If $u \in \mathscr{L}^1(\lambda^n), v\in \mathscr{L}^\infty (\lambda^n)$, then $u \star v$ is bounded and continuous.
Below is the proof I found. As you can see below, we use the boundedness of $v$ by $\Vert v\Vert_\infty$. However, this is only essential boundedness, and thus may not be bounded for a set of measure zero. Hence, don't we need the terms almost surely in front of bounded and continuous in the above theorem?
Proof. For all $x,x' \in \mathbb{R}^n$ we see that $$|u\star v(x)-u\star v(x')| = \int |v(y)u(x-y)-v(y)u(x'-y)| dy \le \Vert v\Vert_\infty \Vert u(x-\cdot) - u(x'-\cdot)\Vert_1 = \Vert v\Vert_\infty \Vert u(x-x' + \cdot)-u\Vert_1 $$ and continuity follows from the continuity of the map $$x \mapsto \int |u(x_y)-u(y)|dy.$$ The boundedness follows exactly with only considering $u \star v(x)$.
The point is that, \begin{align*} |v(y)u(x-y)-v(y)u(x'-y|\leq\|v\|_{\infty}|u(x-y)-u(x'-y)|~~~~\text{a.e.}, \end{align*} then taking integrals both sides, note that a.e. difference in doing integration does not matter.