What is the requirement on the instrumental density in importance sampling? 
*

*From Rubinstein's Simulation Monte Carlo Method

I was wondering why $g$ is required to dominate $Hf$?

*From Wikipedia

The basic idea of importance sampling is to change the probability $P$ so that the estimation of $E[X;P]$ is easier. Choose a random
  variable $L\geq 0$ such that $E[L;P]=1$ and that $P$-almost
  everywhere $L(\omega)\neq 0$. The variate $L$ defines another
  probability $P^{(L)}=L\, P$ that satisfies $$
    \mathbf{E}[X;P] = \mathbf{E}\left[\frac{X}{L};P^{(L)}\right].  $$

The conditions on the instrumental density $L$ are $L\geq 0$ such
that $E[L;P]=1$ and that $P$-almost everywhere $L(\omega)\neq 0$. I
think I can understand them, because they are equivalent to say that
$L$ is a density of $P^{(L)}$ wrt $P$, and the denominator in
$\frac{X}{L}$ is zero only on a subset of probability measure zero
so that it won't affect the expectation/integration.
Given that the two sources seem to give different answers, I wonder what conditions should be put on the instrumental density? Thanks and regards!
 A: We need the condition
$$g(x) = 0 \Rightarrow H(x)f(x)=0 $$
else if $g(x)$ is 0, we would be dividing by 0, but if $f(x)$ is zero too, then we are safe. (handwavey arguments). It's also known as: $f$ is absolutely continuous w.r.t $g$.
To compare this to the wikipedia idea, $L := g(X) \ge 0$ where $X$ is sampled according to $g$. I'm not sure why $E[L;P] = 1$ though.
A: The essential requirement is to chose a proposal distribution that is not 0 or too small in areas where the target distribution could have significant mass. 
The idea is that you're approximating one distribution with a another, and to the extent that they're similar it will take fewer samples to obtain a good approximation. Let $p(z)$ be some intractable target distribution and $q(z)$ be some convenient distribution that your favorite language has a built in sampler for. You can visualize your goal as trying to use $N$ samples from $q(z)$ to make a histogram that looks like $p(z)$. So, obviously you would like your samples to concentrate in regions where $p(z)$ is large. But if $q(z)$ is very small in such regions, then you're unlikely to obtain very many samples there. And if $q(z)=0$ in such regions then you will never obtain samples there. 
Here's the relevant excerpt from Bishop's Pattern Recognition and Machine Learning, page 534 

"If, as is often the case, $p(z)f(z)$ is strongly varying and has a
  significant proportion of its mass concentrated over relatively small
  regions of z space, then the set of importance weights {rl} may be
  dominated by a few weights having large values, with the remaining
  weights being relatively insignificant. Thus the effective sample size
  can be much smaller than the apparent sample size L. The problem is
  even more severe if none of the samples falls in the regions where
  $p(z)f(z)$ is large. In that case, the apparent variances of $r_l$ and
  $rlf(z(l))$ may be small even though the estimate of the expectation
  may be severely wrong. Hence a major drawback of the importance
  sampling method is the potential to produce results that are
  arbitrarily in error and with no diagnostic indication. This also
  highlights a key requirement for the sampling distribution $q(z)$,
  namely that it should not be small or zero in regions where $p(z)$ may
  be significant."

A: The condition "$H f$ is dominated by $g$" in Rubenstein corresponds to the condition "$P$-almost everywhere $L(\omega) \neq 0$" (i.e., greater than zero) in the Wikipedia article. Rubenstein's condition is slightly weaker, because we actually don't need (in Wikipedia-notation) $L$ to have mass where $X$ has no mass. The condition that $\mathbf E[L;P]=1$ in Wikipedia is not actually necessary, it just simplifies the math a bit.
So, why do you need the dominating condition? Otherwise, the random variable $X$ (using Wikipedia's notation) will have positive mass in areas where $L$ will not. Since you can't ever sample those areas under $L$, that part of $X$ will be ignored by the importance sampling, which means that your results will be biased in unpredictable ways.
As jerad notes, although not theoretically required, in practice it is good for $L$ to have reasonably high mass everywhere $X$ does (it helps reduce the variance of the estimator). Heavy-tailed distributions are often used for $L$ because of this.
