For any $x \in X$, define the set $\mathcal{F}(X) = \overline{span \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ $Lip_0(X)$.
The set Lip$_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on Lip$_0(X)$.
Define $\| \mu \|_{\mathcal{F}} = \sup_{\| f \|_{Lip}\leq1}{\int f d\mu}$.
If $X$ and $Y$ are Banach spaces and $i : X \rightarrow Y$ is the canonical embedding, then $\hat{i} : \mathcal{F}(X) \rightarrow > \mathcal{F}(Y)$ is an isometric embedding.
Proof: Suppose $\mu$ is a finitely supported measure on $X$. Choose $f \in Lip_0(X)$ with $\int f d\mu = \| u \| _{\mathcal{F}(X)}$, and $\| f \|_{Lip}=1$. Then $f$ has an extension to $g \in Lip_0(Y)$ with $\| g \|_{Lip}=1$. It follows that $\| \hat{i} \mu \|_{\mathcal{F}(Y)} = \| \mu \|_{\mathcal{F}(X)}$
The statement and proof is Lemma $2.3$.
Question:
$(1)$ Why there exists $f \in Lip_0(X)$ such that $\int f d\mu = \| u \| _{\mathcal{F}(X)}$, and $\| f \|_{Lip}=1$?
$(2)$ Why $f$ has an extension? Is it because $X$ is isometrically embedded into $Y$?