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

$$\operatorname{Prob}(\hspace{.02 in}M \text{ accepts}) \; = \; \operatorname{Prob}\hspace{-0.04 in}\big(\hspace{-0.02 in}(\exists n)(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})\big)$$
$$=$$

$$\displaystyle\sum_n \operatorname{Prob}(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})$$

$$=$$
$$\displaystyle\lim_N \left(\displaystyle\sum_{n\leq \hspace{.02 in}N} \operatorname{Prob}(\hspace{.02 in}M \text{ accepts after exactly } n \text{ steps})\hspace{-0.04 in}\right)$$

$$=$$

$$\displaystyle\lim_N \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } N \text{ steps}) \;\;\; = \;\;\; \displaystyle\lim_n \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})$$

Since $$\: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps}) \:$$ is non-decreasing in $$n$$,

$$\frac12 \;\; < \;\; \operatorname{Prob}(\hspace{.02 in}M \text{ accepts})$$

$$\iff$$

$$\frac12 \;\; < \;\; \displaystyle\lim_n \: \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})$$

$$\iff$$

$$(\exists n)\left(\frac12 < \operatorname{Prob}(\hspace{.02 in}M \text{ accepts in at most } n \text{ steps})\hspace{-0.02 in}\right)$$

 answered Dec 9 '14 at 23:48 user12859