\delta_{x_i}$ is $H$-invariant as well.\
Choose some $k$ in $I$, and let $y = h (x_k)$ be some element in $H(x_k)$. Then $$\mu_k = \sigma(x_k) = \sigma(h^{-1}(y)) = h_{\#} \sigma(y) = \sigma(y) = \sum_{i \in I} \mu_i \delta_{x_i}(y) ,$$ hence $x_i = y$ for some $i \in I$. This gives a contradiction, since $I$ is at most countable, meanwhile the orbit of each point in $X$ is uncountable by hypothesis. It follows that $$\mu \llcorner |u_n|^{p^{\ast}} \to \mu \llcorner |u|^{p^{\ast}}$$ in $\mathcal{M}(X)$; but this is equivalent to say that $$\label{M(X)}
\| \phi u_n \|_{L^{p^{\ast}}(X,\mu)} \to \| \phi u \|_{L^{p^{\ast}}(X,\mu)}$$ whenever $\phi \in C_c(X)$.\
Since $X$ is compact, we can use $\phi = 1$ in (\[M(X)\]), to conclude that if $\{u_n \}$ is a bounded sequence in $M^p_{1,H}(X,\mu)$ converging weakly to some $u$ in $M^p_{1,H}(X,\mu)$, then $$\| u_n \|_{L^{p^{\ast}}(X,\mu)} \to \|  u \|_{L^{p^{\ast}}(X,\mu)}$$ for some subsequence. But $L^{p^{\ast}}(X,\mu)$ is uniformly convex, hence $u_n \to u$ in $L^{p^{\ast}}(X,\mu)$.

A useful consequence of Theorem \[main2\] is:

\[corollary\] Using the same hypotheses as in Theorem \[main2\], define the constant $C$ by $$C:= \inf \{\ A > 0\ :\ \| u \|_{L^{p^{\ast}}(X,\mu)} \leq A \| u \|_{M^p_1(X,\mu)}\ \text{whenever}\ u \in  M^p_{1,H}(X,\mu)\      \}.$$ Then there exists 