[proofplan]
Fix an element $\xi\in\mathfrak g$ and compute the derivative of the function $\mu^\xi$ along an arbitrary trajectory of $X_H$. The moment map identity rewrites this derivative as the symplectic pairing $\theta(\xi_M,X_H)$, and skew-symmetry of $\theta$ converts it to $-dH(\xi_M)$ using the Hamiltonian vector field convention. Finally, $G$-invariance of $H$ implies that $H$ has zero derivative in every infinitesimal action direction, so the derivative of $\mu^\xi$ along the Hamiltonian flow vanishes wherever the flow is defined.
[/proofplan]
[step:Compute the derivative of a moment map component along a Hamiltonian trajectory]
Fix $\xi\in\mathfrak g$ and $p\in M$. Define the open interval
\begin{align*}
I_p=\{t\in\mathbb R:(t,p)\in D\}
\end{align*}
and define the trajectory $\gamma_p:I_p\to M$ by
\begin{align*}
\gamma_p(t)=\varphi_t(p).
\end{align*}
By the defining property of the flow of $X_H$,
\begin{align*}
\frac{d}{dt}\gamma_p(t)=X_H(\gamma_p(t))
\end{align*}
for every $t\in I_p$. Since $\mu^\xi:M\to\mathbb R$ is smooth, the chain rule gives
\begin{align*}
\frac{d}{dt}\mu^\xi(\gamma_p(t))=d\mu^\xi_{\gamma_p(t)}\bigl(X_H(\gamma_p(t))\bigr).
\end{align*}
[/step]
[step:Use the moment map identity and the Hamiltonian convention to identify the derivative]
Let $x\in M$ be arbitrary. The moment map convention gives
\begin{align*}
d\mu^\xi_x\bigl(X_H(x)\bigr)=\theta_x\bigl(\xi_M(x),X_H(x)\bigr).
\end{align*}
Since $\theta_x$ is skew-symmetric,
\begin{align*}
\theta_x\bigl(\xi_M(x),X_H(x)\bigr)=-\theta_x\bigl(X_H(x),\xi_M(x)\bigr).
\end{align*}
The Hamiltonian vector field convention $dH=\iota_{X_H}\theta$ gives
\begin{align*}
\theta_x\bigl(X_H(x),\xi_M(x)\bigr)=dH_x\bigl(\xi_M(x)\bigr).
\end{align*}
Therefore
\begin{align*}
d\mu^\xi_x\bigl(X_H(x)\bigr)=-dH_x\bigl(\xi_M(x)\bigr).
\end{align*}
[guided]
We now translate the derivative of $\mu^\xi$ into an expression involving the derivative of $H$. Fix an arbitrary point $x\in M$. The moment map convention in the statement says that the one-form $d\mu^\xi$ is the contraction of $\theta$ with the fundamental vector field $\xi_M$. Concretely, this means that for every vector $v\in T_xM$,
\begin{align*}
d\mu^\xi_x(v)=\theta_x\bigl(\xi_M(x),v\bigr).
\end{align*}
Apply this equality with $v=X_H(x)$ to obtain
\begin{align*}
d\mu^\xi_x\bigl(X_H(x)\bigr)=\theta_x\bigl(\xi_M(x),X_H(x)\bigr).
\end{align*}
The next move is only a sign change, but it is exactly where the convention matters. Since $\theta_x$ is an alternating [bilinear form](/page/Bilinear%20Form) on $T_xM$, it is skew-symmetric, so
\begin{align*}
\theta_x\bigl(\xi_M(x),X_H(x)\bigr)=-\theta_x\bigl(X_H(x),\xi_M(x)\bigr).
\end{align*}
The Hamiltonian vector field $X_H$ is defined in the statement by $dH=\iota_{X_H}\theta$. This means that for every $v\in T_xM$,
\begin{align*}
dH_x(v)=\theta_x\bigl(X_H(x),v\bigr).
\end{align*}
Using this with $v=\xi_M(x)$ gives
\begin{align*}
\theta_x\bigl(X_H(x),\xi_M(x)\bigr)=dH_x\bigl(\xi_M(x)\bigr).
\end{align*}
Combining these three equalities yields
\begin{align*}
d\mu^\xi_x\bigl(X_H(x)\bigr)=-dH_x\bigl(\xi_M(x)\bigr).
\end{align*}
This identity is the algebraic heart of the theorem: conservation of $\mu^\xi$ will follow once we know that $H$ has zero derivative in the infinitesimal symmetry direction $\xi_M$.
[/guided]
[/step]
[step:Differentiate the invariance of $H$ in the infinitesimal action direction]
Let $x\in M$ be arbitrary. Define the smooth curve $c_x:\mathbb R\to M$ by
\begin{align*}
c_x(s)=\exp(s\xi)\cdot x.
\end{align*}
By definition of the fundamental vector field,
\begin{align*}
c_x'(0)=\xi_M(x).
\end{align*}
Since $H$ is $G$-invariant, the function $H\circ c_x:\mathbb R\to\mathbb R$ is constant with value $H(x)$. Differentiating at $s=0$ gives
\begin{align*}
0=\left.\frac{d}{ds}\right|_{s=0}H(c_x(s))=dH_x\bigl(c_x'(0)\bigr)=dH_x\bigl(\xi_M(x)\bigr).
\end{align*}
Thus
\begin{align*}
dH_x\bigl(\xi_M(x)\bigr)=0
\end{align*}
for every $x\in M$.
[/step]
[step:Conclude that the moment map component is constant along the flow]
For every $t\in I_p$, apply the identity from the preceding steps at the point $x=\gamma_p(t)$. We obtain
\begin{align*}
\frac{d}{dt}\mu^\xi(\varphi_t(p))=d\mu^\xi_{\gamma_p(t)}\bigl(X_H(\gamma_p(t))\bigr)=-dH_{\gamma_p(t)}\bigl(\xi_M(\gamma_p(t))\bigr)=0.
\end{align*}
Since $\xi\in\mathfrak g$, $p\in M$, and $t\in I_p$ were arbitrary, every component $\mu^\xi$ is conserved along every trajectory of the Hamiltonian vector field $X_H$ on its flow domain.
[/step]
