Cotype property of measures 
Let $K$ be a compact Hausdorff space and $M(K)=C(K)^{*}$ the set of all bounded complex Radon measures on $K.$ Is it true that $M(K)$ is of cotype 2?

I think the answer is true, and to prove this it is enough to check that $$\Bigg(\sum\|\mu_k\|^2\Bigg)^{\frac{1}{2}}\leq C\Bigg(\int\|\sum_k\epsilon_k(\omega)\mu_k\|^2\Bigg)^{\frac{1}{2}},$$for any measure $(\mu_k),$ which are Dirac mass.
$(\epsilon_k)$ is the sequence of Rademacher functions.
 A: I wanted to comment that I am concerned by @Mathbuff 's conclusion made in the comment, and I hope that this suffices for a complete answer.
We do not have that $$ \text{cl}\big(\text{span}\{ \delta_x : x \in K \}\big) = \mathscr{M}(K),$$ I believe. This can be seen by recognizing that the Lebesgue measure lies distance 2 from any finite combination of point measures in $\mathscr{M}([0,1])$. So I'm uncertain about their argument that such a map extends to an isometric isomorphism, and I found the conclusion particularly concerning. One point against it is that $\mathscr{M}(K)$ is not separable, as any two point masses lie distance 2 apart whilst $l_1(K)$ is separable for $K = [0,1].$
I think that we can show that $\mathscr{M}(K)$ has cotype 2 by recognising that any separable subspace can be isometrically embedded into a $L_1(\mu)$ space. Then from here, we can recall that a $L_1(\mu)$ space has cotype 2 and then as each separable space has cotype 2 with a UNIFORM bound, we have that $\mathscr{M}(K)$ has cotype 2.
We can embed any separable subspace $X$ into an $L_1(\mu)$ space, by the fact that if $(\mu_i)_{i=1}$ is a dense sequence in $X$ of norm-one measures, then any measure $\nu$ is absolutely continuous with respect to $\mu = \sum_{i=1}^\infty 2^{-i} \mu_i$, a finite measure.
This is as:
If $\mu(A)= 0$, then $\mu_i(A) = 0  \quad  \forall i \in \mathbb{N}.$
Then given $\varepsilon > 0$, there exists $(\alpha_i)_{i=1}^\infty$ a finite sequence with $||\nu - \sum_{i=1}^\infty \alpha_i \mu_i || < \varepsilon$ and so $|\nu(A)| < \varepsilon$.
Hence $\nu(A) = 0 \implies \nu << \mu.$
Thus by the Radon-Nikodym Theorem, we have that X embeds into $L_1(\mu)$. Thus the cotype of $\mathscr{M}(K)$ is 2 as required.
