= Solution
The <stress-energy tensor> is already first order, so its energy flux can be evaluated on the unperturbed <Schwarzschild event horizon> $r=2M_0$, with the infinity-normalized <Killing vector field> $\xi=\partial_v$. On this background $g^{rv}=1$, and hence
$$
T^r{}_b\xi^b=T^r{}_v=T_{vv}=\frac{\dot\mu}{4\pi(2M_0)^2}.
$$
The conserved <stress-energy current from a Killing vector> is $J^a=-T^a{}_b\xi^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^2\,dv\,d\Omega$ gives
$$
\Delta E=\int_{-\infty}^{\infty}\int_{S^2}T^r{}_v(2M_0)^2\,d\Omega\,dv
=\int_{-\infty}^{\infty}\dot\mu(v)\,dv
=\boxed{\Delta M}.
$$
The initial and final <event horizon> areas are $A_0=16\pi M_0^2$ and $A_1=16\pi(M_0+\Delta M)^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M)^2)$. With <Schwarzschild surface gravity> $\kappa_0=1/(4M_0)$, the <Physical-process first law of black-hole mechanics> follows:
$$
\boxed{\Delta E=\Delta M=\frac{\kappa_0}{8\pi}\Delta A+O((\Delta M)^2/M_0).}
$$
The flux is <Killing energy>, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the <event horizon>.
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