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For an integrable function on $[0,1]$, average $f$ on each dyadic interval of length $2^{-n}$. The resulting step function $f_n$ converges to $f$ almost everywhere and in the L1 norm. If $U$ has the uniform distribution on $[0,1]$, then $f_n(U)=\mathbb{E}[f(U)\mid \sigma (b_n(U))]$, where $b_n$ rounds down to a dyadic endpoint. Refinement makes these conditional expectations a uniformly integrable martingale. The uniformly integrable martingale convergence theorem gives almost-sure and convergence in L1 to a limit $Y$.

Since $b_n(U)\to U$, the generated limiting sigma-algebra contains $U$ up to null sets. For every event $A$ in any finite-stage sigma-algebra, convergence in the L1 norm gives $\mathbb{E}[Y{1}_A]=\mathbb{E}[f(U)1_A]$. The pi-lambda theorem extends this equality to the limiting sigma-algebra, identifying $Y=f(U)$. The uniform distribution of $U$ translates the two convergence conclusions into the assertions for $f_n$. Choosing a Borel representative of $f$ and assigning arbitrary values at the endpoint one makes all random variables well defined without changing either conclusion.

For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

For an $L$-Lipschitz function $f$ on $[0,1]$, its slopes on the dyadic partition, evaluated at a uniform distribution input, form a bounded martingale for the dyadic filtration. The uniformly integrable martingale convergence theorem yields a bounded terminal random variable $g(X)$, with $g$ chosen as a Borel measurable function by the Doob-Dynkin lemma. The terminal conditional expectation identity says that each slope is the average of $g$ over the corresponding dyadic interval. Integrating over adjacent dyadic intervals and taking a telescoping sum proves the displayed representation first at dyadic endpoints. The Lipschitz condition for $f$ and the bound $|g|\le L$ extend it to every endpoint by continuity. This proves absolute continuity of a function without presupposing a derivative.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.

Put $Z_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$. It is a conditional-expectation martingale. We justify why its limit is $X_{\infty }$, rather than just quoting convergence to an unspecified random variable.

First the family is uniformly integrable. For $A=\{Z_n>K\}\in {\mathcal{F}}_n$, nonnegativity and the conditional expectation identity giveThus its tail expectations are at most twice the integrable tail of $X_{\infty }$, uniformly in $n$. The uniformly integrable martingale convergence theorem gives a limit $Z$ almost surely and in $L^1$. For any $A\in {\mathcal{F}}_m$ and $n\ge m$,Pass to the $L^1$ limit and extend from $\mathcal{A}$ to all Borel sets by the Monotone class theorem. The equality of these integrals forces $Z=X_{\infty }$ almost surely. Combining this with the preceding convergence of $X_n$ proves

A filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ on a probability space is an increasing sequence of sigma-algebras contained in $\mathcal{F}$: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. A real stochastic process $(X_n)$ is a martingale for this filtration if $X_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|X_n|<\infty$, andThe first condition is adaptedness; the tower property of conditional expectation then gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$ whenever $m\le n$.

The almost-sure martingale convergence theorem states that if a real martingale satisfies ${\sup }_n\mathbb{E}|X_n|<\infty$, then some integrable random variable $X_{\infty }$ satisfies $X_n\to X_{\infty }$ almost surely. The more general almost sure submartingale convergence theorem needs only ${\sup }_n\mathbb{E}[X_n^{+}]<\infty$ for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.

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.

Write $z^{-}=\max (-z,0)$. For $a<b$, let $U_N[a,b]$ be the upcrossing count obtained by successively buying at an observation at or below $a$ and selling at an observation at or above $b$, with the first purchase allowed at time zero. The Doob upcrossings lemma for an integrable discrete-time martingale isThe weaker last bound is also the usual Doob upcrossing inequality.

To prove it, hold one unit between each purchase and sale, and zero units otherwise. If $H_k$ records the holding over $(k,k+1]$, the decision is ${\mathcal{F}}_k$-measurable, with $H_k\in \{0,1\}$. The resulting martingale transform has gainEach completed upcrossing earns at least $b-a$. An unfinished holding loses at most $(X_N-a{)}^{-}$. If a purchase occurs at zero, its price is $X_0\le a$, giving the additional discount $(X_0-a{)}^{-}$, whether that holding has been sold or remains open. If no purchase occurs at zero, this extra term is zero. Therefore, path by path,Taking expectations proves the bound.

The martingale convergence theorem, stated without proof, says that a discrete-time martingale satisfying ${\sup }_n\mathbb{E}|X_n|<\infty$ converges almost surely to a finite integrable random variable $X_{\infty }$. In particular every nonnegative martingale converges almost surely to a finite limit. Under uniform integrability, the uniformly integrable martingale convergence theorem additionally gives convergence in L1 and $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.

The martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

Use the same dyadic slope martingale $G_n$ and dyadic filtration as in (c). Each $G_n$ is integrable, since it takes finitely many finite values. On each dyadic cell, the absolute slope is $2^n$ times the absolute endpoint increment. Consequently the hypothesis in the PDF is exactlyThus $(G_n)$ is uniformly integrable. It is also bounded in L1 norm: choose a finite ${\lambda }_0>0$ at which the supremum of the tails is finite, and use $\Vert G_n{\Vert }_1\le {\lambda }_0+{\sup }_j{V}_j(f,{\lambda }_0)$. The uniformly integrable martingale convergence theorem supplies $g\in L^1([0,1])$ with $G_n\to g$ in L1 norm.

The functions $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$ are again the dyadic linear interpolations of $f$. Since $f$ is continuous on a compact interval, it is uniformly continuous, and $\Vert f_n-f{\Vert }_{\infty }\le {\omega }_f(2^{-n})\to 0$, where ${\omega }_f$ is its modulus of continuity. On the other hand, the integral of $G_n-g$ is uniformly bounded in absolute value by $\Vert G_n-g{\Vert }_1\to 0$. Hence the dyadic slope-tail criterion for absolute continuity givesNo boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.

A sequence of integrable random variables $(X_n)$ has uniform integrability exactly whenThe uniformly integrable martingale convergence theorem states that a uniformly integrable discrete-time martingale $(M_n)$ has an integrable random variable $M_{\infty }$ withIndeed, uniform integrability implies ${\sup }_n\mathbb{E}|M_n|<\infty$, so the martingale convergence theorem gives almost sure convergence. The combination of uniform integrability and almost sure convergence gives convergence in L1. For $m\ge n$, the martingale identity $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ passes to the limit by the L1 contraction of conditional expectation.

Conversely, uniform integrability of conditional expectations shows that a martingale of the form $M_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for one integrable random variable $Y$ is uniformly integrable.

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.