= Solution
Diagonalize the source average $\pi=\sum_j\lambda_j|j\rangle\langle j|$, using its positive eigenvalues. For fixed $\delta>0$, let $\Pi_{n,\delta}$ project onto product eigenvectors whose eigenvalues obey
$$
2^{-n(S(\pi)+\delta)}\leq\lambda_{j_1}\cdots\lambda_{j_n}
\leq2^{-n(S(\pi)-\delta)}.
$$
The <typical subspace theorem> states that, for every $\epsilon>0$ and all sufficiently large $n$,
$$
\operatorname{Tr}(\pi^{\otimes n}\Pi_{n,\delta})\geq1-\epsilon,
$$
$$
2^{-n(S(\pi)+\delta)}\Pi_{n,\delta}
\leq\Pi_{n,\delta}\pi^{\otimes n}\Pi_{n,\delta}
\leq2^{-n(S(\pi)-\delta)}\Pi_{n,\delta},
$$
$$
(1-\epsilon)2^{n(S(\pi)-\delta)}
\leq\dim\mathcal T_{n,\delta}\leq2^{n(S(\pi)+\delta)}.
$$
The probability statement is the <weak law of large numbers> applied to $-\log_2\lambda_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the <quantum typical subspace> retains almost all probability using about $nS(\pi)$ qubits.
For $R>S(\pi)$ choose $0<\delta<R-S(\pi)$. Measure $\{\Pi_{n,\delta},I-\Pi_{n,\delta}\}$. On success, encode the projected state isometrically into a space of dimension $\dim\mathcal T_{n,\delta}$; on failure, output a separate fixed flag. Decode the successful sector by the inverse <isometry>, and map the flag to a fixed state $|\varphi_0\rangle$. The compressed dimension is at most $2^{n(S(\pi)+\delta)}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving <quantum channels>, not merely successful postselected operations.
Writing $\Pi=\Pi_{n,\delta}$, the composite channel is
$$
\mathcal N_n(\tau)=\Pi\tau\Pi+
\operatorname{Tr}[(I-\Pi)\tau]|\varphi_0\rangle\langle\varphi_0|.
$$
For a pure source signal $|\Psi_k\rangle$, set $a_k=\langle\Psi_k|\Pi|\Psi_k\rangle$. Its squared <quantum fidelity> after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is $\pi^{\otimes n}$. Convexity of the square gives
$$
\sum_kp_k^{(n)}\langle\Psi_k|\mathcal N_n(|\Psi_k\rangle\langle\Psi_k|)|\Psi_k\rangle
\geq\sum_kp_k^{(n)}a_k^2
\geq\left[\operatorname{Tr}(\pi^{\otimes n}\Pi)\right]^2
\geq(1-\epsilon)^2.
$$
The <Kraus formula for entanglement fidelity> gives the same lower bound for $F_e(\pi^{\otimes n},\mathcal N_n)$, because one <Kraus operator> is $\Pi$ and the other terms are nonnegative. Taking $\epsilon\to0$ proves \b[reliable compression at every $R>S(\pi)$], in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is <typical-subspace compression with a failure flag>.
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