Importance sampling with a not necessarily strictly positive importance density Let


*

*$(E,\mathcal E,\lambda)$ be a measure space

*$p,q:E\to[0,\infty)$ be $\mathcal E$-measurale with $$\int p\:{\rm d}\lambda=\int q\:{\rm d}\lambda=1\tag1$$

*$P:=\{p\ne0\}$ and $Q:=\{q\ne0\}$

*$\mu:=p\lambda$ (measure with density $p$ wrt $\lambda$) and $\nu:=q\lambda$
Obviously, $$\mu(P^c)=\nu(Q^c)=0.\tag2$$

Now I'm unsure whether I make any mistake in my reasoning or not: Since $\nu(Q^c)=0$, $$\tilde \mu:=\frac pq\nu$$ should be well-defined (since the density should only need to be well-defined outside a $\nu$-null set or am I missing something?)
Next I would like to show that if $P\subseteq Q$, then $\mu=\tilde\mu$, but I'm struggling to see at which point in the following equation chain we really use $P\subseteq Q$: $$\mu(B)=\mu(B\cap P)\stackrel{(a)}=\int_{B\:\cap\:P}q\frac pq\:{\rm d}\lambda\stackrel{(b)}=\int_{B\:\cap\:P}\frac pq\:{\rm d}\nu=\tilde\mu(B\cap P)=\tilde \mu(B)\tag3$$ for all $B\in\mathcal E$. I suppose it is at equation $(a)$, but on the other hand, shouldn't $\nu(Q)=1$ already guarantee this equation?

Remark: Regarding the first part of the question, it's clear that $\tilde\mu$ is well-defined if we assume that $P\subseteq Q$. The question is whether we need to assume this.
 A: Strictly speaking, you are not allowed to divide by $q$ outside of $Q$, so if you are defining
$$
\tilde\mu (A) := \int_A \frac{p}{q}\, \mathrm d\nu,
\qquad
A\in\mathcal E,
$$
you are implicitly using the convention $0\cdot\infty := 0$. A cleaner way to define $\tilde \mu$ would be
$$
\tilde\mu (A) := \int_{A\cap Q} \frac{p}{q}\, \mathrm d\nu,
\qquad
A\in\mathcal E.
$$
Note that in any case $\tilde\mu$ might no longer be a probability measure, since $\tilde \mu(E)=\int_Q p\, \mathrm d\lambda <1$ is possible.
Now, if $P\not\subseteq Q$ and you are using the $0\cdot\infty = 0$ convention, then step (a) in your computation (which uses $p=q\frac{p}{q}$) is not allowed: For $x\in P\setminus Q$ you have $p(x) \neq 0$, but $q(x)\frac{p(x)}{q(x)} = 0$ by the convention. For $P\subseteq Q$ the step is completely fine since you are using the identity $p(x) = q(x)\frac{p(x)}{q(x)}$ only for $x\in Q$.
If you are using the "cleaner" definition, then the same arguments hold true and in addition you will need to use $B\cap P = B\cap P\cap Q$, which is clear for $P\subseteq Q$.
