Prove that if $\lim_{n}\mu(A)=\mu(A)$ uniformly for all measurable set, then $\lim_{n}||\mu_n-\mu||=0$. $\newcommand{\scra}{\mathscr{A}}$
$\newcommand{\limm}[2]{\underset{#1\to#2}{\lim}}$
$\newcommand{\pl}{\parallel}$
$\newcommand{\iy}{\infty}$

Let $\mu$ and $\mu_1,\,\mu_2,\cdots$ be finite signed or complex measures on $(X,\,\scra)$. Show that $\limm{n}{\iy}\pl\mu_n-\mu\pl=0$ holds if and only if $\mu_n(A)$ converges to $\mu(A)$ uniformly in $A$ as $n$ approaches infinity.

The proof of "$\Longrightarrow$" is easy. But how to prove the reversed assertion?
By definition, 
$$
|\mu-\mu_n|(X)=\sup\lbrace\sum_{i=1}^{k}\left|(\mu-\mu_n)(A_i)\right|\{A_i\}_{i=1}^k\;\text{is a partition of}\;X\rbrace.
$$
The difficulty is that $k$ varies as soon as $n$ does.
 A: Hint: 


Prop If $\mu$ is a complex measure on $X$ then there exists $E\subset X$ with  $|\mu(E)|\ge ||\mu||/\pi$.


Getting $|\mu(E)|\ge||\mu||/c$ for $c>\pi$ is trivial from Lemma 6.3 in Rudin, which says


Lemma If $z_1,\dots z_n\in\Bbb C$ there exists $S\subset\{1,\dots,n\}$ with $\left|\sum_{j\in S}z_j\right|\ge \frac 1\pi\sum_{j=1}^n|z_j|$.


It's not so  clear how to  get the Prop as stated from  the  lemma. But one can simply adapt the proof of the lemma:
Proof of the Prop, cribbed  from 6.3 in Rudin:
There  exist a finite positive measure $\nu$ and a real-valued function $\phi$ such that $$d\mu=e^{i\phi}d\nu.$$For $\theta\in\Bbb R$ let $$E(\theta)=\{x\in  X:\cos(\phi(x)-\theta)\ge0\}$$and define $$F(\theta)=|\mu(E(\theta))|.$$Note that $$F(\theta)=\left|\int_{E(\theta)}e^{-i\theta}d\mu\right|\ge\Re \int_{E(\theta)}e^{i(\phi-\theta)}d\nu=\int_X\cos^+(\phi(x)-\theta)\,d\nu(x).$$
Hence $F$ is continuous, so it has a maximum $F(\theta_0)$. And Fubini shows that $$F(\theta_0)\ge\frac1{2\pi}\int_0^{2\pi}F(\theta)\,d\theta
=\frac1{2\pi}\int_X\int_0^{2\pi}\cos^+(\phi(x)-\theta)\,d\theta d\nu(x)=\frac1\pi\int_Xd\nu=\frac1\pi||\mu||.$$
QED.
And $1/\pi$ is  best possible:
Define a complex Borel measure $\mu$ on $[-\pi/2,\pi/2)$ by $$\mu(E)=\int_Ee^{it}\,dt,$$so $||\mu||=2\pi$. It's not hard to show that $|\mu(E)|\le 2$ for every $E$.
A: Lemma: Let $(X,\mathcal{M})$ be a measurable space. Let $\mu$
be a complex measure on $(X,\mathcal{M})$, then there exists $E\subseteq X$
such that $||\mu||\leq8|\mu(E)|$.
Proof: If $||\mu||=0$, we are done. Suppose that $||\mu||>0$. For
$\varepsilon=\frac{1}{2}||\mu||>0$, there exists a partition $X=\cup_{j=1}^{n}E_{j}$,
where $E_{j}\in\mathcal{M}$ such that $||\mu||-\varepsilon<\sum_{j=1}^{n}|\mu(E_{j})|$.
For each $j$, write $\mu(E_{j})=\alpha_{j}+i\beta_{j}$, where $\alpha_{j},\beta_{j}\in\mathbb{R}$.
Define subsets $I_{1},I_{2},I_{3},I_{4}$ of $\{1,\ldots,n\}$ by
$I_{1}=\{j\mid\alpha_{j}\geq0\}$, $I_{2}=\{j\mid\alpha_{j}<0\}$,
$I_{3}=\{j\mid\beta_{j}\geq0\}$, and $I_{4}=\{j\mid\beta_{j}<0\}$.
Note that $I_{1}\cap I_{2}=I_{3}\cap I_{4}=\emptyset$. Further write
$\alpha_{j}=\alpha_{j}^{+}-\alpha_{j}^{-}$, $\beta_{j}=\beta_{j}^{+}-\beta_{j}^{-}$,
where for any $x\in\mathbb{R}$, $x^{+}=\max(x,0)$ and $x^{-}=\max(-x,0)$.
Let $s_{1}=\sum_{j=1}^{n}\alpha_{j}^{+}$, $s_{2}=\sum_{j=1}^{n}\alpha_{j}^{-}$,
$s_{3}=\sum_{j=1}^{n}\beta_{j}^{+}$, and $s_{4}=\sum_{j=1}^{n}\beta_{j}^{-}$.
Observe that 
\begin{eqnarray*}
 &  & \sum_{j=1}^{n}|\mu(E_{j})|\\
 & = & \sum_{j=1}^{n}|(\alpha_{j}^{+}-\alpha_{j}^{-})+i(\beta_{j}^{+}-\beta_{j}^{-})|\\
 & \leq & \sum_{j=1}^{n}\alpha_{j}^{+}+\alpha_{j}^{-}+\beta_{j}^{+}+\beta_{j}^{-}\\
 & = & s_{1}+s_{2}+s_{3}+s_{4}\\
 & \leq & 4\max(s_{1},s_{2},s_{3},s_{4}).
\end{eqnarray*}
Consider the case that $s_{1}=\max(s_{1},s_{2},s_{3},s_{4})$. Define
$E=\cup_{j\in I_{1}}E_{j}$, then 
\begin{eqnarray*}
|\mu(E)| & = & |\sum_{j\in I_{1}}\mu(E_{j})|\\
 & = & |s_{1}+i\gamma_{1}|\\
 & \geq & s_{1}\\
 & = & \max(s_{1},s_{2},s_{3},s_{4}),
\end{eqnarray*}
for some $\gamma_{1}\in\mathbb{R}$ (which is not important).
If $s_{2}=\max(s_{1},s_{2},s_{3},s_{4})$, we define $E=\cup_{j\in I_{2}}E_{j}$,
then $|\mu(E)|=|-s_{2}+i\gamma_{2}|\geq s_{2}=\max(s_{1},s_{2},s_{3},s_{4})$.
If $s_{3}=\max(s_{1},s_{2},s_{3},s_{4})$, we define $E=\cup_{j\in I_{3}}E_{j}$,
then $|\mu(E)|=|\gamma_{3}+is_{3}|\geq s_{3}=\max(s_{1},s_{2},s_{3},s_{4})$.
If $s_{4}=\max(s_{1},s_{2},s_{3},s_{4})$, we define $E=\cup_{j\in I_{4}}E_{j}$,
then $|\mu(E)|=|\gamma_{4}-is_{4}|\geq s_{4}=\max(s_{1},s_{2},s_{3},s_{4}).$
Hence, we obtain 
\begin{eqnarray*}
 &  & \frac{1}{2}||\mu||\\
 & = & ||\mu||-\varepsilon\\
 & < & \sum_{j=1}^{n}|\mu(E_{j})|\\
 & \leq & 4\max(s_{1},s_{2},s_{3},s_{4})\\
 & \leq & 4|\mu(E)|.
\end{eqnarray*}
That is, $||\mu||\leq8|\mu(E)|$.
/////////////////////////////////////////////////////////////////////
Now, we go back to the original question. Suppose that $\mu_{n}$
and $\mu$ are complex measures. For each $n$, define $\nu_{n}=\mu_{n}-\mu$.
Then $\nu_{n}(A)\rightarrow0$ uniformly on $A\in\mathcal{M}$, in
the sense that: For any $\varepsilon>0$, there exists $N$ such that
$|\nu_{n}(A)|<\varepsilon$ whenever $n\geq N$ and $A\in\mathcal{M}$.
Let $\varepsilon>0$ be given. Choose $N$ such that $|\nu_{n}(A)|<\varepsilon/8$
whenever $n\geq N$ and $A\in\mathcal{M}$. For each $n$, choose
$E_{n}\in\mathcal{M}$ such that $||\nu_{n}||\leq8|\nu_{n}(E_{n})|$.
Now, for any $n\geq N$, we have 
\begin{eqnarray*}
 &  & ||\nu_{n}||\\
 & \leq & 8|\nu_{n}(E_{n})|\\
 & < & 8\cdot\frac{\varepsilon}{8}\\
 & = & \varepsilon.
\end{eqnarray*}
Therefore $||\nu_{n}||\rightarrow0$.
