Uniform Law of large numbers

Let $G(\cdot)$ be bounded, continuous, strictly increasing function on $\mathbb{R}.$ Let ${\xi_t},\, t\in\mathbb{Z}_+$ be i.i.d random variables.

Would it be true to say that we have something like the uniform Law of large numbers here? Why? $$\sup_{x\in\mathbb{R}}\;\left|\;n^{-1}\sum_{t=1}^n G(x+\xi_t) - \mathbf{E}G(x+\xi_1)\;\right| \stackrel{P}\longrightarrow 0, \quad n\to\infty.$$

• This thread mathoverflow.net/questions/121238/… could interest you. – Davide Giraudo Jul 9 '13 at 19:41
• Thanks, but unfortunately they're talking about compacts there – Dan Jul 10 '13 at 11:31
• An idea could be the following: after having rescaled $G$, we can assume that $G$ is a cumulative distribution function. Consider $(\eta_i,i\in\Bbb Z)$ an i.i.d. sequence of random variables independent of $(\xi_i,i\in\Bbb Z)$, say on an other probability space $(\Omega',\mathcal F',\mu')$. Then we can use Glivenko-Cantelli's theorem for the product space and the i.i.d. sequence $(\xi_i-\eta_i,i\in\Bbb Z)$. – Davide Giraudo Jul 10 '13 at 13:02

Assume that the $\xi_t$ are defined on a probability space $(\Omega,\mathcal F,\mu)$. We can assume that $G$ is the cumulative distribution function of a real valued random variable. Let $(\eta_t,t\in\Bbb Z)$ be a collection of independent random variables with cumulative distribution function $G$, defined on $(\Omega',\mathcal G,\nu)$. Using Glivenko-Cantelli's theorem for the product space and the family of i.i.d. random variables $(\xi_t-\eta_t,t\in\Bbb Z)$ gives $$\sup_{u\in\Bbb R}\left|\frac 1n\sum_{i=1}^n\chi_{\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}}-\mu\otimes\nu\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}\right|\to 0$$ for $\mu\otimes\nu$ almost every $(\omega,\omega')$. Then integrate with respect to $\nu$.
which is available here. We consider $\mathcal F:=\{t\mapsto G(x+t), x\in\Bbb R)\}$, which is a Glivenko-Cantelli class as $G$ is bounded. But it seems quite overkill for this problem, since their result is more general.