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Set $U(x)=U_0({\delta }_x)$. On a finite measurable space, points in the same atom of a sigma-algebra give the same Dirac measure, so $U$ is a measurable function and constant on atoms. Every probability measure is a finite mixture of representative Dirac measures with its atom masses as weights. Affineness then proves the expected-utility representation, even when the sigma-algebra is smaller than the power set.

A partition atom is a member of a measurable partition. For a finite or countable partition, each cell is an atom of a sigma-algebra generated by that partition, whose measurable sets are unions of cells. A partition atom need not be an atom of a measure: an interval cell with positive Lebesgue measure can be split into smaller positive-measure sets that lie outside the generated partition sigma-algebra.

Values at the endpoint one do not affect any of the conclusions, since $\mathbb{P}(U=1)=0$. Set $b_n(1)=1$ to define each random variable everywhere, and choose a Borel measurable function agreeing with $f$ almost everywhere if necessary. These choices leave all integrals and almost-sure assertions unchanged.

Every dyadic interval at level $n+1$ lies in a dyadic interval at level $n$. Equivalently, $b_n(U)$ is a measurable function of $b_{n+1}(U)$, so ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$: these sigma-algebras form a filtration. Since the uniform distribution of $U$ gives $\mathbb{E}|f(U)|={\int }_0^1|f(u)|du<\infty$, the conditional expectations $X_n=\mathbb{E}[f(U)\mid {\mathcal{F}}_n]$ are integrable and adapted. The tower property of conditional expectation givesThus $(X_n)$ is a martingale, and part (a) proves that it is uniformly integrable.

For $I_{n,k}=[k{2}^{-n},(k+1)2^{-n})$, the event $\{U\in I_{n,k}\}$ is an atom of a sigma-algebra ${\mathcal{F}}_n$, with probability $2^{-n}$. The value of the conditional expectation there is thereforeConsequently $\boxed{X_n=f_n(U)\text{ almost surely}.}$ This also shows directly that ${\int }_0^1|f_n|\le {\int }_0^1|f|$.

By the uniformly integrable martingale convergence theorem, there is an integrable $Y$ such that $X_n\to Y$ almost surely and in the L1 norm. To identify this limit, put ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $0\le U-b_n(U)<2^{-n}$ away from the harmless endpoint, $U={\lim }_n{b}_n(U)$, and $U$ is ${\mathcal{F}}_{\infty }$-measurable. Conversely every $b_n(U)$ is a measurable function of $U$. Thus ${\mathcal{F}}_{\infty }=\sigma (U)$, and both $Y$ and $f(U)$ have ${\mathcal{F}}_{\infty }$-measurable versions.

For $A\in {\mathcal{F}}_m$, the defining property of conditional expectation gives $\mathbb{E}[X_n{1}_A]=\mathbb{E}[f(U)1_A]$ whenever $n\ge m$. Passing to the limit in the L1 norm yieldsThe increasing union ${\bigcup }_m{\mathcal{F}}_m$ is an algebra of sets generating ${\mathcal{F}}_{\infty }$. The class of events for which this integral identity holds is a Dynkin system, so the pi-lambda theorem extends the identity to every $A\in {\mathcal{F}}_{\infty }$. Since both random variables are measurable there, uniqueness of conditional expectation gives $Y=f(U)$ almost surely. We have proved that the dyadic conditional averages recover integrable functions:because a probability-one assertion for $U$ is a full-Lebesgue measure assertion for $x$. Likewise, the uniform distribution of $U$ turns convergence in L1 intoNo pointwise continuity of $f$ is required.

Define $U(x)=U_0({\delta }_x)$, where ${\delta }_x$ is the Dirac measure at $x$. For the full sigma-algebra of subsets of a finite $E$, affineness directly gives $U_0(\lambda )={\sum }_{x\in E}\lambda (\{x\})U(x)$.

The statement also holds for an arbitrary sigma-algebra on a finite set. List each atom of a sigma-algebra as $A_1,\dots ,A_r$ and choose $x_j\in A_j$. Points in one atom have identical Dirac measures on this sigma-algebra, so $U$ is constant on each atom and is measurable. Every probability measure decomposes as $\lambda ={\sum }_j\lambda (A_j){\delta }_{x_j}$, whenceThis is the expected utility representation on a finite measurable space. It does not silently assume that each singleton is measurable.

The finite sigma-algebra ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$ is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is ${\mathcal{F}}_n$-measurable and integrable. Every $A\in {\mathcal{F}}_n$ is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.