An independent finite random index has the weighted expectation formula

Define $\tilde{\mathrm{\Omega }}=\mathrm{\Omega }\times \{1,\dots ,N\}$ and let $\tilde{\mathcal{F}}$ consist of the sets $E={\bigcup }_{k=1}^N{A}_k\times \{k\}$ with every $A_k\in \mathcal{F}$. Each such decomposition is unique, since $A_k=\{\omega :(\omega ,k)\in E\}$. The empty set and whole space have this form; complements replace each $A_k$ by $\mathrm{\Omega }\setminus A_k$; and countable unions replace each $A_k$ by the corresponding countable union in $\mathcal{F}$. Thus $\tilde{\mathcal{F}}$ is a sigma-algebra.

For $E={\bigcup }_k{A}_k\times \{k\}$ define $\tilde{P}(E)={\sum }_k{p}_{k}P(A_k)$. This is nonnegative and $\tilde{P}(\tilde{\mathrm{\Omega }})={\sum }_k{p}_k=1$. If $E_j={\bigcup }_k{A}_{j,k}\times \{k\}$ are pairwise disjoint, then for each fixed $k$ the sets $A_{j,k}$ are pairwise disjoint. Countable additivity of $P$ from [F2] gives $\tilde{P}({\bigcup }_j{E}_j)={\sum }_k{p}_k{\lim }_{m\to \mathrm{\infty }}{\sum }_{j=1}^{m}P(A_{j,k})$. Because the sum over $k$ is finite, its limit commutes with this finite sum; rearranging each finite double sum gives $\tilde{P}({\bigcup }_j{E}_j)={\lim }_{m\to \mathrm{\infty }}{\sum }_{j=1}^m\tilde{P}(E_j)={\sum }_j\tilde{P}(E_j)$. Hence $\tilde{P}$ is a probability measure.

Let $R(\omega ,k)=k$. It is measurable because each inverse image is $\mathrm{\Omega }\times \{k\}$, and $\tilde{P}(R=k)=p_k$. For every $A\in \mathcal{F}$, its lift $\tilde{A}=A\times \{1,\dots ,N\}$ satisfies $\tilde{P}(\tilde{A}\cap \{R=k\})=p_{k}P(A)=\tilde{P}(\tilde{A})\tilde{P}(R=k)$, so $R$ is independent of all lifted original events.

First suppose each $H_k$ is a nonnegative simple function. Write it as a finite sum of nonnegative constants times indicators of measurable sets. On the $k$th slice, the lifted indicator of $A\in \mathcal{F}$ is ${\mathbf{1}}_{A\times \{k\}}$, whose integral is its probability $p_{k}P(A)$ by [F4] and step 2.1. Finite linearity [F5], followed by summing the finitely many slices, gives $\tilde{\mathbb{E}}[H_R]={\sum }_k{p}_k\mathbb{E}[H_k]$.

For arbitrary nonnegative measurable $H_k$, choose increasing simple approximations $H_{k,m}\uparrow H_k$ by [F6]. Since there are finitely many $k$, the lifted simple functions $H_m(\omega ,k)=H_{k,m}(\omega )$ increase pointwise to $H_R$. Apply [F7] on the extension and on the original space to the simple-function identity of step 3.2; the finite sum over $k$ commutes with the increasing limit. Therefore $\tilde{\mathbb{E}}[H_R]={\sum }_k{p}_k\mathbb{E}[H_k]$ for all nonnegative measurable $H_k$.

For integrable real $H_k$, apply step 4.1 separately to $H_k^{+}$ and $H_k^{-}$. Their weighted expectations are finite because ${\sum }_k{p}_k\mathbb{E}\mathrm{\mid }{H}_k\mathrm{\mid }<\mathrm{\infty }$. Subtracting the two nonnegative identities using [F5] proves both integrability of $H_R$ and the stated expectation formula. The extension and selector are explicitly finite-slice constructions, so this proof uses no choice axiom.