Almost-sure martingale convergence does not preserve expectations by itself. Uniform integrability is the exact condition preventing a fixed amount of \(L^1\) mass from escaping to rarer and larger values. It characterizes martingales that genuinely possess an integrable terminal value.
Uniform integrability criteria
Definition. A family \(\mathcal X\subset L^1(P)\) is uniformly integrable if \[ \lim_{K\to\infty}\sup_{X\in\mathcal X} \mathbb E[|X|1_{\{|X|>K\}}]=0. \] Uniform integrability implies \(\sup_{X\in\mathcal X}\mathbb E|X|<\infty\), but the converse is false.
Proposition. Each of the following is sufficient for uniform integrability: \[ \sup_{X\in\mathcal X}\mathbb E|X|^p<\infty \quad\text{for some }p>1; \] or there is an increasing convex function \(\Phi:[0,\infty)\to[0,\infty)\) with \(\Phi(x)/x\to\infty\) and \[ \sup_{X\in\mathcal X}\mathbb E\Phi(|X|)<\infty. \]
Proof. In the first case, \[ |X|1_{\{|X|>K\}} \le K^{1-p}|X|^p. \] In the second, \[ |X|1_{\{|X|>K\}} \le \sup_{x>K}\frac{x}{\Phi(x)}\,\Phi(|X|), \] and the scalar supremum tends to zero. The second criterion is the de la Vallee Poussin criterion; its converse also holds.
Theorem (Vitali). If \(X_n\to X\) in probability, then \[ X_n\to X\text{ in }L^1 \quad\Longleftrightarrow\quad \{X_n:n\ge1\}\text{ is uniformly integrable}. \]
A proof, including the finite-measure qualifications and tail decomposition, appears in real analysis. Here it supplies the passage from almost-sure martingale limits to \(L^1\) limits.
Terminal-value characterization
Theorem (uniformly integrable martingales). For a martingale \(M_n\), the following are equivalent:
- \(\{M_n:n\ge0\}\) is uniformly integrable;
- there is \(M_\infty\in L^1\) such that \(M_n\to M_\infty\) almost surely and in \(L^1\);
- there is \(Y\in L^1\) such that \[ M_n=\mathbb E[Y\mid\mathcal F_n]\quad\text{for all }n. \]
Proof. Assume (1). Uniform integrability implies boundedness in \(L^1\), so the martingale convergence theorem gives an almost-sure finite limit \(M_\infty\). Vitali's theorem gives \(L^1\)-convergence, proving (2). Assume (2). For \(A\in\mathcal F_n\) and \(m\ge n\), \[ \mathbb E[1_AM_m]=\mathbb E[1_AM_n]. \] Let \(m\to\infty\). Since \(M_m\to M_\infty\) in \(L^1\), \[ \mathbb E[1_AM_\infty]=\mathbb E[1_AM_n]. \] Hence \(M_n=\mathbb E[M_\infty\mid\mathcal F_n]\), proving (3). Assume (3). To prove uniform integrability, choose \(K>0\) and split \(Y=Y1_{\{|Y|\le R\}}+Y1_{\{|Y|>R\}}\). Conditional contraction gives \[ |M_n| \le R+\mathbb E[|Y|1_{\{|Y|>R\}}\mid\mathcal F_n]. \] For \(K>2R\), the event \(\{|M_n|>K\}\) forces the second term above to exceed \(K-R\), and a standard conditional Markov estimate yields \[ \sup_n\mathbb E[|M_n|1_{\{|M_n|>K\}}] \le C\,\mathbb E[|Y|1_{\{|Y|>R\}}]. \] First choose \(R\) large and then \(K>2R\); the right side tends to zero. Thus (1) holds.
A martingale satisfying these conditions is called closed by its terminal variable. The representation says the entire process is obtained by revealing more information about one integrable object.
Levy's upward theorem
Theorem. Let \(\mathcal F_n\uparrow\mathcal F_\infty :=\sigma(\bigcup_n\mathcal F_n)\). For \(X\in L^1\), \[ \mathbb E[X\mid\mathcal F_n] \longrightarrow \mathbb E[X\mid\mathcal F_\infty] \quad\text{almost surely and in }L^1. \]
Proof. The conditional expectations form a uniformly integrable martingale, so they converge almost surely and in \(L^1\) to some \(Y\in L^1(\mathcal F_\infty)\). If \(A\in\bigcup_n\mathcal F_n\), then for all sufficiently large \(m\), \[ \mathbb E[1_A\mathbb E(X\mid\mathcal F_m)] =\mathbb E[1_AX]. \] Passing to the \(L^1\) limit gives \(\mathbb E[1_AY]=\mathbb E[1_AX]\). A monotone-class argument extends the identity to every \(A\in\mathcal F_\infty\), identifying \(Y=\mathbb E[X\mid\mathcal F_\infty]\).
Corollary (\(L^p\) closure). If \(X\in L^p\), \(1<p<\infty\), then \[ \mathbb E[X\mid\mathcal F_n] \to\mathbb E[X\mid\mathcal F_\infty] \quad\text{in }L^p. \]
Proof. Conditional Jensen gives uniform \(L^p\)-boundedness. Approximate \(\mathbb E[X\mid\mathcal F_\infty]\) in \(L^p\) by a bounded \(\mathcal F_m\)-measurable variable \(Z\). For \(n\ge m\), \[ \|\mathbb E[X\mid\mathcal F_n]-Z\|_p = \|\mathbb E[X-Z\mid\mathcal F_n]\|_p \le\|X-Z\|_p, \] and refine the approximation using the \(L^p\)-density of \(\bigcup_mL^p(\mathcal F_m)\) in \(L^p(\mathcal F_\infty)\).
An \(L^1\)-bounded martingale that is not closed
Example. Let \(S_n\) be simple symmetric random walk, and \(\tau=\inf\{n:S_n=1\}\). Then \(M_n=S_{n\wedge\tau}\) is a martingale and \(M_n\to1\) almost surely. Nevertheless \(\mathbb EM_n=0\) for every \(n\), whereas the limit has mean \(1\). Hence convergence is not in \(L^1\), and \(\{M_n\}\) is not uniformly integrable. This is the same missing-mass mechanism that invalidates optional stopping at \(\tau\).