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Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i}h(X_i)$ for the expectation and empirical measure. A sufficient bracketing of a function class condition is that, for every $\epsilon >0$, finitely many brackets $[l_j,u_j]$ cover $\mathcal{H}$, with measurable integrable endpoints and $P(u_j-l_j)<\epsilon$. The uniform strong law from finite L1 bracketing then givesHere the observations are independent and identically distributed. For an uncountable class, one either assumes a measurable supremum, for example through a pointwise separable function class, or formulates the conclusion as a pathwise bound on a common probability-one event.

For the parameterized class, the Heine-Borel theorem makes $\Theta$ compact. Put $M(x)={\sup }_{\theta \in \Theta }|q(\theta ,x)|$. A countable dense subset of $\Theta$ gives the same supremum because of continuity, so $M$ is measurable. Define the modulusThe supremum is measurable by taking a countable dense subset of the compact set of admissible pairs. For each $x$, uniform continuity on $\Theta$ gives $w_{\delta }(x)\to 0$. Also $w_{\delta }\le 2M$, and $PM<\infty$. Thus the dominated convergence theorem gives $P{w}_{\delta }\to 0$.

Choose a finite net of radius $\delta$ ${\theta }_1,\dots ,{\theta }_N$ in $\Theta$. The lower and upper envelopes of $q(\theta ,\cdot )$ over each closed ball $\Theta \cap \overline{B}({\theta }_j,\delta )$ are measurable integrable function brackets; the same countable-dense-set argument applies within each such compact ball. Their widths are at most $w_{2\delta }$. They cover the whole class, so its bracketing of a function class condition follows. Moreover, dominated convergence shows that $\theta \mapsto Pq(\theta ,\cdot )$ is continuous. Both empirical and population functions therefore have a supremum over a common countable dense parameter set. Applying the uniform strong law from finite L1 bracketing proves uniform almost-sure convergence over the entire compact parameter set, rather than merely convergence at each fixed parameter.

For the exponential family,where $A={\sup }_{\Theta }|\theta |$ and $B={\sup }_{\Theta }|K(\theta )|$. Thus a weak sufficient condition is bounded $K$ on $\Theta$, $f_0(X)>0$ almost surely, and integrability of $|X|+|\log f_0(X)|$ under the sampling law. No positive lower bound on $f_0$ over the whole real line is required. Zeros off the sampling support are harmless: choose arbitrary finite versions of the log-densities on that common null set when applying the function-class theorem.

If the sampling law is $f({\theta }_0,\cdot )$, one convenient assumption is that $\Theta$ is a compact subset of the interior of the finite domain of the cumulant function of an exponential family, andOn that interior, $K$ is continuous, hence bounded on $\Theta$. Finiteness of $K$ at ${\theta }_0\pm a$ for some $a>0$ implies ${\mathbb{E}}_{{\theta }_0}{e}^{a|X|}<\infty$, hence ${\mathbb{E}}_{{\theta }_0}|X|<\infty$. The displayed integral gives the remaining integrability. If the observations come from an arbitrary unrelated law, assumptions on $K$ and $f_0$ alone cannot control that law's tails; the sampling-law integrability must then be stated explicitly.

Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i=1}^{n}h(X_i)$. A function bracket $[\ell ,u]$ contains the measurable functions $h$ satisfying $\ell (z)\le h(z)\le u(z)$ for every $z\in T$. Require its endpoints to be integrable and call $P(u-\ell )$ its $L^1(P)$ width. A sufficient condition is that, for every $\epsilon >0$, finitely many brackets of width at most $\epsilon$ cover the whole class $\mathcal{H}$. Under this condition the uniform strong law from finite L1 bracketing statesThe conclusion holds outside a common measurable null set. If the supremum is not initially known to be measurable, this formulation means pathwise convergence on a measurable probability-one event; a pointwise separable function class, including the application below, has a measurable supremum. Pointwise brackets also ensure the sample inequalities hold simultaneously over the class.

To prove the result, choose a finite $\epsilon$-cover $[{\ell }_r,u_r]$, $1\le r\le N$. If $h$ belongs to bracket $r$, monotonicity of the empirical measure and of expectation givesandConsequentlyApply the strong law of large numbers to these finitely many integrable endpoints. On a probability-one event, the maximum tends to zero. Repeat with $\epsilon =1/q$, $q\in \mathbb{N}$, and intersect the countably many probability-one events. The limiting supremum is bounded by $1/q$ for every $q$, hence is zero. This proves the uniform law of large numbers without a boundedness assumption on the class itself.

For the moment-generating function, use the empirical measure estimatorIf $t=0$, this estimator and $m$ are both identically one. Otherwise, $X\ge 0$ makes $e^{sX}$ increasing in $s$, and $e^{sX}\le e^{tX}$ provides an integrable envelope. The dominated convergence theorem shows that $m$ is continuous on $[0,t]$, hence uniformly continuous.

For any $\epsilon >0$, choose a partition $0=s_0<s_1<\cdots <s_N=t$ so that $m(s_r)-m(s_{r-1})\le \epsilon$ for every $r$. If $s\in [s_{r-1},s_r]$, then $e^{s_{r-1}z}\le e^{sz}\le e^{s_{r}z}$ for every $z\ge 0$. These endpoint functions form finitely many integrable function brackets with the required $L^1(P)$ widths. The just-proved uniform law of large numbers therefore givesBoth functions of $s$ are continuous; their supremum equals the supremum over a countable dense subset, so it is measurable. This proves uniform consistency of an empirical moment-generating function using only the observed sample.

If $X\ge 0$ and $\mathbb{E}{e}^{tX}<\infty$ for some finite $t\ge 0$, the estimator ${\hat{m}}_n(s)=n^{-1}{\sum }_i{e}^{s{X}_i}$ satisfies ${\sup }_{s\in [0,t]}|{\hat{m}}_n(s)-\mathbb{E}{e}^{sX}|\to 0$ almost surely. The dominated convergence theorem makes the population moment-generating function uniformly continuous on this interval. Parameter subintervals give function brackets whose endpoints are ordered exponentials and whose widths are the corresponding increments of the population mean. Apply the uniform strong law from finite L1 bracketing. Continuity in the parameter makes the supremum measurable.