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$L^p$ boundedness means ${\sup }_n\Vert X_n{\Vert }_p<\infty$, with the essential-supremum norm for $p=\infty$. For finite $p>1$, let $C={\sup }_n\mathbb{E}|X_n{|}^p<\infty$. On $|X_n|>M$, $|X_n|\le M^{1-p}|X_n{|}^p$, soThis proves uniform integrability. For an $L^{\infty }$ bounded family the tails are identically zero beyond the common essential bound.

The assertion is false. On $([0,1],\mathcal{B},\text{Lebesgue measure})$, put $X_n=n{1}_{(0,1/n)}$. Then $\mathbb{E}|X_n|=1$ for every $n$, so the family is $L^1$ bounded. For every $M$, choose $n>M$; then $\mathbb{E}[|X_n|1_{|X_n|>M}]=1$. The tail criterion for uniform integrability fails.

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.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

From convergence in probability, choose a subsequence $X_{n_k}\to X$ almost surely. The Fatou lemma then givesso $X\in L^2$.

The bounded second moments make $(X_n)$ uniformly integrable, and $X$ is integrable. Convergence in probability together with uniform integrability therefore givesConvergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, letThen $X_n\to 0$ in probability and $\mathbb{E}|X_n{|}^2=1$, but $\Vert X_n{\Vert }_2=1$ for every $n$. This is the bounded second moments do not upgrade convergence in probability to convergence in L2 example.

Now assume $X_n\to X$ almost surely and $\mathbb{E}[X_n^2]\to \mathbb{E}[X^2]$. In particular, $(X_n)$ is bounded in $L^2$, so it is uniformly integrable in $L^1$. Almost-sure convergence and uniform integrability therefore give

We next show that $X_n$ converges weakly to $X$ in the L2 space. For $Y\in L^2$, let $Y_M=Y{1}_{\{|Y|\le M\}}$. Thenwhile the Cauchy-Schwarz inequality gives, uniformly in $n$,Letting first $n\to \infty$ and then $M\to \infty$ proves $\mathbb{E}[X_{n}Y]\to \mathbb{E}[XY]$. Taking $Y=X$ and expanding the square now givesConsequentlyThis proves the almost-sure convergence and convergence of second moments criterion.

For $K>0$, the Cauchy-Schwarz inequality and part a giveThus $(Y_n^{-})$ is uniformly integrable. Combining this with the weak convergence of random variables from part b yields convergence of the first moments:By symmetry of the standard normal density,

By the Reflection invariance of Brownian motion, $-B$ is Brownian motion, and the first time $B_t-t$ reaches $b<0$ is the first time $-B_t+t$ reaches $-b>0$. Part b therefore gives

Now $L_t=\exp (B_t-t/2)$ is the exponential Brownian martingale. At $S_b$, the identity $B_{S_b}=b+S_b$ gives $L_{S_b}=e^{b+S_b/2}$, so $\mathbb{E}{L}_{S_b}=1$. The nonnegative stopped martingale $(L_{t\wedge S_b})$ therefore loses no mass at infinity and is uniformly integrable. The optional sampling theorem at any stopping time $R$ gives

False without a boundedness or uniform integrability condition. Take the upper half-planeand $u(x,y)=y$. This function is harmonic and continuous on $\overline{D}$, with boundary value $f=0$. The second coordinate of planar Brownian motion is one-dimensional Brownian motion, so its first hitting time $\tau$ of zero is finite almost surely. NeverthelessThe stopped local martingale $u(B_{t\wedge \tau })$ is not uniformly integrable, which is exactly why optional stopping fails in the limit.

The martingale convergence theorem states that a discrete-time martingale $(X_n)$ with uniform integrability has an integrable random variable $X_{\infty }$ such thatalmost surely and in $L^1$. Moreover, the martingale is closed by its limit:

For bounded stopping times $S\wedge n\le T\wedge n$, the optional sampling theorem for a supermartingale givesA stopped family drawn from a uniformly integrable martingale is uniformly integrable. Since $X_{T\wedge n}\to X_T$ and $X_{S\wedge n}\to X_S$ almost surely, uniform integrability upgrades both convergences to $L^1$. Passing to the limit yields

Put ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$ and $\tau =T_0\wedge T_y$. Before $\tau$, the integer-valued increment belongs to $\{-1,0,1\}$. The martingale property givesTheir sum is at least $1/2$, so each conditional probability is at least $1/4$. From any state in $\{1,\dots ,y-1\}$, a run of at most $y$ upward moves reaches $y$ and has conditional probability at least $4^{-y}$. Applied in successive blocks of $y$ steps, this givesso $\tau <\infty$ almost surely.

The stopped process $X_{n\wedge \tau }$ takes values in $[0,y]$, hence is a bounded martingale and has uniform integrability. The optional sampling theorem for a supermartingale givesbecause $X_{\tau }$ is zero or $y$. Therefore

Fix $t>0$ and define $w(s,y)=u(t-s,|y|)$ for $0\le s\le t$. The assumed smooth extension and the Neumann boundary condition at zero make $w$ a $C^{1,2}$ function. The heat equation givesThe Itô formula therefore makes $w(s,B_s)$ a local martingale. Stop first when $|B|$ leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on $[0,t]$ give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.

Almost-sure convergence plus uniform integrability gives $X_n\to X$ in $L^1$. Conditional expectation is an $L^1$ contraction, so

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.

The Skorokhod embedding theorem represents a centered probability distribution $\mu$ on the real numbers with finite variance as $B_T\sim \mu$ for a stopping time $T$ of Brownian motion. One may choose $T$ so that the stopped process $(B_{t\wedge T})$ is uniformly integrable and $\mathbb{E}T=\int x^2\mu (dx)$.