Capacity of a disjoint sum of channels ($C={\log }_2{\sum }_i{2}^{C_i}$)

Combine finitely many discrete memoryless channels with disjoint input alphabets and disjoint output alphabets into a block-diagonal channel. The block label $B$ is determined by both input and output. If it has law $\alpha$, the mutual information is at most $H(\alpha )+{\sum }_i{\alpha }_i{C}_i$, attained by using each block's capacity-achieving conditional input law. Put $S={\sum }_i{2}^{C_i}$ and $r_i=2^{C_i}/S$. The expression equals ${\log }_{2}S-d(\alpha ,r)$, so Gibbs inequality proves the displayed capacity, attained for block probabilities ${\alpha }_i=r_i$. Even blocks of capacity zero contribute distinguishable block choices; three disjoint zero-capacity blocks give capacity ${\log }_{2}3$.