= Solution
Let $(f_n)$ be a <singular Weyl sequence> for $L$ at $\lambda\in\Sigma_{\mathrm e}(L)$. Since $K$ is compact and $f_n\rightharpoonup0$, the preceding result gives $Kf_n\to0$. Therefore
$$
(L+K-\lambda I)f_n=(L-\lambda I)f_n+Kf_n\to0.
$$
The norms remain one and the weak limit remains zero. The <singular Weyl sequence> criterion yields $\lambda\in\Sigma_{\mathrm e}(L+K)$. Apply the same argument to $L+K$ and the compact <self-adjoint operator> $-K$ for the reverse inclusion. Both operators are bounded and self-adjoint. Thus the <Weyl theorem for compact self-adjoint perturbations> is
$$
\boxed{\Sigma_{\mathrm e}(L+K)=\Sigma_{\mathrm e}(L).}
$$
The corrected shifted-range definition is necessary. For the diagonal example in the preceding solutions, a rank-one perturbation changing the entry $2$ to $0$ removes $2$ from the spectrum. The unshifted printed definition had classified $2$ as essential merely because the original range was not closed, so it would make this invariance false.
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