Definition: Let $(X, \Sigma)$ be a measurable space. Then, $$Meas(X, \Sigma) = \{ \mu | \mu \text{is a finite signed measure on} (X,\Sigma)\} .$$
Then, we know that $(Meas(X, \Sigma), ||\cdot||_{meas})$ is a normed linear space.
Question: Show that it is Banach space.
I am able to show that cauchy sequence, $\mu_n$, in $(Meas(X, \Sigma), ||\cdot||_{meas})$ is also cauchy in $\mathbb{R}$. So, for $E \in \Sigma$, $(\mu_n(E))$ converges uniformly to $\mu(E)$. I can also show that $\mu(E)$ is a finite signed measure.
But, I am having trouble to show that $\lim_{n \to \infty}||\mu_n - \mu||_{meas} = 0$.
I think that we have to show something like: for $n \ge N$, $$||\mu_n - \mu||_{meas} \le \cdot \le |\mu_n(E) - \mu(E)| < \varepsilon.$$
But, I am not sure what should be in this middle section ($\cdot$).
I appreciate if you give some help.
I do not know what definition f norm of a signed measure you are using but the Cauchy property implies that given $\epsilon>0$ there exists $N$ such that $|\mu_n(A)-\mu_k(A)| <\epsilon$ for all $A$ for $n,k \geq N$. Just let $k \to \infty$ to get $|\mu_n(A)-\mu(A)| <\epsilon$ for all $A$ an dthis implies that $\mu_n \to \mu$ in the norm. [ If you have defined the norm using partitions then $\sup_A |\mu(A)| \leq \|\mu\| \leq 2\sup_A |\mu(A)|$ for all real valued countable additive measures $\mu$. This follows immediately from Hahn decomposition of signed measures].