Say that an $x\in 2^{\omega}$ is Solovay random if for all computably enumerable collections of intervals $\{I_n\}$ such that $\sum_n\mu(I_n)<\infty$, then $x\in I_n$ for at most finitely many $n$.
Also say an $x\in 2^{\omega}$ is Martin-Lof random if for all computable collections $\{U_n\}$ of c.e. open sets with $\mu(U_n)\leq2^{-n}$ (these are called Martin-Lof tests) then $x\notin\bigcap_nU_n$ (this is called passing the ML-test). So $x$ is ML-random if it passes all ML-tests.
I'm trying to show that Solovay Randomness is equivalent to ML Randomness. I've thought about each direction but am not able to finish either off.
For one direction, suppose $x$ is Solovay Random and $\{U_n\}$ is a ML-test. We know each $U_n=\bigcup I_{n,m}$, $\{I_{n,m}\}$ is a c.e. collection of intervals, and $x\notin\bigcap I_k$ since $x$ is Solovay random. But that doesn't get us that $x\notin\bigcap U_n$
For the other way suppose $x$ isn't Solovay Random; so there a c.e. collection of intervals $\{I_n\}$ such that $\sum\mu(I_n)<\infty$ but $x\in I_n$ for infinitely many $n$. I want to use these to get a ML-test that is failed by $x$. But I need to be able to computably get a collection of open sets so $x$ is in all of them (so just taking the $I_n$ that include $x$ doesn't work) and I need to computably thin them out to get the measures small enough (the $\mu(I_n)\rightarrow 0$, so that helps, but I need the open sets I end up with to have measure $\leq 2^{-n}$)