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!

Proof:

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 $\mu\in M$.