 Solution

Choose $0<\epsilon <R-S(\pi )$. The preceding construction and the typical subspace theorem give, for all sufficiently large $n$,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 $\mathrm{Tr}({\pi }^{\otimes n}{P}_{\epsilon }^{(n)})\ge 1-\delta$. The proved fidelity bound yieldsSince $\delta$ is arbitrary, $F_n\to 1$. ThereforeThis 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.