The busy beaver function eventually dominates every total computable function

By [L1], choose a deterministic machine $F$ that, on input $\mathrm{bin}(n)$, halts with output $\mathrm{bin}(f(n))$. For each $n\in \mathbb{N}$, build a new machine $P_n$ that has the binary digits of $n$ hard-wired into a straight-line prefix of its finite control, using one additional state per written digit and then handing control to a fixed simulation of $F$. On blank input, $P_n$ writes the binary numeral for $n$ onto its tape, runs $F$ on that input until $F$ halts with output $\mathrm{bin}(f(n))$, and then runs a fixed binary-counter routine for $f(n)+1$ iterations before halting. Therefore $P_n$ halts on the blank input, and its total running time is strictly greater than $f(n)$.

Write $L_n:=\mathrm{\mid }\mathrm{bin}(n)\mathrm{\mid }$. The construction in step 1.1 uses at most $a+b{L}_n$ states and a fixed tape alphabet, for constants $a,b$ depending only on $F$ and the fixed routines. In the concrete code of [L3], this gives $O(L_n)$ transition triples, and every state numeral in those triples has length $O({\log }_2(L_n+2))$. The self-delimiting tuple encoding therefore gives constants $A,B\in \mathbb{N}$ such that $\mathrm{\mid }<mi\; mathvariant="normal">⌜</mi>P_n<mi\; mathvariant="normal">⌝</mi>\mathrm{\mid }\le A+B{L}_n{\log }_2(L_n+2)$. Since $L_n\le 1+{\log }_2(n+1)$, the right-hand side is $O(\log n\log \log (n+2))$, so there exists $N$ such that $\mathrm{\mid }<mi\; mathvariant="normal">⌜</mi>P_n<mi\; mathvariant="normal">⌝</mi>\mathrm{\mid }\le n$ whenever $n\ge N$. For those $n$, the machine $P_n$ is among the competitors counted by $BB(n)$.

If $n\ge N$, then step 2.1 puts $P_n$ inside the size budget for $BB(n)$, and step 1.1 gives a halting time larger than $f(n)$. By [L2], the maximum $BB(n)$ is therefore at least that halting time, so $BB(n)\ge f(n)$. Thus $BB$ eventually dominates $f$.