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Integrability and Uniform Integrability · Tutorial 152 of 946

L1 Convergence Implies Uniform Integrability

See why \(L^1\) convergence controls tail expectations uniformly, and apply the result to truncations, perturbations, and rare-event approximations.

Intermediate 10 min read

What You'll Learn

  • State uniform integrability using tail expectations.
  • Prove that an \(L^1\)-convergent sequence and its limit form a uniformly integrable family.
  • Identify the role of uniform integrability of a fixed integrable variable on events of small probability.
  • Apply the result to truncated variables and explicit \(L^1\) approximations.
  • Calculate tail expectations for a sequence of shrinking rare-event spikes.

Why \(L^1\) Convergence Controls Tails

Convergence in \(L^1\) says that the expected absolute error between a random variable and its limit becomes small. Uniform integrability asks for a different kind of control: the expected contribution from large values must become small uniformly across an entire family.

The connection between these ideas is useful because a sequence can have occasional large values even while approaching its limit. The key is that \(L^1\) convergence makes the errors small in expectation, while the fixed limit controls what remains.

Definition and a Small-Probability Estimate

Definition: A family \(\mathcal{Y}\) of integrable random variables is uniformly integrable if, for every \(\varepsilon>0\), there exists \(K>0\) such that $$ \sup_{Y\in\mathcal{Y}}\mathbb{E}\bigl[|Y|\mathbf{1}_{\{|Y|>K\}}\bigr]<\varepsilon. $$

Here \(\mathbf{1}_A\) denotes the indicator of an event \(A\). The tail expectation measures not just how often \(Y\) is large, but how much expected absolute value lies on the event where it is large.

We will use the fact that an integrable random variable has small expected contribution on events whose probabilities are small. This is stronger than merely saying that the probability of such events is small.

Lemma: If \(X\) is integrable, then for every \(\varepsilon>0\), there exists \(\delta>0\) such that every event \(A\) with \(\mathbb{P}(A)<\delta\) satisfies $$ \mathbb{E}\bigl[|X|\mathbf{1}_A\bigr]<\varepsilon. $$
Proof. Choose \(M>0\) large enough that $$ \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>M\}}\bigr]<\frac{\varepsilon}{2}. $$ Such an \(M\) exists by dominated convergence, since \(|X|\mathbf{1}_{\{|X|>M\}}\) tends pointwise to zero and is bounded by the integrable variable \(|X|\). For any event \(A\), $$ \mathbb{E}\bigl[|X|\mathbf{1}_A\bigr] \leq \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>M\}}\bigr]+M\mathbb{P}(A). $$ Set \(\delta=\varepsilon/(2M)\). If \(\mathbb{P}(A)<\delta\), the right-hand side is strictly less than \(\varepsilon\).
\(\square\)

A Preliminary Consequence of \(L^1\) Convergence

Let \(X_n\) converge to \(X\) in \(L^1\), meaning that all these random variables are integrable and $$ \mathbb{E}|X_n-X|\longrightarrow 0. $$ In particular, the sequence has bounded \(L^1\) norms, and its tail probabilities vanish uniformly.

Proposition: If \(X_n\to X\) in \(L^1\), then $$ \sup_{n\geq 1}\mathbb{E}|X_n|<\infty \qquad\text{and}\qquad \sup_{n\geq 1}\mathbb{P}(|X_n|>K)\longrightarrow 0 \quad\text{as }K\longrightarrow\infty. $$
Proof. For each \(n\), the triangle inequality gives $$ \mathbb{E}|X_n|\leq \mathbb{E}|X|+\mathbb{E}|X_n-X|. $$ The second term tends to zero, so it is bounded for all sufficiently large \(n\). The remaining finitely many expectations are finite. Thus \(C=\sup_{n\geq1}\mathbb{E}|X_n|\) is finite. Markov's inequality gives $$ \sup_{n\geq1}\mathbb{P}(|X_n|>K) \leq \frac{C}{K}, $$ which tends to zero as \(K\) tends to infinity.
\(\square\)

Uniformly small tail probabilities alone do not establish uniform integrability: a rare event can still carry a substantial expected absolute value. The theorem below uses \(L^1\) convergence to control the error in expectation as well.

The Main Theorem

Theorem: If \(X_n\to X\) in \(L^1\), then the family $$ \{X\}\cup\{X_n:n\geq1\} $$ is uniformly integrable.
Proof. Fix \(\varepsilon>0\). By \(L^1\) convergence, choose \(N\) such that $$ \mathbb{E}|X_n-X|<\frac{\varepsilon}{2} \qquad\text{for every }n\geq N. $$ By the proposition, \(C=\sup_{n\geq1}\mathbb{E}|X_n|\) is finite. For \(K>0\), let \(A_{n,K}=\{|X_n|>K\}\). Markov's inequality gives $$ \mathbb{P}(A_{n,K})\leq \frac{C}{K}. $$ Apply the lemma to the integrable variable \(X\), with tolerance \(\varepsilon/2\). It supplies \(\delta>0\) such that \(\mathbb{P}(A)<\delta\) implies \(\mathbb{E}[|X|\mathbf{1}_A]<\varepsilon/2\). Choose \(K\) large enough that \(C/K<\delta\). Then, for every \(n\geq N\), $$ \begin{aligned} \mathbb{E}\bigl[|X_n|\mathbf{1}_{A_{n,K}}\bigr] &\leq \mathbb{E}|X_n-X| +\mathbb{E}\bigl[|X|\mathbf{1}_{A_{n,K}}\bigr]\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \end{aligned} $$ This controls all indices \(n\geq N\) with the same \(K\). For each of the finitely many variables \(X_1,\ldots,X_{N-1}\), integrability and dominated convergence imply $$ \mathbb{E}\bigl[|X_j|\mathbf{1}_{\{|X_j|>K\}}\bigr]\longrightarrow 0 \quad\text{as }K\longrightarrow\infty. $$ The same is true for \(X\). Increase \(K\), if necessary, until the tail expectations of these finitely many variables and of \(X\) are all less than \(\varepsilon\). The tail expectations for \(n\geq N\) remain below \(\varepsilon\), since increasing the threshold can only decrease them. Therefore the whole family is uniformly integrable.
\(\square\)

The proof separates the sequence into two parts. For large indices, \(X_n\) is close to \(X\) in expected absolute value, and the limit controls the contribution on events of small probability. The finitely many early terms are handled individually; finiteness of that set lets one choose a common threshold for them.

Worked Example: Truncating an Exponential Random Variable

Worked Example: Truncating an Exponential Random Variable

Let \(X\) have the exponential distribution with density \(e^{-x}\) on \(x\geq0\), and define \(X_n=X\wedge n\), where \(a\wedge b=\min(a,b)\). Each \(X_n\) is bounded, and $$ |X-X_n|=(X-n)_+, $$ where \(u_+=\max(u,0)\). Its expected value is $$ \begin{aligned} \mathbb{E}|X-X_n| &=\int_n^\infty (x-n)e^{-x}\,dx\\ &=e^{-n}\int_0^\infty u e^{-u}\,du\\ &=e^{-n}. \end{aligned} $$ In the second line, \(u=x-n\); integration by parts gives \(\int_0^\infty u e^{-u}\,du=1\). Thus \(X_n\to X\) in \(L^1\). The theorem implies that the family consisting of \(X\) and all its truncations \(X_n\) is uniformly integrable. This conclusion includes the limit, even though \(X\) itself is not bounded.

Worked Example: Approximating by Small Perturbations

Worked Example: Approximating by Small Perturbations

Let \(X\) again have the exponential distribution, and put $$ Y_n=X+\frac{1}{n}\sin(nX). $$ Since \(|\sin(nX)|\leq1\), $$ \mathbb{E}|Y_n-X| =\frac{1}{n}\mathbb{E}|\sin(nX)| \leq\frac{1}{n}. $$ Therefore \(Y_n\to X\) in \(L^1\). Each \(Y_n\) is integrable because \(|Y_n|\leq |X|+1\), and \(X\) is integrable. The theorem shows that \(\{X\}\cup\{Y_n:n\geq1\}\) is uniformly integrable. No independence assumption about the perturbation is needed: it is a function of \(X\), and the deterministic bound on its size already proves the required \(L^1\) convergence.

Worked Example: Rare Spikes with Vanishing \(L^1\) Error

Worked Example: Rare Spikes with Vanishing \(L^1\) Error

Take \([0,1]\) with Lebesgue probability measure, let \(A_n=[0,n^{-2}]\), and set \(Z_n=n\mathbf{1}_{A_n}\). Then $$ \mathbb{E}|Z_n|=n\mathbb{P}(A_n)=n\cdot\frac{1}{n^2}=\frac{1}{n}, $$ so \(Z_n\to0\) in \(L^1\). For a threshold \(K>0\), the variable \(Z_n\) exceeds \(K\) precisely on \(A_n\) when \(n>K\). Consequently, $$ \mathbb{E}\bigl[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}\bigr] = \begin{cases} 1/n,& n>K,\\ 0,& n\leq K. \end{cases} $$ Hence the supremum over \(n\geq1\) is \(1/(\lfloor K\rfloor+1)\), which tends to zero as \(K\to\infty\). This calculates uniform integrability directly, in agreement with the theorem applied to \(Z_n\to0\). The spike height grows, but its expected contribution shrinks.

What the Result Does—and Does Not—Say

The theorem is a sufficient condition: \(L^1\) convergence guarantees uniform integrability of the sequence together with its limit. It does not say that every uniformly integrable sequence converges in \(L^1\); uniform integrability is a property of a family, while \(L^1\) convergence also specifies a particular limit and convergence of expected absolute errors.

A useful way to remember the proof is to examine the decomposition \(X_n=X+(X_n-X)\). On the event where \(X_n\) is large, the limit \(X\) contributes little in expectation once that event has small probability, and the error contributes little because its \(L^1\) norm is small for all sufficiently large \(n\). The finitely many remaining terms cannot obstruct uniformity.

Takeaway: \(L^1\) convergence controls both the size of the approximation error and, through the integrable limit, the expected contribution on rare events. Together these controls give uniform tail-expectation bounds.

Check Your Understanding

Use the definitions and proof strategy above to answer the following questions.

  1. What quantity must become small uniformly over a family for that family to be uniformly integrable?
  2. Why does \(L^1\) convergence imply that \(\sup_n\mathbb{E}|X_n|\) is finite?
  3. In the main proof, why is it necessary to treat the finitely many indices before \(N\) separately?
  4. For the exponential truncations, what is the exact value of \(\mathbb{E}|X-X_n|\)?
  5. For the rare-spike example, why can the spike height grow while its \(L^1\) contribution tends to zero?

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