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The Uniform Law of Large Numbers states that under certain conditions regarding the compactness of $\Theta$ and continuity of $f$, we have that:

$$ \sup_{\theta\in\Theta} \left\| \frac1n\sum_{i=1}^n f(X_i,\theta) - \operatorname{E}[f(X,\theta)] \right\| \xrightarrow{\mathrm{a.s.}} \ 0. $$

I am wondering, from this result, can we say that:

$$ \frac1n\sum_{i=1}^n f(X_i,\theta) = \operatorname{E}[f(X,\theta)] \ \ \text{(a.s.) ?} $$

I am a bit confused how to interpret the supremum outside. It seems to be that I can re-write as:

$$ \operatorname{P}\left(\lim_{n \to \infty}\sup_{\theta\in\Theta} \left\| \frac1n\sum_{i=1}^n f(X_i,\theta) - \operatorname{E}[f(X,\theta)] \right\| =0\right) = 1 $$

Is there a way to re-write this without the supremum part?

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    The answer to your first question is yes. Because: $$0 \leq || \frac{1}{n}\sum_{i=1}^n f(X_i,\gamma)-E[f(X,\gamma)]|| \leq \sup_{\theta \in \Theta}|| \frac{1}{n}\sum_{i=1}^n f(X_i,\theta)-E[f(X,\theta)]||$$ for all $\gamma \in \Theta$, since the supremum considers a set that includes $\gamma$. I am assuming $||\cdot||$ is a norm and $f$ is a vector-valued function. – 2017-02-09

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