Deform a complex structure fixing marked points Let $\Sigma$ be a closed orientable surface of genus $g$ with $m$ marked points $x=\{x_1, \ldots, x_m\}$ and $j_0$ denote a complex structure on $\Sigma$. Take a neighborhood $U$ of the isomorphism class of $(\Sigma, x, j_0)$ in the moduli space $\mathcal{M}_{g,m}$ of genus $g$ Riemann surfaces with $m$ marked points.
My question is the following:
Can one take a compact subset $K$ of $\Sigma$ such that $K$ does not contain any marked point, and a representative of each element of a neighbourhood $U$ of $j_0$ can be obtained from $j_0$ by deforming it only on $K$?
Any comment will be helpful. Thank you in advance.
 A: This is true if $U$ satisfies an extra boundedness condition: There exists $\delta>0$ such that for all $(\Sigma',x', j_0') \in U$, the distance between any two marked points $x_i',x_j'$ is at least $\delta$.
This boundedness condition follows from the mild boundedness condition that the closure of $U$ is compact. (If so, then the closure of $U$ remains closed in  $\overline{\mathcal M}_{g,n}$, hence does not intersect the boundary, but if the distance between two points goes to zero then in the limit those points collide and bubble off, forming a Riemann surface on the boundary.)
To check this, take $\epsilon < \delta /2$ and define $K$ to be $\Sigma$ minus the open ball of radius epsilon around each marked point $x_i$. Then $\Sigma \setminus K$ is a union of $n$ balls of radius $\epsilon$, with a marked point in the center.
Every other surface $\Sigma'$ in $U$ has the same form: We can write it as the union of $(\Sigma \setminus K)$ and a varying compact surface $K'$, that being $\Sigma'$ minus an open ball of radius $\epsilon$ around each part. Thus, it is obtained from $\Sigma$ by deforming $K$ and leaving $|Sigma \setminus K$ fixed.
