Showing a subset of tempered distributions equals the space of complex Radon measures

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The following question is part a) of exercise 29 in the Distributions chapter of Folland's Real Analysis.

Let $S'$ denote the space of tempered distributions on $\mathbb{R}^n$. That is, $S'$ is the space of distributions that act on the space of Schwartz functions $S$.

For $1 \leq p < \infty$, let $C_p$ be the set of all $F \in S'$ for which there exists $C \geq 0$ such that $||F \ast \phi||_p \leq C||\phi||_p$ for all $\phi \in S$, so that the map $\phi \mapsto F \ast \phi $ extends to a bounded operator on $L^p$

a) Show that $C_1 = M(\mathbb{R}^n)$ where $M(\mathbb{R}^n)$ is the space of complex Radon measures on $\mathbb{R}^n$.

Folland provides a hint which reads: "If $F \in C_1$, consider $F \ast \phi_t$ where $\{\phi_t\}$ is an approximate identity, and apply Alaoglu's theorem."

So, I am actually pretty stuck on where to even begin. It seems as though the Riesz representation theorem should be used, but I am not completely sure.

Appealing to the hint, I have noticed that if we let $F_t = F \ast \phi_t$, then we have that $F_t \longrightarrow F$ in $S'$:

$$ \langle F_t, \psi \rangle = \langle F \ast \phi_t, \psi \rangle = \langle F, \widetilde{\phi_t} \ast \psi \rangle \longrightarrow \langle F, \psi \rangle \text{ as } t \longrightarrow 0 $$

Where the convergence follows by properties of approximations to the identity.

If anyone could provide a step in the right direction, that'd be very helpful. I don't see how Alaoglu's theorem applies.

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In fact $$||F*\phi_t||_{C_0(\Bbb R^n)^*} =||F*\phi_t||_1\le C||\phi_t||_1=C;$$apply Alaoglu in $C_0^*$.