= Solution
Choose $0<\varepsilon<R-S(\pi)$. The preceding construction and the <typical subspace theorem> give, for all sufficiently large $n$,
$$
\dim\mathcal K_n\leq2^{n(S(\pi)+\varepsilon)}+1\leq2^{\lceil nR\rceil}.
$$
Embed this code into $m_n=\lceil nR\rceil$ qubits. The unused subspace can be decoded to a fixed source state, making the extended decoder trace preserving; encoded states never enter it. Hence its rate is $m_n/n\to R$.
For every $\delta>0$, sufficiently large $n$ has $\operatorname{Tr}(\pi^{\otimes n}P_\varepsilon^{(n)})\geq1-\delta$. The proved fidelity bound yields
$$
1\geq F_n\geq1-2\delta.
$$
Since $\delta$ is arbitrary, $F_n\to1$. Therefore
$$
\boxed{R>S(\pi)\ \Longrightarrow\ \text{a reliable quantum source code of rate }R\text{ exists}.}
$$
This establishes the achievability part of <Schumacher compression>, including the failure flag and without presuming the source ensemble is orthogonal. If $R$ exceeds one, the same rate-budget construction simply uses spare code qubits; no claim of reducing the physical register size is then needed.
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