 Solution

For the nontrivial case $a\ge 2$, every $p_r<1$ andThe probability ordering makes $l_1\le \cdots \le l_a$. At a dyadic number, choose the terminating binary expansion followed by zeros. Let $c_r$ be the first $l_r$ bits of $t_r$, and letbe its binary cylinder interval.

Suppose $r<s$ and $c_r$ were a prefix of $c_s$. Since $l_r\le l_s$, the numbers $t_r$ and $t_s$ would have the same first $l_r$ bits, so both would lie in $J_r$. In particular $t_s-t_r<2^{-l_r}$. Buta contradiction. A later word cannot be a shorter prefix of an earlier word because the lengths are nondecreasing; equal-length coincidences were also excluded. Thus the cumulative Shannon code is a prefix code.

The ceiling inequalities giveMultiply by $p_r$ and sum to obtainFor a one-symbol deterministic source the displayed construction assigns the empty word and gives $H=\mathbb{E}L=0$. This zero-bit convention is valid when the number of source emissions is supplied separately. It is not an unframed uniquely decodable code: concatenating different numbers of empty words gives the same bit string. If positive-length unframed words are required, the shortest deterministic code has length $1$, so the strict upper bound $\mathbb{E}L<H+1$ requires the nontrivial-source assumption.