Prove that if $\lim_{n}\mu(A)=\mu(A)$ uniformly for all measurable set, then $\lim_{n}||\mu_n-\mu||=0$.

Question

$\newcommand{\scra}{\mathscr{A}}$ $\newcommand{\limm}[2]{\underset{#1\to#2}{\lim}}$ $\newcommand{\pl}{\parallel}$ $\newcommand{\iy}{\infty}$

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.

Answer 1

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 $E$.

Answer 2

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, $||\mu||\leq8|\mu(E)|$.

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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 $||\nu_{n}||\rightarrow0$.