= Solution
Write $Ph=\mathbb Eh(X)$ and $P_nh=n^{-1}\sum_{i=1}^nh(X_i)$. 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 $\varepsilon>0$, finitely many brackets of width at most $\varepsilon$ cover the whole class $\mathcal H$. Under this condition the <uniform strong law from finite L1 bracketing> states
$$
\boxed{\sup_{h\in\mathcal H}|P_nh-Ph|\longrightarrow0\quad\text{almost surely}.}
$$
The 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 $\varepsilon$-cover $[\ell_r,u_r]$, $1\le r\le N$. If $h$ belongs to bracket $r$, monotonicity of the <empirical measure> and of expectation gives
$$
P_nh-Ph\le P_nu_r-Pu_r+P(u_r-h)\le |P_nu_r-Pu_r|+\varepsilon,
$$
and
$$
Ph-P_nh\le P(h-\ell_r)+P\ell_r-P_n\ell_r\le\varepsilon+|P_n\ell_r-P\ell_r|.
$$
Consequently
$$
\sup_{h\in\mathcal H}|P_nh-Ph|\le\varepsilon+\max_{r\le N}\max\{|P_nu_r-Pu_r|,|P_n\ell_r-P\ell_r|\}.
$$
Apply the <strong law of large numbers> to these finitely many integrable endpoints. On a probability-one event, the maximum tends to zero. Repeat with $\varepsilon=1/q$, $q\in\mathbb N$, 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> estimator
$$
\boxed{\widehat m_n(s)=\frac1n\sum_{i=1}^n e^{sX_i},\qquad 0\le s\le t.}
$$
If $t=0$, this estimator and $m$ are both identically one. Otherwise, $X\ge0$ 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 $\varepsilon>0$, choose a partition $0=s_0<s_1<\cdots<s_N=t$ so that $m(s_r)-m(s_{r-1})\le\varepsilon$ for every $r$. If $s\in[s_{r-1},s_r]$, then $e^{s_{r-1}z}\le e^{sz}\le e^{s_rz}$ for every $z\ge0$. 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 gives
$$
\boxed{\sup_{s\in[0,t]}|\widehat m_n(s)-m(s)|\longrightarrow0\quad\text{almost surely}.}
$$
Both 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.
Back to article page
