In some lecture notes I am reading, there is the following;
Consider $X_{1},...,X_{n}$, each with pdf $g$ (the instrumental distribution). Our aim is to estimate $E_{f}[h(X)]$ where $h(X)$ is some measurable function and $f$ a pdf (the target distribution).
Then:
$\large\frac{1}{n}\displaystyle\sum\limits_{i=1}^n\frac{f(X_{i})h(X_{i})}{g(X_{i})} \rightarrow E_{g}\left[\frac{f(X_{i})h(X_{i})}{g(X_{i})}\right] = E_{f}[h(X)]$ as $n\rightarrow \infty$,
provided
$E_{g}\left[\left|\large\frac{f(X_{i})h(X_{i})}{g(X_{i})}\right|\right] < \infty$ and supp($g$) $\supset$ supp($f.h$).
Now, it is the condition on the last line I am not happy with. I understand that the expectation given must be finite to be able to apply the law of large numbers. However, I really do not understand how the last statement comes into play.
Is it to ensure that the expectation is finite? Is it to ensure that the series above converges?
It makes sense to me that we should want supp($g$) $\supset$ supp($f$), since $g$ is the pdf we are using to aquire information from the target distribution $f$.
Thanks for your insight and thoughts.
Suppose that there was $x$ such that $g(x)=0$ but $f(x)h(x)\neq 0$. Then this $x$ contributes to the $E_f(h)$ expectation but not the $E_g(\cdots)$ one. Hence the two contributions aren't the same.
To put it another way, it doesn't matter if $g$ misses out a point with $f\neq 0$ so long as this is a point with $h=0$.