Let $\mathcal{J}$ be the collection of intervals from $[0,1]$. By intervals, I include singletons, emptyset, etc. So $\mathcal{J}$ looks like
$$\mathcal{J} = \{ (0, 0.7), 0.5 ,\emptyset, [0,1], \cdots \}$$
I have two different $\sigma$-algebras constructed from $\mathcal{J}$. One is $\sigma(\mathcal{J})$, which is the Borel $\sigma$-algebra of subsets of $[0,1]$ and usually denoted as $\mathcal{B}([0,1])$. Another one is the $\sigma$-algebra you get when you apply the Caratheodory extension theorem to $\mathcal{J}$. Let's call it $\mathcal{M}$.
Apparently we have $\mathcal{B} \subset \mathcal{M}$, by definition. However the following comments on my book confuse me and I don't know why they are true.
It can be shown that $\mathcal{M}$ is in fact much bigger than $\mathcal{B}$; it even has larger cardinality (why?). Furthermore, it turns out that the Lebesgue measure restricted to $\mathcal{B}$ is not complete (why?), though on $\mathcal{M}$ it is.
Why those two statements are true? No proof is provided in the book but I am just curious. For the first statement I simply have no idea how to prove it. For the second statement, by definition we must find some $A \in \mathcal{B}$ with $\mathbb{P}(A) = 0$ and some $B \subset A$ such that $B \notin \mathcal{B}$, but cannot think of an example...
Answer to both questions: there are $2^{\aleph_0}$ Borel subsets of $[0, 1]$. On the other hand, the Cantor set has measure zero, so every subset of it is in $\mathcal{M}$. There are $2^{2^{\aleph_0}}$ such subsets, so some must not be Borel, although the Cantor set is Borel.