= Solution
{c}
The <importance weight> for the two normal densities is
$$
w_\sigma(x)=\frac1\sigma
\exp\left[\frac12(1-\sigma^{-2})x^2\right].
$$
Use $\hat\mu_N=N^{-1}\sum_iw_\sigma(X_i)f(X_i)$ with $X_i\sim N(0,1)$. Its second moment before averaging is
$$
E_q(w_\sigma^2f^2)=\frac1{\sigma^2\sqrt{2\pi}}
\int_{\mathbb R}f(x)^2
e^{-(\sigma^{-2}-1/2)x^2}\,dx.
$$
For the source's general assumption $E_pf^2<\infty$, a guaranteed range is
$$
\boxed{0<\sigma\leq1.}
$$
Indeed $w_\sigma\leq1/\sigma$ there, so $E_q(w_\sigma^2f^2)=E_p(w_\sigma f^2)\leq\sigma^{-1}E_pf^2<\infty$.
This range cannot be enlarged uniformly over all such $f$. If $\sigma>1$, choose $f^2=e^{ax^2}$ with
$$
\max(0,\sigma^{-2}-1/2)<a<\frac1{2\sigma^2}.
$$
Then $E_pf^2<\infty$ but the displayed weighted second-moment integral diverges. Thus no variance range depending only on $\sigma$ can characterize every individual integrand beyond the guaranteed range.
If an additional boundedness assumption on $f$ is intended, the familiar sufficient range becomes $0<\sigma<\sqrt2$. For $f=1$ this condition is also necessary, but a compactly supported $f$ can have finite weighted variance for every $\sigma$. Distinguishing these assumptions prevents imposing an unstated restriction on the source integrand.
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