Martin-Löf randomness implies computable randomness

A successful martingale has $d(\epsilon )>0$, since a nonnegative martingale starting at $0$ is identically $0$. Enumerate the pairs $(n,\sigma )$ satisfying $d(\sigma )>2^{n}d(\epsilon )$ and let $U_n$ be the union of the cylinders of strings enumerated with first coordinate $n$. Strict comparison of computable reals is semidecidable uniformly in $(n,\sigma )$, so one algorithm enumerates all these pairs and $(U_n)$ is uniformly effectively open.

The prefix-minimal threshold-crossing strings have the same union as $U_n$ and are prefix-free. For any finite set $S$ of them, the martingale conservation equation Computable martingales on binary strings gives ${\sum }_{\sigma \in S}{2}^{-\mathrm{\mid }\sigma \mathrm{\mid }}d(\sigma )\le d(\epsilon )$: extend the strings to one common length, where their descendant cylinders are disjoint. Since every $d(\sigma )>2^{n}d(\epsilon )$, it follows that ${\sum }_{\sigma \in S}{2}^{-\mathrm{\mid }\sigma \mathrm{\mid }}\le 2^{-n}$. Taking increasing finite subsets gives $\mu (U_n)\le 2^{-n}$, so $(U_n)$ is a Martin-Löf test.

Success puts the sequence in every level of that test, contradicting Martin-Löf randomness as defined in Martin-Löf tests and random sequences.