"this class of problems lie"s in **RE**, so its name is "**RE**".

\begin{align*}\operatorname{Prob}(M \text{ accepts}) &= \operatorname{Prob}\big((\exists n)(M \text{ accepts after exactly n steps})\big)\\ &=\sum_n \operatorname{Prob}(M \text{ accepts after exactly n steps})\\ &=\lim_N \sum_{n\leq N} \operatorname{Prob}(M \text{ accepts after exactly n steps})\\ &=\lim_N \;\operatorname{Prob}(M \text{ accepts in at most N steps}) \\ &= \lim_n \; \operatorname{Prob}(M \text{ accepts in at most n steps})\,. \end{align*}

For all $m$ and $n$ with $m\leq n$, and for all randomness strings $r$,
$M$ accepts in at most $m$ steps if and only if it accepts in exactly some $t\leq m$ steps. But then $t\leq n$ so this occurs whenever $M$ accepts in at most $n$ steps with randomness string $r$.

For all $m$ and $n$, with $m\leq n$, the probability that $M$ accepts in at most $m$ steps does not exceed the probability that it accepts in at most $n$ steps.

\begin{align*} &\tfrac12 < \operatorname{Prob}(M \text{ accepts}) \\ &\iff \tfrac12 < \lim_n \; \operatorname{Prob}(M \text{ accepts in at most n steps})\\ &\iff (\exists n)\left(\tfrac12 < \operatorname{Prob}(M \text{ accepts in at most n steps})\right)\,. \end{align*}

Therefore, a machine that loops over the positive integers $n$ and accepts
if and only if $\tfrac12 < \operatorname{Prob}(M \text{ accepts in at most$n$steps})$ will accept
exactly the inputs that $M$ has a probability greater than $\tfrac12$ of accepting.