Compactness of Convolutions of Compact Measures This regards measures on $d$-dimensional Euclidean space $\mathbb R^d$ and their associated densities.
A super-level set of a density $f : \mathbb R^d \to \mathbb R^+$ at level $t$ is the set $\{x \in \mathbb R^d \mid f(x) \geq t\}$.
Say a density is super-compact if for every $t>0$ the super-levelset at level $t$ of $f$ is compact.  
Given two super-compact densities $f,g$, is the convolution $f*g$ also super-compact?
More specifically, I am interested in the case where one of the two super-compact densities $g$ is a Gaussian kernel, and the other $f$ may have compact support, that is even for $t=0$ the super-levelset of $f$ is compact.  But I am hoping there is a more general known result about preserving compactness on convolutions.
If this is not true, then what properties does $f$ need so that $f*g$ is super-compact?  
 A: If $f$ and $g$ are both integrable and square-integrable, and $f$ is super-compact, and $f*g$ is continuous, then $f*g$ is super-compact. (This covers your specific case, at least if the compactly supported $f$ is continuous.) Note we don't even need $f$ and $g$ to be nonnegative.
We need to prove that the $t$-level sets are both closed and bounded. First, closed is easy: the $t$-level set is the inverse image of the closed interval $[t,\infty)$ under the map $f*g\colon \mathbb R^d\to\mathbb R$, and hence it is automatically closed (by the topological definition of continuity).
As for bounded: given $\varepsilon>0$, choose $R=R(\varepsilon)$ so large that $|f(x)|<\varepsilon$ for $|x|>R$ (by super-compactness) and that the integral of $|g(x)|^2$ over the complement of the ball of radius $R$ is less than $\varepsilon$ (by square-integrability). Then for $|x|>2R$,
$$
(f*g)(x) = \int_{\mathbb R^d} f(y)g(x-y) \,dy = \int_{|y|\le R} f(y)g(x-y) \,dy + \int_{|y|> R} f(y)g(x-y) \,dy.
$$
In the first integral, use Cauchy-Schwarz:
\begin{align*}
\bigg| \int_{|y|\le R} f(y)g(x-y) \,dy \bigg| &\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|y|\le R} |g(x-y)|^2 \,dy \bigg)^{1/2} \\
&\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|z|> R} |g(z)|^2 \,dy \bigg)^{1/2} < \|f\|_2 \varepsilon^{1/2}.
\end{align*}
(Here we've used the fact that $|x|>2R$ and $|y|\le R$ imply $|z|>R$ in the change of variables $z=x-y$.)
For the second integral,
$$
\bigg| \int_{|y|> R} f(y)g(x-y) \,dy \bigg| < \varepsilon \int_{|y|> R} |g(x-y)| \,dy \le \|g\|_1 \varepsilon.
$$
Thus $|(f*g)(x)|$ is bounded by $\|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$ for all $|x|>2R(\varepsilon)$. In other words, the $t$-level set is bounded for all $t \ge \|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$, since $\varepsilon$ can be taken arbitrarily small, so can $t$.
