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# A Short Note on Uniform Integrability

## Introduction

A sequence of random variables $(X_n)_{n \ge 1} \sub L^1$ is called $L^1$-convergent if there exists some limit $X \in L^1$ such that $\E|X_n - X| \to 0$ as $n \to \infty$. In this post, we briefly discuss a necessary and sufficient condition for $L^1$-convergence called uniform integrability.

## Uniform Integrability

Definition. A collection of random variables $\mathcal{F}$ is called uniformly integrable if

$$\lim_{K \to \infty} \sup\,\{ \E[\ind_{|X| \ge K} |X|] : X \in \mathcal{F} \} = 0.$$

Noting that $\E[\ind_{\abs{X} \ge K} \abs{X}] = \E\abs{X - \ind_{\abs{X} < K} X}$, this condition can also be written as

$$\lim_{K \to \infty} \sup\,\{ \E\abs{X - \ind_{\abs{X} < K} X} : X \in \mathcal{F} \} = 0.$$

In other words, if $\mathcal{F}$ is uniformly integrable, then you can choose a single value of $K > 0$ such that, uniformly over $X \in \mathcal{F}$, the random variable $\ind_{\abs{X} < K} X$ is a good approximation of $X$ in terms of the $L^1$-norm. Crucially, every $\ind_{\abs{X} < K} X$ is a bounded random variable, which is often a desirable property. Therefore, you could aptly call a family which is uniform integrable a family which allows a uniform bounded approximation.

But what about the name uniform integrability? For a single variable $X$, it is true that $$\E\abs{X} < \infty \iff \lim_{K \to \infty} \E[\ind_{|X| \ge K} |X|] = 0.$$ Hence, you could call a family of random variables uniformly integrable if the limit on the RHS, which is equivalent to integrability, converges uniformly over the family.

The bounded approximation given by uniform integrability can be made a bit nicer. Instead of bounding $X$ by applying the function $f_K(x) = \ind_{\abs{x} < K} x$, which exhibits a discontinuity at $\abs{x} = K$, uniform integrability allows us to bound $X$ by applying the nicer function $g_K(x) = \max(\min(x, K), -K)$, which is a fully continuous function: $$\E\abs{g_K(X) - X} = \E[\ind_{\abs{X} \ge K}\abs{\abs{X} - K}] \le \E[\ind_{\abs{X} \ge K}\abs{X}] + \E[\ind_{\abs{X} \ge K} K] \le 2 \E[\ind_{\abs{X} \ge K} \abs{X}],$$ which uniformly converges to zero as $K \to \infty$. Henceforth, for any random variable $X$, denote by $X^K =\max(\min(X, K), -K)$ the trunction of $X$ at level $K$. Since $g_K$ is continuous, trunctions in this way preserves limits.

Finally, to check that a family of random variables is uniformly integrable, the following two facts are very useful:

1. If $\sup\,\{ \E[\abs{X}^{p}] : X \in \mathcal{F}\} < \infty$ for some $p > 1$, then $\mathcal{F}$ is uniformly integrable.

2. Every family $\{ \E[X \cond \mathcal{G}] : \mathcal{G} \text{ is a sub-}\sigma\text{-algebra}\}$ is uniformly integrable.

## A Necessary and Sufficient Condition for $L^1$-Convergence

A standard way to prove that a sequence of random variables $(X_n)_{n \ge 1}$ is $L^1$-convergent to some limit is to use bounded convergence, an instance of the dominated convergence theorem. Recall that a sequence of random variables $(X_n)_{n \ge 1}$ is called convergent in probability if there exists a limit $X$ such that $\P(\abs{X - X_n} \ge \e) \to 0$ for every $\e > 0$.

Theorem (bounded convergence). If $(X_n)_{n \ge 1}$ and $X$ are bounded by some $K > 0$ and $X_n \to X$ in probability, then $X_n \to X$ in $L^1$.

Proof. Without loss of generality, assume that $X = 0$, so it remains to demonstrate that $\E\abs{X_n} \to 0$. Let $\e > 0$. Using the assumption that $\abs{X_n} \le K$, the idea is to consider the cases $\abs{X_n} \in [0, \e]$ and $\abs{X_n} \in (\e, K]$:

$$\E\abs{X_n} = \E[\abs{X_n} \ind_{\abs{X_n} \in [0, \e]}] + \E[\abs{X_n} \ind_{\abs{X_n} \in (\e, K]}] \le \e + K\, \E[\ind_{\abs{X_n} \in (\e, K]}] \le \e + K\, \P(\abs{X_n} \ge \e).$$

Using that assumpion that $\P(\abs{X_n} \ge \e) \to 0$ as $n \to \infty$, hence $\limsup_{n \to \infty} \E\abs{X_n} \le \e$. Since $\e > 0$ was arbitrary, this proves that $\lim_{n \to \infty} \E\abs{X_n}=0$. $$\blacksquare$$

Bounded convergence is an incredibly useful tool, but the assumption that $(X_n)_{n \ge 1}$ and $X$ are bounded can be too strong. A looser assumption is that $(X_n)_{n \ge 1}$ and $X$ uniformly allow a bounded approximation, i.e. that $(X_n)_{n \ge 1}$ (and therefore the union of $(X_n)_{n \ge 1}$ and $X$) are uniformly integrable. This looser condition turns out to not just be sufficient but also necessary.

Theorem (Vitali’s). Let $(X_n)_{n \ge 1}$ be a sequence of random variables and let $X$ be a random variable. Then (a) $(X_n)_{n \ge 1} \sub L^1$, $X \in L^1$, and $X_n \to X$ in $L^1$ if and only if (b) $(X_n)_{n \ge 1} \sub L^1$ is uniformly integrable and $X_n \to X$ in probability.

Proof. We only show the hard direction, which is that (b) implies (a). Assume that $(X_n)_{n \ge 1} \sub L^1$ is uniformly integrable and $X_n \to X$ in probability. To begin with, it is true1 that $X \in L^1$. Since $X \in L^1$, $(X_n - X)_{n \ge 1}$ is uniformly integrable and $X_n - X \to 0$ in probability in any case, so without loss of generality assume that $X = 0$.

Uniform integrability gives a uniform bounded approximation of the sequence:

$$\label{eq:uniform-approx} \lim_{K \to \infty} \sup_{n \ge 1}\, \E\abs{X_n - \ind_{\abs{X_n} < K} X_n} = 0.$$

For every $K>0$, the sequence $(\ind_{\abs{X_n} < K} X_n)_{n \ge 1}$ is bounded and $\ind_{\abs{X_n} < K} X_n \to 0$ in probability, so $\ind_{\abs{X_n} < K} X_n \to 0$ in $L^1$ by bounded convergence. The idea is to then take $K \to \infty$ to show that also $X_n \to 0$ in $L^1$. To wit, by the triangle inequality,

$$\limsup_{n \to \infty} \E\abs{X_n} \le \sup_{n \ge 1}\, \E\abs{X_n - \ind_{\abs{X_n} < K} X_n} + \limsup_{n \to \infty} \E\abs{\ind_{\abs{X_n} < K} X_n} \overset{\text{(i)}}{=} \sup_{n \ge 1}\, \E\abs{X_n - \ind_{\abs{X_n} < K} X_n}$$

where (i) follows from that $\ind_{\abs{X_n} < K} X_n \to 0$ in $L^1$ by bounded convergence. Taking $K \to \infty$ and using \eqref{eq:uniform-approx} then shows the result. $$\blacksquare$$

## Application: Strengthening of Convergence in Distribution

A sequence of random variables $(X_n)_{n \ge 1}$ is called weakly convergent if there exists a limit $X$ such that, for every $f \colon \R \to \R$ continuous and bounded, it is true that $\E[f(X_n)] \to \E[f(X)]$. A limitation of weak convergence is that it only handles bounded $f$; for example, weak convergence does not imply that $\E[X_n] \to \E[X]$. As we illustrate now, the assumption of uniform integrability can be used to strengthen the conclusion of weak convergence to include $\E[X_n] \to \E[X]$.

The key observation is as follows: if $(X_n)_{n \ge 1}$ and $X$ were bounded by some $K > 0$, then we can apply the truncation function $g_K$, which is a continuous and bounded function, to conclude that $$\E[X_n] = \E[g_K(X_n)] \to \E[g_K(X)] = \E[X].$$ Instead of assuming boundedness, now assume that $(X_n)_{n \ge 1}$ is only uniformly integrable. For all $K > 0$, consider the uniform bounded approximations $(X^K_n)_{n \ge 1}$ and $X^K$. Because the trunction operation is continuous, every $(X_n^K)_{n \ge 1}$ is still weakly convergent to $X^K$. Morever, $(X_n^K)_{n \ge 1}$ and $X^K$ are bounded by $K > 0$. The foregoing argument then shows that $\lim_{n \to \infty} \E[X_n^K] = \E[X^K].$ Therefore, $$\lim_{n \to \infty} \E[X_n] = \lim_{n \to \infty} \lim_{K \to \infty} \E[X_n^K] = \lim_{K \to \infty} \lim_{n \to \infty} \E[X_n^K] = \lim_{K \to \infty} \E[X^K] = \E[X],$$ where the interchange of limits is allowed by uniformity of the bounded approximation.

## Summary

A family of random variables is called uniformly integrable if it allows a uniform bounded approximation. Allowing a uniform bounded approximation turns out to be the right characterisation of $L^1$-convergence: a sequence is $L^1$-convergent if and only if it is uniformly integrable. Uniformly integrability is generally useful tool: if you can prove a result for bounded random variables, then you might be able to prove the result for the greater class of uniformly integrable random variables by considering a uniform bounded approximation.

Thanks to Jiri Hron for helpful comments on a draft of this post.

1. Since $X_n \to X$ in probability, $X_{n_k} \to X$ almost surely along some subsequence $(X_{n_k})_{k \ge 0}$. Therefore, using Fatou’s lemma,

$$\E\abs{X} = \E[\lim_{k \to \infty} \abs{X_{n_k}}] \le \liminf_{k \to \infty} \E[\abs{X_{n_k}}] < \infty,$$

where the right hand side is bounded because any uniformly integrable family is uniformly bounded in $L^1$.

Published on 5 August 2021.