset \mathbb{R}$, 
$$\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c).$$
The collection of all such sets $E$ are what we call Lebesgue-measurable sets. Non-measurable sets do not satisfy this equality. After we restrict $\mu^*$ to these sets, we obtain a bona fide measure $\mu$, and $\mu$ is not defined on all subsets of $\mathbb{R}$. 
