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11 — Uniform Integrability (UI)

Lecture 11 continues the theme of Uniform Integrability (UI), showing:

  • Why integrability of a single dominating function implies UI,
  • A classical counterexample showing UI does not always come from DCT hypotheses,
  • The main theorem relating UI + a.s. convergence to convergence of expectations,
  • Characterizations of UI and how to check it,
  • A very important sufficient condition: bounded in $L^p$ for some $𝑝>1$ implies UI,
  • A two-part characterization: (i) bounded expectations, (ii) tight tails on sets of small probability.

We work on a probability space $(\Omega, \mathcal{F}, P)$ with $P(\Omega)=1$.

1. Basic Claim: Integrable Domination Implies Tail Vanishing

Suppose: \(Y \ge 0, \quad \mathbb{E}[Y] < \infty.\)

Claim:
\(\mathbb{E}\big( Y \, \mathbf{1}_{\{Y > M\}} \big) \xrightarrow[M\to\infty]{} 0.\)

Reason (page 1):

  • The sets ${Y > M}$ decrease to the empty set.
  • Thus: \(Y\mathbf{1}_{\{Y > M\}} \downarrow 0 \quad \text{a.e.}\)
  • By the Monotone Convergence Theorem (or Dominated Convergence since $Y \mathbf{1}_{{Y>M}} \le Y$): \(\mathbb{E}[Y\mathbf{1}_{\{Y>M\}}] \to 0.\)

This is the fundamental mechanism behind UI.

2. Definition of Uniform Integrability (UI)

A family ${X_n}_{n\ge 1}$ is uniformly integrable if: \(\phi(M) = \sup_{n\ge 1} \mathbb{E}\big[ \vert X_n\vert \mathbf{1}_{\{\vert X_n\vert >M\}} \big] \xrightarrow[M\to\infty]{} 0.\)

The graphical note on page 3 shows that UI prevents the “mass” of $X_n$ from drifting to arbitrarily large values at high magnitude.

3. Dominating Function Implies UI

Observation (page 1):
If $\vert X_n\vert \le Y$ for all $n$ and $\mathbb{E}[Y] < \infty$, then: \(\vert X_n\vert \mathbf{1}_{\{\vert X_n\vert > M\}} \le Y \mathbf{1}_{\{Y > M\}}.\)

Thus \(\phi(M) \le \mathbb{E}[Y\mathbf{1}_{\{Y>M\}}] \to 0.\)

Hence: \(\boxed{ \vert X_n\vert \le Y\in L^1 \quad \Longrightarrow \quad \{X_n\} \text{ is UI}. }\)

4. A Counterexample: UI Without DCT Domination

On page 1–2, the notes give a classical counterexample showing UI does not imply existence of a dominating integrable $Y$, nor DCT.

Let $\Omega=[0,1]$ with Lebesgue measure.
Define the disjoint intervals: \(I_n = \left(\frac{1}{2^{n+1}}, \frac{1}{2^n}\right].\)

Define: \(X_n(x) = \frac{2^n}{n} \, \mathbf{1}_{I_n}(x).\)

Properties:

  1. Almost sure convergence:
    \(X_n(x) \to 0 \quad \text{a.s.}\) because the intervals move toward 0 and shrink.

  2. Expectations:
    \(\mathbb{E}[X_n] = \frac{2^n}{n} \cdot \vert I_n\vert = \frac{2^n}{n} \cdot \frac{1}{2^{n+1}} = \frac{1}{2n} \to 0.\)

  3. UI holds:
    One checks that $\sup_n \mathbb{E}[X_n] < \infty$ and the tail mass goes to zero; hence ${X_n}$ is UI.

  4. But no DCT domination:
    A smallest dominating function would be: \(Y(x) = \sum_{n=1}^\infty \frac{2^n}{n} \mathbf{1}_{I_n}(x),\) and \(\mathbb{E}[Y] = \sum_{n=1}^\infty \frac{2^n}{n}\cdot \frac{1}{2^{n+1}} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} = \infty.\) So no integrable $Y$ dominates them.

Thus UI does not reduce to DCT.

5. Main Theorem: UI + a.s. Convergence ⇒ $L^1$ Convergence

As written on pages 2–3:

Theorem.
If:

  1. $X_n \to X$ a.s.,
  2. ${X_n}$ is UI,

then: \(\mathbb{E}\vert X_n - X\vert \to 0.\)

Consequences:

\[\mathbb{E}[X_n] \to \mathbb{E}[X].\]

This is the central reason UI is so important in probability theory.

6. Sufficient Condition: $L^p$–Boundedness Implies UI for $p>1$

From page 2:

If \(\sup_n \mathbb{E}\vert X_n\vert ^p < \infty \quad\text{for some } p>1,\) then ${X_n}$ is UI.

Proof (as in notes):

For any $M>0$, by Hölder/Markov: \(\mathbb{E}\big(\vert X_n\vert \mathbf{1}_{\{\vert X_n\vert >M\}}\big) \le \mathbb{E}\left(\frac{\vert X_n\vert ^p}{M^{p-1}}\right) = \frac{\mathbb{E}\vert X_n\vert ^p}{M^{p-1}} \le \frac{C}{M^{p-1}} \to 0.\)

Thus: \(\boxed{ L^p \text{ boundedness for } p>1 \quad \Longrightarrow \quad \text{UI}. }\)

This condition appears frequently in martingale theory and in proving tightness of random variables.

7. A Deep Characterization of UI

From pages 2–3, we have the following equivalence:

A family ${X_n}\subset L^1$ is UI iff:

  1. Uniformly bounded expectations:
    \(\sup_n \mathbb{E}\vert X_n\vert < \infty.\)

  2. Tight control on small-probability sets:
    For every $\varepsilon>0$ there exists $\delta>0$ such that for every measurable set $A$ with $P(A)<\delta$, \(\sup_n \mathbb{E}\big(\vert X_n\vert \mathbf{1}_A\big) \le \varepsilon.\)

This version is extremely useful in probability theory (e.g., Prokhorov-type ideas and martingale convergence).

8. Corollaries

From page 3:

  • If ${X_n}$ and ${Y_n}$ are UI, then ${X_n + Y_n}$ is UI.
  • If $X_n \to X$ in $L^1$, then ${X_n}$ is UI.
  • If $\sup_n \mathbb{E}\vert X_n\vert < \infty$ and $X=0$ with $\mathbb{E}\vert X_n\vert \to 0$, then ${X_n}$ is UI (detailed proof given in the notes).

9. Final Result of Lecture 11

The lecture concludes with:

If $X_n \to X$ a.s. and ${X_n}$ is UI, then not only
$\mathbb{E}\vert X_n - X\vert \to 0$,
but also
\(\lim_{M\to\infty} \phi(M) = 0,\) confirming the family is uniformly integrable.

This completes the key structural properties of UI necessary for advanced probability.

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