Total variation inequality liminf I'm trying to understand the proof, given by t.b. here Space of Complex Measures is Banach (proof?)
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In the course of the proof, the author uses the equality
$\left|\left| \mu - \mu_{m} \right|\right| \leq \liminf\limits_{n \to \infty} \left|\left|\mu_{n}-\mu_{m}\right|\right|$, 
where $\left|\left| \cdot \right|\right|$ denotes the total variation norm and
$\mu(A):=\lim\limits_{n \to \infty} \mu_{n}\left(A\right)$
the limit of a sequence of measures $\mu_{n}$ with bounded total variation.
I would like to know, why the above inequality holds. I tried several thinks using the definition of the total variation and I came up with nothing.
Any help is most appreciated.
 A: It suffices to show that $\|\mu\| \le \liminf_n \|\mu_n\|$. The general case follows quickly from that.
Take any finite disjoint collection $\{E_k\}_{k=1}^m$ of measurable sets. Then $$\sum_{k=1}^m \mu(E_k) = \lim_{n \to \infty} \sum_{k=1}^m |\mu_n(E_k)|.$$ But 
$$\sum_{k=1}^m |\mu_n(E_k)| \le \|\mu_n\|$$ by the definition of variation. Take the limit inferior on both sides to get $$\lim_{n \to \infty} \sum_{k=1}^m |\mu_n(E_k)| \le \liminf_{n \to \infty} \|\mu_n\|.$$  Now take the supremum over all such collections $\{E_k\}$ to obtain $\|\mu\| \le \liminf \|\mu_n\|$.
A: We will use the notation from the proof that was mentioned.
By defintion, $| \mu_n(E) - \mu_m(E) | \le || \mu_n - \mu_m ||$, hence $\liminf_n | \mu_n(E) - \mu_m(E) | \le \liminf_n || \mu_n - \mu_m ||$.
But $\liminf_n | \mu_n(E) - \mu_m(E) | = \lim_n | \mu_n(E) - \mu_m(E) | = | \mu(E) - \mu_m(E) |$.
It follows that  $| \mu(E) - \mu_m(E) | \le \liminf_n || \mu_n - \mu_m ||$.
Taking supremum over E, we have
$|| \mu - \mu_m || \le \liminf_n || \mu_n - \mu_m ||$, q.e.d.
