In the chapter 11 of PRML, equation 11.73. The equation is trying to use samples obtained from Markov chain to define the important-sampling distribution.
$$ P_G(z) = \frac{1}{Z_G}exp(-G(z)) = \frac{1}{L}\sum^L_{l=1}T(z^{(l)},z) $$ where $T(z,z')$ is the transition markov chain probability, and the sample set is given by $z^{(1)},...,z^{(L)},$
According to Markov chain, a distribution is said to be invariant if $$ p(x) = \sum_{x'} T(x',x)p(x') $$
But I don't understand why sum of transition probability could use for estimation of proposed probability. Is there any proof or relevant theorem?
In the invariant expression, $$p(x) = \sum_{x'} T(x', x)p(x')$$ we can think that $p(x)$ is the average value of $T(x', x)$ according to the probability $p(x')$. That is, $$p(x) = \sum_{x'} T(x', x)p(x') \approx \frac{1}{L}\sum_{l=1}^{L}T(x'^{(l)},x) \,.$$ So, we can get the desired result by $p(x) = P_G(x)$.