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I have a question about the proof below:

Let $\mathcal{C}:=\mathcal{C}(\mathbb{R})$ be the space of continuous functions on $\mathbb{R}$ and $\mathcal{C}_b$ its subspace consisting of bounded elements.

Let $M=M(\mathbb{R})$ be the space of bounded measures on $\mathbb{R}$. Let $M$ be endowed with the topology generated by the sets

$$ U_{f_1, f_2,\dots, f_n, r}(\mu)=\left\{ \nu\in M: \left|\int_{\mathbb{R}}f_id\mu-\int_{\mathbb{R}}f_id\nu\right|<r|\right\} $$

where $\mu\in\mathbf{M}$, $f_i\in\mathcal{C}_b$ and $r>0$.

The dual space of $M$ can be identified by $M^{\ast}=\mathcal{C}_b$. My question is about the following proof presented in Lemma 3.2 : I do not understand the argument indicated below on bold black. If someone knows about this argument, please let me know, thanks a lot!


For each $f\in\mathcal{C}_b$,

\begin{eqnarray} f: M&\to&\mathbb{R} \\ \mu&\mapsto&\int_{\mathbb{R}}fd\mu \end{eqnarray}

determines a unique element in $M^{\ast}$. Let $F\in M^{\ast}$ and define $f(x):=F(\delta_{\{x\}})$ where $\delta$ denotes the dirac measure concentrated on $x$. Thus $f$ is continuous.

Moreover, because of the way in which the topology on $M$ is defined, we can find a finite set $\{f_j\}_{1\le j\le J}\subset\mathcal{C}_b$ such that

$$ |F(\mu)|\le \sum_{j=1}^J\left|\int_{\mathbb{R}}f_jd\mu\right|,~ \forall\mu\in M $$

and from this it is clear that $f$ is bounded. Finally, it is ovbious that $F(\mu)=\int_{\mathbb{R}}fd\mu$ if $\mu$ is a linear combination of dirac measures; and, because such $\mu$'s are dense in $M$, it follows that this equation holds for all $\mu\in M$.

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The functional $F$ is continuous at $\nu:=0$, the null measure. The set $F^{-1}(-1,1)$ is open. Therefore, the exists a positive $r$, an integer $J$ and $g_1,\dots,g_J\in\mathcal C_b$ such that $$V:=\left\{\nu\in\mathcal M:\forall j\in [J],\left|\int_{\mathbb R}g_j\mathrm d\nu\right|\lt r\right\}\subset F^{-1}(-1,1).$$ We define $f_j:=g_j/r$. If $\nu\in\mathcal M$ and $\int_{\mathbb R}f\mathrm d\mu\neq 0$ for some $j\in [J]$, define $$\nu':=\frac{1}{\sum_{j=1}^J\left|\int_{\mathbb R} f_j\mathrm d\nu\right|}\nu.$$ We have $F(\nu)\lt 1$ which gives the wanted result.

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I am so happy. Thanks a lot! – CodeGolf Jan 20 '14 at 10:31
You are welcome! – Davide Giraudo Jan 20 '14 at 10:32

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