Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $d\in\mathbb N$, $\Lambda\in\mathcal B(\mathbb R)^{\otimes d}$ with $\lambda^{\otimes d}(\Lambda)\in(0,\infty)$ and $f:\Lambda\to\mathbb C$ be Borel measurable with $$\int_\Lambda|f|\:{\rm d}\lambda^{\otimes d}<\infty.$$ Suppose we want to estimate $$I:=\int_\Lambda f\:{\rm d}\lambda^{\otimes d}$$ using the estimator $$\tilde I:=\frac1k\sum_{i=1}^kf(x_i),$$ where $k\in\mathbb N$ and $x_1,\ldots,x_k\in\Lambda$. Now observe the following: If we define $$\sigma(x):=\frac1k\sum_{i=1}^k\delta(x_i-x)\;\;\;\text{for }x\in\mathbb R^d,$$ then we can write $$\tilde I=\int\hat\sigma\overline{\hat f}\:{\rm d}\lambda^{\otimes d}\tag1,$$ where $\hat g$ denotes the Fourier transform of $g$.
Now, $\left(\Lambda,\mathcal B(\Lambda),\frac{\left.\lambda^{\otimes d}\right|_\Lambda}{\lambda^{\otimes d}(\Lambda)}\right)$ is clearly a probability space. Assume we replace $x_1,\ldots,x_k$ by random variables on this probability space. I wonder whether we can bound the variance of $\tilde I$ in terms of $\operatorname E\left[\left|\hat\sigma\right|^2\right]$?
Yes, you can bound the variance of $\tilde{I}$ in terms of $\operatorname{E}\left[|\hat{\sigma}|^2\right]$. To do this, use the fact that the random variables $x_1, \ldots, x_k$ are i.i.d. with respect to the probability measure $\frac{\left.\lambda^{\otimes d}\right|_\Lambda}{\lambda^{\otimes d}(\Lambda)}$.
First, let's compute the expectation of $\tilde{I}$: $$\operatorname{E}[\tilde{I}] = \operatorname{E}\left[\frac{1}{k} \sum_{i=1}^k f(x_i)\right] = \frac{1}{k} \sum_{i=1}^k \operatorname{E}[f(x_i)] = \int_\Lambda f \, {\rm d}\left(\frac{\left.\lambda^{\otimes d}\right|_\Lambda}{\lambda^{\otimes d}(\Lambda)}\right) = I$$ Now, let's compute the variance of $\tilde{I}$:
$$\operatorname{Var}[\tilde{I}] = \operatorname{E}\left[|\tilde{I} - I|^2\right] = \operatorname{E}\left[\left|\int (\hat{\sigma}\overline{\hat{f}} - \hat{f}\overline{\hat{f}}) \, {\rm d}\lambda^{\otimes d}\right|^2\right]$$
Applying Cauchy-Schwarz inequality, we get:
$$\operatorname{Var}[\tilde{I}] \leq \operatorname{E}\left[\int |\hat{\sigma} - \hat{f}|^2 \, {\rm d}\lambda^{\otimes d}\right] = \int \operatorname{E}\left[|\hat{\sigma} - \hat{f}|^2\right] \, {\rm d}\lambda^{\otimes d}$$
Now, using the linearity of the Fourier transform and Parseval's theorem, we have:
$$\operatorname{E}\left[|\hat{\sigma} - \hat{f}|^2\right] = \operatorname{E}\left[|\hat{\sigma}|^2\right] - 2\operatorname{E}\left[\operatorname{Re}\{\hat{\sigma}\overline{\hat{f}}\}\right] + |\hat{f}|^2$$
Since $-2\operatorname{E}\left[\operatorname{Re}{\hat{\sigma}\overline{\hat{f}}}\right] \leq 0$, we have: $$\operatorname{Var}[\tilde{I}] \leq \int \left(\operatorname{E}\left[|\hat{\sigma}|^2\right] + |\hat{f}|^2\right) \, {\rm d}\lambda^{\otimes d}$$ Thus, we can bound the variance of $\tilde{I}$ in terms of $\operatorname{E}\left[|\hat{\sigma}|^2\right]$.