Several domination results have been obtained for maximal outerplanar graphs (mops). The classical domination problem is to minimize the size of a set $S$ of vertices of an $n$-vertex graph $G$ such that $G - N[S]$, the graph obtained by deleting the closed neighborhood of $S$, contains no vertices. In the proof of the Art Gallery Theorem, Chvátal showed that the minimum size, called the domination number of $G$ and denoted by $γ(G)$, is at most $n/3$ if $G$ is a mop. Here we consider a modification by allowing $G - N[S]$ to have a maximum degree of at most $k$. Let $ι_k(G)$ denote the size of a smallest set $S$ for which this is achieved. If $n \le 2k+3$, then trivially $ι_k(G) \leq 1$. Let $G$ be a mop on $n \ge \max\{5,2k+3\}$ vertices, $n_2$ of which are of degree $2$. Upper bounds on $ι_k(G)$ have been obtained for $k = 0$ and $k = 1$, namely $ι_{0}(G) \le \min\{\frac{n}{4},\frac{n+n_2}{5},\frac{n-n_2}{3}\}$ and $ι_1(G) \le \min\{\frac{n}{5},\frac{n+n_2}{6},\frac{n-n_2}{3}\}$. We prove that $ι_{k}(G) \le \min\{\frac{n}{k+4},\frac{n+n_2}{k+5},\frac{n-n_2}{k+2}\}$ for any $k \ge 0$. For the original setting of the Art Gallery Theorem, the argument presented yields that if an art gallery has exactly $n$ corners and at least one of every $k + 2$ consecutive corners must be visible to at least one guard, then the number of guards needed is at most $n/(k+4)$. We also prove that $γ(G) \le \frac{n - n_2}{2}$ unless $n = 2n_2$, $n_2$ is odd, and $γ(G) = \frac{n - n_2 + 1}{2}$. Together with the inequality $γ(G) \le \frac{n+n_2}{4}$, obtained by Campos and Wakabayashi and independently by Tokunaga, this improves Chvátal's bound. The bounds are sharp.