Let $(r_n)_{n = 1}^{\infty}$ be an enumeration of the rationals. Define $\mu = \sum_{n \ge 1} 2^{-n} \delta_{r_n}$ so that $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu)$ is a measure space, where $\delta_{r_n}(A) = 1$ if $r_n \in A$, else $0$ and $\mathcal{B}(\mathbb{R})$ is the Borel algebra.
How do I find the completion of this measure space, $(\mathbb{R}, \mathbb{B}, \overline{\mu})$? In particular, I want to prove that $\mathbb{B} = P(\mathbb{R})$.
Attempt:
Suppose $(\mathbb{R}, P(\mathbb{R}), \overline{\mu})$ is not a complete measure space. This means there exists some null set $N \in \mathcal{B}(\mathbb{R})$ such that $\mu(N)=0$ and $N$ contains a subset $S$ which is not measurable. By the formula of $\mu$, $N$ must consist of only irrational numbers. The same applies to $S$. Finally, (here is the part which I am unclear on) this means $S$ has measure $0$ and is therefore measurable, contradiction.
Why are we allowed to do this?