Q:

Real Analysis, Folland Problem 5.1.8 Normed Vector Spaces

Background information:
Proposition 3.14 - If $\nu_1,\nu_2$ are complex measures on $(X,M)$, then $|\nu_1 + \nu_2|\leq |\nu_1| + |\nu_2|$
Theorem 5.1 - A normed vector space $X$ is complete if and only if every absolutely convergent series in $X$ converges.
Let $(X,M)$ be a measurable space, and let $M(X)$ be the space of complex measures on $(X,M)$. Then $\lVert \mu \rVert = |\mu|(X)$ is a norm on $M(X)$ that makes $M(X)$ into a Banach space. (Use Theorem 5.1.)
Attempted proof: For each $\nu\in M(X)$ there is a finite measure $\mu$ on $X$ and a function $f\in L^1(\mu)$ such that $d\nu = fd\mu$. Note that $\lVert \nu \rVert = |\nu|(X) = \int |f|d\nu$.
If $a\in\mathbb{C}$ then $\lVert a\nu\rVert = |a\nu|(X) = \int|af|d\nu = |a|\int |f|d\mu = |a||\nu|(X) = |a|\lVert \nu\rVert$. The triangle inequality follows from proposition 3.13. Therefore $\lVert \cdot \rVert$ is a norm on $M(X)$.
I fail to see how we use theorem 5.1 and show that the norm on $M(X)$ makes $M(X)$ into a Banach space. Any suggestions is greatly appreciated. 

A:

Suppose that $\{\mu_n\}$ is a sequence of measures such that $\sum_{n=1}^{\infty}||\mu_n||<\infty$, and define a new measure $\mu$ by
$$ \mu(E)=\sum_{n=1}^{\infty}\mu_n(E) $$
This sum converges absolutely, because
$$ \sum_{n=1}^{\infty}|\mu_n(E)|\leq \sum_{n=1}^{\infty}|\mu_n|(E)\leq \sum_{n=1}^{\infty}||\mu_n||<\infty $$
Moreover $\mu(\emptyset)=0$, and if $E=\cup_{j=1}^{\infty}E_j$ with the $E_j$ pairwise disjoint, then
$$ \mu(E)=\sum_{n=1}^{\infty}\mu_n(E)=\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\mu_n(E_j)=\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\mu_n(E_j)=\sum_{j=1}^{\infty}\mu(E_j)$$
by Fubini's theorem.
Therefore $\mu$ is a measure. Finally,
$$ ||\mu-\sum_{n=1}^N\mu_n||\leq \sum_{n>N}||\mu_n||$$
which tends to zero as $N\to\infty$. Therefore $\sum_n\mu_n$ converges to $\mu$, which proves that $M(X)$ is complete.

