= Solution
Take $k>0$, so the proposal is the <uniform distribution> on $[0,k]$, with <probability density function> $g(x)=1/k$. Symmetry of the standard <Cauchy distribution> gives
$$
\mu=\frac12-\int_0^kf(x)\,dx
=\frac12-k\mathbb E_g[f(X)].
$$
Thus the proposed expression is exactly the unbiased <importance sampling> estimator of this complementary <integral>:
$$
\boxed{\widehat\mu=\frac12-\frac{k}{n}\sum_{i=1}^n\frac1{\pi(1+x_i^2)}.}
$$
It has finite <variance> because $f$ is bounded on $[0,k]$. Explicitly,
$$
\begin{aligned}
\mathbb E_gf(X)&=\frac{\arctan k}{\pi k},\\
\mathbb E_gf(X)^2&=\frac1{2\pi^2k}\left(\arctan k+\frac{k}{1+k^2}\right),\\
\operatorname{Var}(\widehat\mu)
&=\frac1{n\pi^2}\left[\frac{k}{2}\left(\arctan k+\frac{k}{1+k^2}\right)-(\arctan k)^2\right].
\end{aligned}
$$
These follow from the antiderivatives of $(1+x^2)^{-1}$ and $(1+x^2)^{-2}$. The target itself is $\mu=1/2-\arctan(k)/\pi$. At $k=0$ it is exactly $1/2$ without simulation; for negative $k$ the printed uniform-interval construction must be changed.
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