Uniform Integrability¶

Martingale Convergence showed that $L^1$-boundedness forces almost sure convergence but not $L^1$ convergence — mass can escape along increasingly rare, increasingly large values. Uniform integrability is the precise fix. It is the condition that characterizes $L^1$-convergent martingales, and it is what extends optional sampling to unbounded stopping times.

Definition¶

Uniform integrability

A family $\{X_\alpha\}_{\alpha \in A}$ is uniformly integrable (UI) if

\[ \lim_{K \to \infty}\, \sup_{\alpha \in A}\, \mathbb{E}\bigl[|X_\alpha|\,\mathbf{1}_{\{|X_\alpha|>K\}}\bigr] = 0. \]

The truncation level $K$ controls the tails of every $X_\alpha$ at once. Taking $K = 0$ gives $L^1$-boundedness: $\sup_\alpha \mathbb{E}|X_\alpha| < \infty$. UI is strictly stronger — it also forbids the concentration of mass on shrinking high-value sets.

The canonical failure

On $\Omega = [0,1]$ with Lebesgue measure, let $X_n = n\,\mathbf{1}_{[0,1/n]}$. Then $X_n \to 0$ a.s. yet $\mathbb{E}[X_n] = 1$. For any $K$, once $n > K$, $\mathbb{E}[X_n \mathbf{1}_{\{X_n>K\}}] = 1$ — the family is not UI, and that is exactly why the expectation does not match the a.s. limit.

Two Equivalent Characterizations¶

Characterization theorem

A family $\{X_\alpha\}$ is UI if and only if it is $L^1$-bounded and equi-absolutely continuous: for every $\varepsilon > 0$ there exists $\delta > 0$ such that

\[ \mathbb{P}(A) < \delta \implies \sup_\alpha \mathbb{E}[|X_\alpha|\,\mathbf{1}_A] < \varepsilon. \]

Proof. ($\Rightarrow$) Given $\varepsilon > 0$, pick $K$ so the UI tail bound is $< \varepsilon/2$. Then for any event $A$,

\[ \mathbb{E}[|X_\alpha|\mathbf{1}_A] \le K\mathbb{P}(A) + \mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}] \le K\mathbb{P}(A) + \varepsilon/2. \]

Take $\delta = \varepsilon/(2K)$. ($\Leftarrow$) Let $C = \sup_\alpha \mathbb{E}|X_\alpha|$. Markov gives $\mathbb{P}(|X_\alpha|>K) \le C/K$, so choosing $K = C/\delta$ makes the tail event small enough for equi-absolute continuity to yield the UI bound. $\square$

A second useful test is de la Vallée Poussin: if there is a convex $\Phi$ with $\Phi(x)/x \to \infty$ and $\sup_\alpha \mathbb{E}[\Phi(|X_\alpha|)] < \infty$, then $\{X_\alpha\}$ is UI. The idea is that $|x| \le \Phi(x) / (\Phi(K)/K)$ for $x > K$, so tails of $|X_\alpha|$ are controlled by tails of $\Phi(|X_\alpha|)$, which are bounded uniformly.

UI Examples¶

UI families

Dominated: if $|X_\alpha| \le Y$ for some $Y \in L^1$, then $\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}] \le \mathbb{E}[Y\mathbf{1}_{\{Y>K\}}] \to 0$.
$L^p$-bounded with $p > 1$: $\mathbb{E}[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}] \le K^{-(p-1)}\mathbb{E}|X_\alpha|^p \to 0$.
Conditional expectations of a single $L^1$ variable: $\{\mathbb{E}[Y\mid \mathcal{G}]: \mathcal{G} \subseteq \mathcal{F}\}$ is UI. Indeed $\mathbb{E}|\mathbb{E}[Y\mid\mathcal{G}]| \le \mathbb{E}|Y|$, and for $A \in \mathcal{F}$ with $\mathbb{P}(A)<\delta$, $\mathbb{E}[\mathbb{E}[|Y|\mid\mathcal{G}]\mathbf{1}_A] = \mathbb{E}[|Y|\mathbf{1}_A] < \varepsilon$ (as in equi-absolute continuity for $|Y|$). This is the decisive example: every UI martingale arises this way.
Finite collection in $L^1$.

Non-examples

$\{n\mathbf{1}_{[0,1/n]}\}$ (mass escape).
The exponential martingale $\{e^{\theta W_t - \theta^2 t/2}\}$ for $\theta \ne 0$: $\mathbb{E}[Z_t] = 1$ but $Z_t \to 0$ a.s., contradicting UI via Vitali.

Vitali: UI Is Exactly the L¹ Bridge¶

Vitali convergence theorem

Suppose $X_n \to X$ in probability. Then $X_n \to X$ in $L^1$ if and only if $\{X_n\}$ is uniformly integrable.

Proof idea. ($\Rightarrow$) $L^1$-convergent sequences are automatically $L^1$-bounded and equi-absolutely continuous.

($\Leftarrow$) Fix $\varepsilon > 0$ and choose $\delta$ from equi-absolute continuity. By Egorov (after passing to an a.s.-convergent subsequence), there is $A$ with $\mathbb{P}(A) < \delta$ such that $X_n \to X$ uniformly on $A^c$. Then

\[ \mathbb{E}|X_n - X| \le \underbrace{\mathbb{E}[|X_n - X|\mathbf{1}_{A^c}]}_{\to 0 \text{ by uniform conv.}} + \underbrace{\mathbb{E}[|X_n|\mathbf{1}_A] + \mathbb{E}[|X|\mathbf{1}_A]}_{<\,2\varepsilon \text{ by equi-abs. cont.}}. \]

$\square$

The UI Martingale Theorem¶

This is the crux of the chapter. Combined with Martingale Convergence, it identifies closed martingales with UI martingales.

UI martingale characterization

For a martingale $(M_n)$, the following are equivalent:

$(M_n)$ is UI.
$M_n \to M_\infty$ in $L^1$.
$M_n \to M_\infty$ a.s. and in $L^1$.
$(M_n)$ is closed: $M_n = \mathbb{E}[X \mid \mathcal{F}_n]$ for some $X \in L^1$.

Proof. $(4) \Rightarrow (1)$: the conditional-expectation example above. $(1) \Rightarrow (3)$: UI $\Rightarrow L^1$-bounded $\Rightarrow$ a.s. convergence (Doob), then Vitali gives $L^1$. $(3) \Rightarrow (4)$: for $A \in \mathcal{F}_n$, $\int_A M_m\,d\mathbb{P} = \int_A M_n\,d\mathbb{P}$ for $m \ge n$; let $m \to \infty$ using $L^1$ convergence to get $\int_A M_\infty = \int_A M_n$, hence $M_n = \mathbb{E}[M_\infty \mid \mathcal{F}_n]$. $\square$

Lévy's upward theorem

If $X \in L^1$ and $\mathcal{F}_n \uparrow \mathcal{F}_\infty$, then $\mathbb{E}[X \mid \mathcal{F}_n] \to \mathbb{E}[X \mid \mathcal{F}_\infty]$ a.s. and in $L^1$. This is the previous theorem applied to the closed martingale $M_n = \mathbb{E}[X \mid \mathcal{F}_n]$.

UI and Optional Sampling¶

UI is also the condition that extends optional sampling to unbounded stopping times.

UI optional sampling

If $(M_t)$ is a UI martingale and $\sigma \le \tau$ are stopping times (possibly unbounded), then

\[ \mathbb{E}[M_\tau \mid \mathcal{F}_\sigma] = M_\sigma \quad \text{a.s.} \]

Why UI. Apply bounded optional sampling to $\tau\wedge T$ and $\sigma\wedge T$ and let $T\to\infty$. UI supplies the $L^1$ convergence needed to pass to the limit.

Without UI the conclusion fails: the doubling strategy produces an $L^1$-bounded, non-UI martingale $S_n$ with $\mathbb{E}[S_0] = 0$ but $\mathbb{E}[S_\tau] = 1$ at $\tau = \inf\{n : S_n = 1\}$.

Practical Summary¶

Sufficient for UI	Reason
$	X_\alpha
$\sup_\alpha \mathbb{E}	X_\alpha
$\sup_\alpha \mathbb{E}[\Phi(	X_\alpha
$X_\alpha = \mathbb{E}[Y \mid \mathcal{G}_\alpha]$, $Y \in L^1$	Closure
Finite $\{X_1,\ldots,X_n\}\subset L^1$	Trivial

UI is the "no mass escape" condition. It plays the role for random variables that tightness plays for probability measures, and it is exactly what closes the gap between a.s. and $L^1$ convergence.

Exercises¶

Exercise 1. Show that $\{W_t : 0 \le t \le T\}$ is UI for every $T < \infty$, while $\{e^{W_t - t/2} : t \ge 0\}$ is not.

Solution to Exercise 1

For the first family, $\sup_{t\le T}\mathbb{E}[W_t^2] = T < \infty$, so the family is $L^2$-bounded, hence UI by the $L^p$ criterion with $p = 2$.

For the second, $Z_t = e^{W_t-t/2}$ is a martingale with $\mathbb{E}[Z_t] = 1$, yet $Z_t \to 0$ a.s. (the law of the iterated logarithm gives $\log Z_t = W_t - t/2 \to -\infty$). If the family were UI, Vitali would force $\mathbb{E}[Z_t] \to 0$, contradicting $\mathbb{E}[Z_t] = 1$. $\square$

Exercise 2. If $X_n \to X$ in probability and $\sup_n \mathbb{E}[X_n^2] < \infty$, show $X_n \to X$ in $L^1$.

Solution to Exercise 2

The $L^2$ bound implies UI (take $p = 2$ in the $L^p$ criterion). Vitali's theorem (which applies to convergence in probability) then gives $X_n \to X$ in $L^1$. $\square$

Exercise 3. Construct a martingale that is $L^1$-bounded but not UI, and verify directly that it fails to converge in $L^1$.

Solution to Exercise 3

Let $\xi_i$ be i.i.d. with $\mathbb{P}(\xi_i = 2) = \mathbb{P}(\xi_i = 0) = 1/2$ and $M_n = \prod_{i=1}^n \xi_i$ (with $M_0 = 1$). Then $(M_n)$ is a martingale, $\mathbb{E}|M_n| = 1$ for all $n$, and $M_n \to 0$ a.s. (some $\xi_i$ eventually equals $0$).

Because $M_n \mathbf{1}_{\{M_n \le K\}} \to 0$ a.s. and is bounded by $K$, dominated convergence gives $\mathbb{E}[M_n\mathbf{1}_{\{M_n\le K\}}] \to 0$, so $\mathbb{E}[M_n\mathbf{1}_{\{M_n>K\}}] \to 1$ for every $K$. Thus $(M_n)$ is not UI, and $\mathbb{E}[M_n] = 1 \ne 0 = \mathbb{E}[\lim M_n]$, so $M_n$ does not converge in $L^1$. $\square$

Exercise 4. Prove that if $\sup_\alpha \mathbb{E}[|X_\alpha|\log(1+|X_\alpha|)] < \infty$, then $\{X_\alpha\}$ is UI.

Solution to Exercise 4

Take $\Phi(x) = x\log(1+x)$. Then $\Phi$ is convex on $[0,\infty)$ (direct check: $\Phi''(x) = (1+x)^{-1} + (1+x)^{-2} > 0$), $\Phi(0) = 0$, and $\Phi(x)/x = \log(1+x) \to \infty$ as $x \to \infty$. The de la Vallée Poussin criterion applies. $\square$

Exercise 5. Let $(M_t)$ be a UI martingale with limit $M_\infty$ and $\tau$ any stopping time (possibly infinite). Prove $M_\tau = \mathbb{E}[M_\infty \mid \mathcal{F}_\tau]$ a.s.

Solution to Exercise 5

Since $(M_t)$ is UI, $M_t = \mathbb{E}[M_\infty \mid \mathcal{F}_t]$. For bounded $\tau_k = \tau \wedge k$, bounded optional sampling gives $M_{\tau_k} = \mathbb{E}[M_\infty \mid \mathcal{F}_{\tau_k}]$, so for $A \in \mathcal{F}_\tau \subseteq \mathcal{F}_{\tau_k}$,

\[ \mathbb{E}[M_{\tau_k}\mathbf{1}_A] = \mathbb{E}[M_\infty \mathbf{1}_A]. \]

UI and a.s. convergence of $M_t$ imply $M_{\tau_k} \to M_\tau$ a.s. and in $L^1$, so $\mathbb{E}[M_\tau \mathbf{1}_A] = \mathbb{E}[M_\infty \mathbf{1}_A]$ for all $A \in \mathcal{F}_\tau$, giving the result. $\square$