= Solution
Choose a proposal <probability density function> $q$ positive wherever $|h|f$ is nonzero, up to null sets, and draw <independent and identically distributed random variables> $Y_1,\ldots,Y_n$ from $q$. The ordinary <importance sampling> estimator is
$$
\boxed{\widehat\mu_{\rm IS}=\frac1n\sum_{i=1}^n\frac{h(Y_i)f(Y_i)}{q(Y_i)}.}
$$
The <support condition for importance sampling> and absolute integrability give $\mathbb E_q[h(Y)f(Y)/q(Y)]=\int hf=\mu$, so this is an <unbiased estimator>. Its <variance>, possibly infinite, is
$$
\operatorname{Var}(\widehat\mu_{\rm IS})=\frac1n\left[\int\frac{h(x)^2f(x)^2}{q(x)}\,dx-\mu^2\right].
$$
Let $A=\int|h|f$. If $A>0$, the <Cauchy-Schwarz inequality> yields
$$
A^2=\left(\int\frac{|h|f}{\sqrt q}\sqrt q\right)^2\leq\int\frac{h^2f^2}{q}.
$$
Equality holds for $q$ proportional to $|h|f$. Thus the <minimum-variance importance distribution> and the minimum <variance> are
$$
\boxed{q_*(x)=\frac{|h(x)|f(x)}{A},\qquad \operatorname{Var}(\widehat\mu_{\rm IS})_{\min}=\frac{A^2-\mu^2}{n}.}
$$
For a constant-sign integrand this is zero <variance>. If $A=0$, the integrand vanishes almost everywhere and the zero estimator already has zero <variance>. Requiring the proposal to cover all of the target's support, including where $h=0$, can exclude $q_*$; in that stricter class, $(1-\epsilon)q_*+\epsilon f$ approaches the same infimum as $\epsilon\downarrow0$. This distinguishes coverage of the integral from coverage of every target event.
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