[proofplan] For each fixed function $f_j$, the random variables $f_j(X_i)$ are i.i.d. and integrable, so the [Strong Law of Large Numbers](/theorems/520) gives almost sure convergence of $P_n f_j$ to $P f_j$. Since the class has only finitely many members, we intersect the finitely many probability-one events on which these convergences hold. On that common event, the supremum over $\mathcal F$ is the maximum of finitely many non-negative sequences converging to $0$, and therefore also converges to $0$. [/proofplan]

[step:Apply the strong law to each fixed function in the class]Fix $j\in\{1,\dots,m\}$. Define, for every $i\in\mathbb N$, the real-valued [random variable](/page/Random%20Variable) \begin{align*} Y_{i,j}:\Omega\to\mathbb R,\qquad Y_{i,j}(\omega):=f_j(X_i(\omega)). \end{align*} Since $f_j$ is $\mathcal S$-measurable and $X_i$ is $\mathcal A/\mathcal S$-measurable, $Y_{i,j}$ is $\mathcal A/\mathcal B(\mathbb R)$-measurable. Because $X_i$ has distribution $P$, \begin{align*} \mathbb E[|Y_{i,j}|]=\int_S |f_j(x)|\,dP(x)<\infty. \end{align*} The random variables $(Y_{i,j})_{i\ge 1}$ are i.i.d. because $(X_i)_{i\ge 1}$ are i.i.d. and each $Y_{i,j}$ is obtained from $X_i$ by the same measurable map $f_j:S\to\mathbb R$. By the [Strong Law of Large Numbers](/theorems/1852) (citing a result not yet in the wiki: Strong Law of Large Numbers), there is an event $A_j\in\mathcal A$ with $\mathbb P(A_j)=1$ such that, for every $\omega\in A_j$, \begin{align*} \frac{1}{n}\sum_{i=1}^{n}Y_{i,j}(\omega)\to \mathbb E[Y_{1,j}]. \end{align*} Since $Y_{i,j}=f_j(X_i)$ and since $X_1$ has distribution $P$, \begin{align*} \mathbb E[Y_{1,j}]=\int_S f_j(x)\,dP(x)=P f_j. \end{align*} Thus, for every $\omega\in A_j$, \begin{align*} P_n f_j(\omega)\to P f_j. \end{align*}[/step]

[guided]Fix one index $j\in\{1,\dots,m\}$. The point is to reduce the statement for the function $f_j$ to the ordinary one-dimensional Strong Law of Large Numbers. For every $i\in\mathbb N$, define the real-valued random variable \begin{align*} Y_{i,j}:\Omega\to\mathbb R,\qquad Y_{i,j}(\omega):=f_j(X_i(\omega)). \end{align*} This is measurable because it is the composition of the measurable map $X_i:(\Omega,\mathcal A)\to(S,\mathcal S)$ with the measurable map $f_j:(S,\mathcal S)\to(\mathbb R,\mathcal B(\mathbb R))$. We must also verify integrability before applying the Strong Law of Large Numbers. Since $X_i$ has distribution $P$, the change-of-law identity for expectations gives \begin{align*} \mathbb E[|Y_{i,j}|]=\mathbb E[|f_j(X_i)|]=\int_S |f_j(x)|\,dP(x)=P|f_j|<\infty. \end{align*} The sequence $(Y_{i,j})_{i\ge 1}$ is i.i.d.: independence follows because the variables $X_i$ are independent and each $Y_{i,j}$ is a [measurable function](/page/Measurable%20Function) of the corresponding $X_i$, and identical distribution follows because all $X_i$ have distribution $P$. Therefore the Strong Law of Large Numbers (citing a result not yet in the wiki: Strong Law of Large Numbers) applies to the integrable i.i.d. sequence $(Y_{i,j})_{i\ge 1}$. Hence there exists an event $A_j\in\mathcal A$ with $\mathbb P(A_j)=1$ such that, for every $\omega\in A_j$, \begin{align*} \frac{1}{n}\sum_{i=1}^{n}Y_{i,j}(\omega)\to \mathbb E[Y_{1,j}]. \end{align*} Substituting the definition of $Y_{i,j}$ gives \begin{align*} \frac{1}{n}\sum_{i=1}^{n}Y_{i,j}(\omega)=\frac{1}{n}\sum_{i=1}^{n}f_j(X_i(\omega))=P_n f_j(\omega). \end{align*} Since $X_1$ has distribution $P$, \begin{align*} \mathbb E[Y_{1,j}]=\mathbb E[f_j(X_1)]=\int_S f_j(x)\,dP(x)=P f_j. \end{align*} Thus, on the event $A_j$, the empirical average $P_n f_j$ converges to the population average $P f_j$.[/guided]

[step:Intersect the finitely many convergence events] Define the event \begin{align*} A:=\bigcap_{j=1}^{m} A_j. \end{align*} Since the intersection is finite and $\mathbb P(A_j)=1$ for every $j\in\{1,\dots,m\}$, \begin{align*} \mathbb P(A)=1. \end{align*} For every $\omega\in A$ and every $j\in\{1,\dots,m\}$, \begin{align*} |P_n f_j(\omega)-P f_j|\to 0. \end{align*} [/step]

[step:Pass from coordinatewise convergence to the finite supremum] Fix $\omega\in A$. For every $n\in\mathbb N$, define the non-negative real number \begin{align*} M_n(\omega):=\max_{1\le j\le m}|P_n f_j(\omega)-P f_j|. \end{align*} Because $\mathcal F=\{f_1,\dots,f_m\}$, including possible repeated functions, the supremum over $\mathcal F$ satisfies \begin{align*} \sup_{f\in\mathcal F}|P_n f(\omega)-P f|=M_n(\omega). \end{align*} Let $\varepsilon>0$. For each $j\in\{1,\dots,m\}$, the convergence on $A$ gives an integer $N_j(\omega,\varepsilon)\in\mathbb N$ such that, whenever $n\ge N_j(\omega,\varepsilon)$, \begin{align*} |P_n f_j(\omega)-P f_j|<\varepsilon. \end{align*} Define \begin{align*} N(\omega,\varepsilon):=\max_{1\le j\le m}N_j(\omega,\varepsilon). \end{align*} Then, for every $n\ge N(\omega,\varepsilon)$ and every $j\in\{1,\dots,m\}$, \begin{align*} |P_n f_j(\omega)-P f_j|<\varepsilon. \end{align*} Therefore, for every $n\ge N(\omega,\varepsilon)$, \begin{align*} M_n(\omega)<\varepsilon. \end{align*} Hence $M_n(\omega)\to 0$ for every $\omega\in A$. [/step]

[step:Conclude almost sure uniform convergence over the finite class] Since $\mathbb P(A)=1$ and, for every $\omega\in A$, \begin{align*} \sup_{f\in\mathcal F}|P_n f(\omega)-P f|=M_n(\omega)\to 0, \end{align*} we obtain \begin{align*} \sup_{f\in\mathcal F}|P_n f-Pf| \xrightarrow{a.s.} 0. \end{align*} This is the asserted finite-class [uniform law of large numbers](/theorems/1855). [/step]