"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 can only occur if $M$ 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.