= Solution
The <support condition for importance sampling> is missing from the printed assertion. Assume $M>1$, $E_f|h|<\infty$, and $Mg(z)>f(z)$ for $f$-almost every $z$. For a proposal $Z\sim g$, rejection has conditional probability $1-f(Z)/(Mg(Z))$. Its unconditional probability is $(M-1)/M$, so the <conditional density> of a rejected draw is
$$
r(z)=\frac{g(z)(1-f(z)/(Mg(z)))}{(M-1)/M}=\frac{Mg(z)-f(z)}{M-1}.
$$
Thus the rejected-draw weight is precisely $f(z)/r(z)$, and
$$
E_r\left[h(Z)\frac{(M-1)f(Z)}{Mg(Z)-f(Z)}\right]
=\int h(z)f(z)\,dz.
$$
Conditionally on any positive rejection count $n$ among $N$ independent proposals, the rejected locations have this iid <conditional density>. Hence
$$
\boxed{\frac1n\sum_{i=1}^nh(Z_i)\frac{(M-1)f(Z_i)}{Mg(Z_i)-f(Z_i)}}
$$
is an <unbiased estimator> conditionally on $n$, and therefore also conditionally on the event $0<n<N$. This is <importance sampling> from the <rejected-proposal distribution>.
The nonstrict assumption $f\leq Mg$ alone does not suffice: the rejected distribution may completely miss part of the target support. For example, let $g$ be uniform on $[0,1]$, $f(z)=2\mathbf1_{[0,1/2]}(z)$, $M=2$, and $h=1$. All rejected proposals lie in $(1/2,1]$, where $f=0$, so the displayed estimator is always $0$, whereas $E_fh=1$. This remains a counterexample after conditioning on $0<n<N$. A strict envelope, such as choosing $M$ larger than the essential supremum of $f/g$, repairs the missing support condition. Near-equality can still give high-<variance> weights, even when it remains an <unbiased estimator>.
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