Let $M(G)$ be the Banach space of all bounded complex regular Borel measures on the locally compact abelian group $G$.Associate to each Borel set $E\subset G$ the set
$$ E_2=\{(x,y)\in G\times G:x+y\in E\}\subset G\times G=G^2 $$
if $\mu$ and $\lambda \in M(G)$, define their convolution $\mu *\lambda$ to the set function
$$
(\mu *\lambda)(E)=(\mu \times\lambda)(E_2)
$$
for every Borel set $E\subset G$, $\mu \times\lambda$ is product of measure. Then $\mu *\lambda \in M(G)$. Mainly, I'm trying to prove that the convolution is regular.
The proof of inner regularity is the following. Let $E=\bigcup_{i=1}^{\infty}E_i$ be a Borel set of $G$. The regularity of $\mu \times \lambda$ given $\epsilon>0$ there exist a compact set $K\subset E_2$ such that $$(\mu\times\lambda)(K)>(\mu\times\lambda)(E_2)-\epsilon.$$Let $C$ be the image of $K$ under the map $(x,y)\to x+y$ (since this is continuous) $C$ is a compact subset of $E$, and $K\subset E_2$ hence, $$(\mu*\lambda)(C)=(\mu\times\lambda)(C_2)\ge (\mu\times \lambda)(K)>(\mu*\lambda)(E)-\epsilon.$$I have a problem proving the outer regularity. A hint is given that using complementation will prove the outer regularity. I'm not sure how to go about doing that.
Let $\mu$ be any finite measure which is inner regular. I claim that $\mu$ is also outer regular.
Indeed, let $A \subset G$ be a Borel set. By inner regularity, there is $K \subset A^c$ satisfying $\mu(K) > \mu(A^c) - \varepsilon$. Hence, $U := K^c \supset A$ and $$ \mu(U) = \mu(G) - \mu(K) < \mu(G) - [\mu(A^c) - \varepsilon] = \mu(A) + \varepsilon. $$ Since $\varepsilon > 0$ was arbitrary, we see that $\mu$ is outer regular.
This easily applies to your case. Note that we crucially used that $\mu$ is a finite measure.
Above, I also used that $\mu$ is a positive measure. But any complex regular measure can be written as $\mu = \mu_+ - \mu_- + i(\mu_+^i - \mu_-^i)$ for suitable regular finite positive measures $\mu_+, \mu_-, \mu_+^i, \mu_-^i$ and then by linearity, we get the claim also for regular finite complex measures.