In this Wikipedia article, they derive the variance of a Monte Carlo estimator $Q_N$ for a function $f: \mathbb R^m \rightarrow \mathbb R$, using $N$ samples drawn uniformly over an integration region $\Omega \subset \mathbb R^m$ (which has volume $V$).
They show that the variance of $Q_N \propto V^2 \frac{\sigma_N^2 }{N}$, and then the standard deviation is the square root of that ($\propto V/\sqrt N$).
However, it says right after that:
This result does not depend on the number of dimensions of the integral, which is the promised advantage of Monte Carlo integration against most deterministic methods that depend exponentially on the dimension
This intuitively strikes me as wrong, because it seems like the more dimensions there are, it seems like there's more "space to search" for the significant values of $f$ that contribute to the integral.
Similarly, they say that it doesn't depend on the number of dimensions, which may be naively true in the expression $Q_N \propto V^2 \frac{\sigma_N^2 }{N}$ (since $m$ doesn't appear), but... doesn't the integration volume scale with $m$?