= Solution
Assume independent samples $X_i\sim g$, $\int f|\theta|<\infty$, and $g>0$ wherever $f\theta$ can contribute. The <importance weight> is $w(x)=f(x)/g(x)$, and the ordinary <importance sampling> <estimator> is
$$
\boxed{\widehat\mu_g=\frac1n\sum_{i=1}^n\theta(X_i)\frac{f(X_i)}{g(X_i)}.}
$$
Its <expectation> is $\int \theta(x)f(x)\,dx=\mu$, by cancellation of $g$ in the <expectation> <integral>. The <strong law of large numbers> gives almost-sure convergence to $\mu$. Common <probability support> is sufficient; the precise <support condition for importance sampling> only requires covering the nonzero integrand for this particular <expectation>.
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