The halting oracle is not computably dominated

With an oracle query to $0^{\mathrm{\prime }}$, define $h(e)=0$ when ${\mathrm{\Phi }}_e^{\mathrm{\varnothing }}(e)$ diverges. Otherwise simulate that program, answering every empty-oracle query negatively, and let $h(e)$ be its least halting stage, counted from its initial configuration. The query guarantees that this simulation terminates when it is undertaken. Thus $h$ is total and $0^{\mathrm{\prime }}$-computable.

If a total computable function $g$ satisfies $h(e)\le g(e)$ for every $e$, an ordinary machine decides $0^{\mathrm{\prime }}$ as follows: compute $g(e)$, simulate ${\mathrm{\Phi }}_e^{\mathrm{\varnothing }}(e)$ through that many transitions (also checking the initial configuration), and answer yes exactly if it halts. Every halting computation is detected because its halting stage is at most $g(e)$; every nonhalting computation is answered no. This would make $0^{\mathrm{\prime }}$ computable from the empty oracle, contrary to [F2].

Therefore this particular total $0^{\mathrm{\prime }}$-computable function $h$ has no total computable pointwise upper bound. By [F3], $0^{\mathrm{\prime }}$ is not computably dominated.