The collection of all charges on $(X, \mathcal{X})$ is a Banach space. 
The collection of all charges on $(X, \mathcal{X})$ is a Banach space under
  the vector operations $(c\mu)(E) = c\mu(E)$, $(\lambda + \mu)(E) = \lambda(E) + \mu(E)$
  and the norm $\|\mu\| = |\mu|(X)$.

I took a Cauchy sequence $(\mu_n)$ and I defined $\mu(E)=\lim_{n\to \infty}\mu_n(E)$, I had proved that $\lim_{n\to \infty}\mu_n(E)$ exists, and that $\mu$ is a charge, but I can't proof that $$\|\mu-\mu_n\|=|\mu-\mu_n|(X)\to 0.$$ 
I know that this question is similar to "Space of Complex Measures is Banach (proof?)" but I don't understand why 
$$\lVert\mu-\mu_m\rVert
\leq \liminf_{n\to\infty}\lVert\mu_n-\mu_m\rVert
\xrightarrow{m\to\infty}0.$$
 A: I made this on this way: Use the fact that, for a charge $\lambda$, we have
$$
    |\lambda|(X)=\sup\left\{\sum_{j=1}^{n}|\lambda(A_j)|: \{A_j\}_{j=1}^n \text{ finite partition of }X\right\}.
$$
Let $\{A_j\}_{j=1}^k$ be a finite partition of $X$, we have:
$$
    \sum_{j=1}^k|\mu_n(A_j)-\mu_m(A_j)|=\sum_{j=1}^k|(\mu_n-\mu_m)(A_j)|\leq \|\mu_m-\mu_n\|<\epsilon,
$$
taking the limit $n\to \infty$,
$$
    \lim_{n\to\infty}\sum_{j=1}^k|\mu_n(A_j)-\mu_m(A_j)| = \sum_{j=1}^k|\lim_{n\to\infty}\mu_n(A_j)-\mu_m(A_j)| = \sum_{j=1}^k|\mu(A_j)-\mu_m(A_j)|\leq\epsilon,
$$
hence
$$
    \|\mu-\mu_n\|\leq\epsilon.
$$
A: I think it needs $\mu(E)\leq\mu(X)$ if E is the subset of X. Then if we have a Cauchy sequence {$\mu_n$} satisfy ||$\mu_n$-$\mu_m$||
A: The full answer is in the link you provided, so I'll just answer your question. A Borel measure is basically what you described -- a charge is just a signed measure, I guess. 
The proof really follows from the following observation: $\sup\{g(y):y\in Y\}\leq \inf\{f(x):x\in X\}$
Let $h(N,P) = \sup_{n \geq N} \sum_{i=1}^{k_P} |\mu-\mu_n|(A_i)$, where $P$ is a partition of $X$, i.e. $P = \{A_i\}_{i=1}^{k_P}$ such that $X = \cup_{i=1}^{k_P} A_i$ and the $A_i$ are measurable and disjoint. We will be taking the sup over all such $P$. Notice that $h(N,P)$ is decreasing in $N$, so
$$\lim_{N\rightarrow\infty} h(N,P) = \inf_N h(N,P).$$
\begin{align*}
\|\mu-\mu_m\| &=  \sup_P\{\lim_{N \rightarrow \infty} h(N,P)\} \\
&= \sup_P \{\inf_N h(N,P)\} \\
&\leq \inf_N  \{\sup_P h(N,P)\} \\
&= \inf_N \{\sup_P\sup_{n \geq N} \sum_{i=1}^{k_P} |\mu_n - \mu_m|(A_i)\}\\
&= \inf_N \sup_{n \geq N} \{\sup_P \sum_{i=1}^{k_P} |\mu_n-\mu_m|(A_i)\} \\
&= \liminf_{n \rightarrow \infty} \|\mu_n-\mu_m\|.
\end{align*}
