Why the space of complex measures is Banach?

It can be justified via Fatou's lemma with varying measures, which can be found here: https://en.wikipedia.org/wiki/Fatou%27s_lemma#Fatou.27s_Lemma_with_Varying_Measures. Express the norm as an integral of an indicator function and apply the lemma to the sequence $\nu_n = \mu_n - \mu_m$.

I think there are easier ways to prove this space is complete. Recall a normed space $X$ is a Banach space iff for all $(v_n)_{n=1}^\infty \subset X$ with $$\sum_{n=1}^\infty ||v_n||_X<\infty$$ we have $$\sum_{n=1}^\infty v_n$$ converges in $X$. This should be a straightforward computation.

There's a slick way to prove completeness with the Radon-Nikodym theorem: let $(\mu_n)_{n=1}^\infty$ be a cauchy sequence of complex measures (which is necessarily bounded). Consider the measure $$\nu = \sum_{n=1}^\infty 2^{-n}|\mu_n|$$ Then each $\mu_n$ is absolutely continuous with respect to $\nu$ and by Radon-Nikodym there is a sequence $(h_n)_{n=1}^\infty \subset L^1(\nu)$ such that $\mu_n = h_n \cdot \nu$ for all $n$. Then show $||h_n \cdot \nu||$ in measure norm is equal to $||h_n||_{L^1(\nu)}$ and exploit completeness of $L^1$...