[proofplan]
The proposed density is obtained by normalizing the non-negative integrand $hp$ by its integral $I$. Once this normalization is checked, the ratio in the importance sampling summand is evaluated by direct substitution on the support where $q^*>0$. The exceptional set where $q^*=0$ has zero probability under the proposal measure, so the equality holds almost surely. Independent averages of an almost surely constant summand are again almost surely constant, which gives zero variance.
[/proofplan]
[step:Verify that the normalized integrand is a probability density]
Since $h:E\to[0,\infty)$ and $p:E\to[0,\infty)$ are measurable, their pointwise product $hp:E\to[0,\infty)$ is measurable. Because $I$ is a finite positive real number, the map
\begin{align*}
q^*:E\to[0,\infty), \qquad q^*(x)=\frac{h(x)p(x)}{I}
\end{align*}
is measurable and non-negative. Its total mass with respect to $\mu$ is
\begin{align*}
\int_E q^*(x)\,d\mu(x)=\int_E \frac{h(x)p(x)}{I}\,d\mu(x)=\frac{1}{I}\int_E h(x)p(x)\,d\mu(x)=1.
\end{align*}
Thus $q^*$ is a probability density with respect to $\mu$.
[guided]
We first check the only two requirements for being a probability density with respect to $\mu$: measurability, non-negativity, and total integral equal to $1$. The functions $h:E\to[0,\infty)$ and $p:E\to[0,\infty)$ are measurable, so the product function $hp:E\to[0,\infty)$ is measurable and non-negative. The hypothesis
\begin{align*}
0<I=\int_E h(x)p(x)\,d\mu(x)<\infty
\end{align*}
says exactly that this product has a finite, strictly positive normalizing constant. Therefore division by $I$ is legitimate, and the function
\begin{align*}
q^*:E\to[0,\infty), \qquad q^*(x)=\frac{h(x)p(x)}{I}
\end{align*}
is again measurable and non-negative.
Now compute its total mass. By linearity of the integral for the positive finite scalar $1/I$,
\begin{align*}
\int_E q^*(x)\,d\mu(x)=\int_E \frac{h(x)p(x)}{I}\,d\mu(x)=\frac{1}{I}\int_E h(x)p(x)\,d\mu(x)=\frac{I}{I}=1.
\end{align*}
This proves that $q^*$ is a probability density with respect to the same reference measure $\mu$.
[/guided]
[/step]
[step:Evaluate the importance sampling ratio on the positive-density set]
Define the positive proposal set
\begin{align*}
A:=\{y\in E:q^*(y)>0\}.
\end{align*}
For every $y\in A$, the denominator $q^*(y)$ is non-zero, and the defining identity $q^*(y)=h(y)p(y)/I$ gives
\begin{align*}
\frac{h(y)p(y)}{q^*(y)}=\frac{h(y)p(y)}{h(y)p(y)/I}=I.
\end{align*}
The set $E\setminus A=\{y\in E:q^*(y)=0\}$ has proposal probability
\begin{align*}
\int_{E\setminus A} q^*(y)\,d\mu(y)=0.
\end{align*}
Hence, if $Y$ has density $q^*$ with respect to $\mu$, then $Y\in A$ almost surely under its law, and therefore
\begin{align*}
\frac{h(Y)p(Y)}{q^*(Y)}=I
\end{align*}
almost surely.
[/step]
[step:Conclude that the importance sampling estimator has zero variance]
Let $n\in\mathbb N$, and let $Y_1,\dots,Y_n$ be independent $E$-valued random variables, each with density $q^*$ with respect to $\mu$. Define the importance sampling average
\begin{align*}
\widehat I_n:=\frac{1}{n}\sum_{k=1}^n \frac{h(Y_k)p(Y_k)}{q^*(Y_k)}.
\end{align*}
By the almost sure identity proved above, each summand equals $I$ almost surely. Therefore
\begin{align*}
\widehat I_n=\frac{1}{n}\sum_{k=1}^n I=I
\end{align*}
almost surely. Since $\widehat I_n$ is almost surely the constant $I$, its variance is
\begin{align*}
\operatorname{Var}(\widehat I_n)=\mathbb E[(\widehat I_n-I)^2]=0.
\end{align*}
This proves the zero-variance property of the proposal density $q^*$.
[/step]
