Let $M(\mathbb R)$ the space of complex Borel measures with the norm $\Vert \mu \Vert = \vert \mu \vert(\mathbb R)$, where $\vert \mu \vert$ is the total variation of $\mu$. Now consider for $\alpha \geq 0$ the norm $$ \Vert \mu \Vert_{\alpha} := \int_{\mathbb R} e^{\alpha \vert s \vert} \, \vert \mu \vert (ds)$$ on the space $M_\alpha(\mathbb R):= \{\mu \in M(\mathbb R): \Vert \mu \Vert_{\alpha} < \infty\}$. I want to show that the set of measures with compact support $M_c(\mathbb R)$ are dense in $M_\alpha(\mathbb R)$ (which is a Banach algebra).
I have no concrete idea to how to get this. I know that $M_c(\mathbb R)$ is dense in $M(\mathbb R)$ and that $\Vert \mu \Vert \leq \Vert \mu \Vert_{\alpha}$ for all $\mu \in M_\alpha(\mathbb R)$. But I don't see how to get the result from this.
Given $\mu\in M_\alpha$ you can construct a sequence of measures with compact support converging to $\mu$ in sort of the most obvious way imaginable: Let $d\mu_n=\chi_{[-n,n]}d\mu$. In other words, define $$\mu_n(E)=\mu(E\cap[-n,n]).$$