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Is $U\subseteq X^{s}(L)$?

Let $X$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristics $0$. Let $G$ be a connected, linearly reductive, affine algebraic group acting regularly on $X$. Let $L$ be $G$-linearised invertible sheaf on $X$. Now, consider the set of semi-stable points $X^{ss}(L)$. Assume that there is a $G$-invariant open set $U\subseteq X^{ss}(L)$ such that the geometric quotient of $U$ exists.

Is it true that $U\subseteq X^{s}(L)$? Or, there is some $G$-invariant open subset $U'\subseteq U$ such that $U'\subseteq X^{s}(L)$?

Note that in the definition of "stable" w.r.t. $L$, Mumford doesn't assume the stabilizer is of dimension $0$. (see page 36 of Mumford's book on GIT)

I'm actually interested in the setting of Quiver representation, related to King's stability conditions.

I feel like the answer would be yes, but having trouble trying to prove it. I'm trying to use the following result from Mumford's book.

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