Let
- $d\in\mathbb N$ with $d>1$
- $\lambda^d$ denote the Lebesuge measure on $\mathcal B\left(\mathbb R^d\right)$
- $f\in C^2(\mathbb R)$ be positive and $$\pi(x):=\prod_{i=1}^df(x_i)\;\;\;\text{for }x\in\mathbb R^d$$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $X:\Omega\to\mathbb R^d$ with $X_\ast\operatorname P=\pi\lambda^d$
Now, let $$g(x):=\sum_{i=2}^d\frac1{d-1}\left|\frac{f'(x_i)}{f(x_i)}\right|^2\;\;\;\text{for }x\in\mathbb R^d.$$
Note that $$\operatorname E\left[g(X)\right]=\int\frac{\left|f'\right|^2}f\:{\rm d}\lambda^1=:I.\tag1$$
How can we apply the weak law of large numbers to obtain $$\operatorname P\left[\left|g(X)-I\right|>\varepsilon\right]\xrightarrow{d\to\infty}0\tag2$$ for all $\varepsilon>0$?