TITLE: Space of Complex Measures is Banach (proof?)
QUESTION [32 upvotes]: How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation.
I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy sequence of measures has the properties of Complex measures.
We are using the norm of total variation :$$\lVert \mu\rVert = \lvert \mu \rvert(X)$$

REPLY [15 votes]: Another proof, using Radon-Nikodym.
Set
$$
\lambda=\sum_{n=1}^\infty \frac{1}{n^2}|\mu_n|.
$$
Then clearly, $\lambda$ is a bounded positive measure, and $\mu_n\ll\lambda$, for all $n\in\mathbb N$. Hence, due to Radon-Nikodym, there exist functions $\{f_n\}\in L^1(\lambda)$, such that $d\mu_n=f_n\,d\lambda$, for all $n\in\mathbb N$, and
$$
\|\mu_m - \mu_n\|=\int_X|f_m(x)-f_n(x)|\,d\lambda(x)=\|f_m-f_n\|_{L^1(\lambda)}
$$
Thus $\{f_n\}$ is a Cauchy sequence, and therefore, convergent  to some $f\in L^1(\lambda)$, since $L^1(\lambda)$ is complete. If the measure $\mu$ is defined as
$d\mu=f\,d\lambda$, then
$$
\|\mu_n-\mu\|=\int_X|f_n(x)-f(x)|\,d\lambda(x)=\|f_n-f\|_{L^1(\lambda)}\to 0.
$$