Chapter 5: Uniform Integrability and the Vitali Convergence Theorem

Date: Oct 2, 2026

When does \(X_n\to X\) imply \(\mathbb E X_n\to\mathbb E X\) ? The escaping spike of Chapter 4 shows that convergence of the variables, even at every sample point, is not enough. The Dominated Convergence Theorem gives a sufficient condition: one integrable variable that bounds every \(\lvert X_n\rvert\) . But domination asks for more than the problem needs. Let \(A_1,A_2,\dots\) be independent events with \(\mathbb P(A_n)=1/n\) and let \(X_n=\sqrt n\,\mathbf 1_{A_n}\) . Then \(\mathbb E X_n=n^{-1/2}\to 0\) , as one would hope for a sequence tending to \(0\) in probability. Yet the events occur infinitely often, so \(\sup_nX_n=\infty\) almost surely, and no integrable variable, indeed no finite one, dominates the sequence. The expectations converge for a reason that the Dominated Convergence Theorem cannot express.

The reason is that no fixed fraction of the expectation of \(X_n\) sits on values so large that they are improbable. In the spike \(n\mathbf 1_{(0,1/n)}\) the whole expectation sits there; in \(\sqrt n\,\mathbf 1_{A_n}\) a vanishing amount does. This chapter makes that distinction into a definition, uniform integrability, and proves that it is exactly the condition that turns convergence in probability into convergence of expectations. The exactness is the point: Vitali’s theorem is an equivalence, so there is nothing weaker to look for.

5.1 The definition and its two faces

Definition 5.1 (uniform integrability). A family \((X_i)_{i\in I}\) of random variables is uniformly integrable if

\[ \lim_{M\to\infty}\ \sup_{i\in I}\ \mathbb E\bigl[\lvert X_i\rvert;\ \lvert X_i\rvert>M\bigr]=0 . \]

For a single integrable variable \(X\) , the quantity \(\mathbb E[\lvert X\rvert;\lvert X\rvert>M]\) tends to \(0\) as \(M\to\infty\) by the Dominated Convergence Theorem: it is the part of the expectation carried by values beyond \(M\) . Uniform integrability asks that this tail contribution be small for all members of the family at once, at a common level \(M\) . The definition involves each \(X_i\) only through its law, so it is a property of a family of laws; the variables need not live on one sample space.

Three kinds of families are uniformly integrable for immediate reasons. A single integrable variable is. So is a finite family of integrable variables, by taking the largest of finitely many levels. And so is a family dominated by one integrable variable: if \(\lvert X_i\rvert\le Y\) for all \(i\) , then \(\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]\le\mathbb E[Y;Y>M]\) , because the event \(\{\lvert X_i\rvert>M\}\) is contained in \(\{Y>M\}\) .

The definition has a second form that is often easier to verify and almost always easier to use.

Proposition 5.2 (two faces of uniform integrability). A family \((X_i)_{i\in I}\) on one probability space is uniformly integrable if and only if both of the following hold.

(a) It is bounded in \(L^1\) : \(\sup_i\mathbb E\lvert X_i\rvert<\infty\) .

(b) It is uniformly absolutely continuous: for every \(\varepsilon>0\) there is \(\delta>0\) such that \(\mathbb P(A)<\delta\) implies \(\sup_i\mathbb E[\lvert X_i\rvert;A]<\varepsilon\) .

Proof. Write \(\eta(M)=\sup_i\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]\) . For any event \(A\) and any level \(M\) ,

\[ \mathbb E[\lvert X_i\rvert;A]\le M\,\mathbb P(A)+\mathbb E\bigl[\lvert X_i\rvert;\lvert X_i\rvert>M\bigr]\le M\,\mathbb P(A)+\eta(M), \]

by splitting \(A\) according to whether \(\lvert X_i\rvert\le M\) . Suppose the family is uniformly integrable. Taking \(A=\Omega\) and \(M\) with \(\eta(M)\le 1\) gives \(\mathbb E\lvert X_i\rvert\le M+1\) for all \(i\) , which is (a). Given \(\varepsilon\) , choose \(M\) with \(\eta(M)<\varepsilon/2\) and put \(\delta=\varepsilon/(2M)\) ; then \(\mathbb P(A)<\delta\) gives \(\mathbb E[\lvert X_i\rvert;A]<\varepsilon\) for all \(i\) , which is (b).

Conversely assume (a) and (b), let \(C=\sup_i\mathbb E\lvert X_i\rvert\) , and fix \(\varepsilon\) with its \(\delta\) . By Markov’s inequality \(\mathbb P(\lvert X_i\rvert>M)\le C/M<\delta\) for every \(i\) as soon as \(M>C/\delta\) . Applying (b) to the event \(A=\{\lvert X_i\rvert>M\}\) gives \(\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]<\varepsilon\) for every \(i\) . So \(\eta(M)\le\varepsilon\) for all \(M>C/\delta\) .

Condition (b) says that no member of the family can hide an appreciable amount of expectation on a set of small probability. Condition (a) alone does not prevent that, as the first example shows.

Example 5.1: Spikes of every height

On \(((0,1),\mathcal B,\lambda)\) let \(X_n=n^{a}\mathbf 1_{(0,1/n)}\) for a fixed \(a>0\) . The variable \(X_n\) exceeds \(M\) exactly when \(n^{a}>M\) , and then all of its expectation is beyond \(M\) :

\[ \mathbb E[X_n;X_n>M]=\begin{cases}n^{a-1},& n^{a}>M,\\ 0,& n^{a}\le M.\end{cases} \]

Case \(a<1\) . The sequence \(n^{a-1}\) is decreasing, so the supremum over \(n\) with \(n>M^{1/a}\) is at most \(M^{(a-1)/a}\) , which tends to \(0\) . The family is uniformly integrable. For \(a=1/2\) and \(M=10\) , the indices concerned are \(n>100\) and \(\eta(10)=101^{-1/2}\approx 0.0995\) .

Case \(a=1\) . Now \(\mathbb E[X_n;X_n>M]=1\) for every \(n>M\) , so \(\eta(M)=1\) for all \(M\) . The family is bounded in \(L^1\) , with \(\mathbb E X_n=1\) , and is not uniformly integrable. Face (b) of Proposition 5.2 fails in the plainest way: the event \(A=(0,1/n)\) has probability \(1/n\) , as small as we please, and \(\mathbb E[X_n;A]=1\) .

Case \(a>1\) . Here \(\mathbb E X_n=n^{a-1}\to\infty\) ; the family is not even bounded in \(L^1\) .

So the threshold for uniform integrability is the same threshold \(a<1\) at which \(\mathbb E X_n\to 0\) . That is no coincidence, as Section 5.3 will show.

5.2 Criteria

Definition 5.1 is rarely checked by computing \(\eta(M)\) . Usually one bounds a slightly stronger moment.

Proposition 5.3 (a bounded higher moment). If \(\sup_i\mathbb E\lvert X_i\rvert^p<\infty\) for some \(p>1\) , the family \((X_i)\) is uniformly integrable.

Proof. On the event \(\{\lvert X_i\rvert>M\}\) we have \(\lvert X_i\rvert=\lvert X_i\rvert^p\,\lvert X_i\rvert^{1-p}\le\lvert X_i\rvert^p\,M^{1-p}\) . Hence \(\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]\le M^{1-p}\sup_j\mathbb E\lvert X_j\rvert^p\) , which tends to \(0\) as \(M\to\infty\) uniformly in \(i\) .

The proposition explains the opening example: \(\mathbb E(\sqrt n\,\mathbf 1_{A_n})^2=1\) for all \(n\) . The power function is only one of many possible gauges, and the following theorem says that some gauge always exists.

Theorem 5.4 (de la Vallée-Poussin criterion). A family \((X_i)_{i\in I}\) is uniformly integrable if and only if there is a nondecreasing convex function \(\Phi:[0,\infty)\to[0,\infty)\) with \(\Phi(x)/x\to\infty\) as \(x\to\infty\) and \(\sup_i\mathbb E\,\Phi(\lvert X_i\rvert)<\infty\) .

Proof. Suppose such a \(\Phi\) exists and let \(K=1+\sup_i\mathbb E\,\Phi(\lvert X_i\rvert)\) , so that \(K>0\) . Given \(\varepsilon>0\) choose \(M\) so that \(\Phi(x)/x\ge K/\varepsilon\) for all \(x>M\) . On \(\{\lvert X_i\rvert>M\}\) we then have \(\lvert X_i\rvert\le(\varepsilon/K)\,\Phi(\lvert X_i\rvert)\) , so \(\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]\le(\varepsilon/K)\,\mathbb E\,\Phi(\lvert X_i\rvert)\le\varepsilon\) for every \(i\) .

Conversely let the family be uniformly integrable, with \(\eta(M)\) as in the proof of Proposition 5.2. Choose levels \(M_1<M_2<\cdots\) tending to infinity with \(\eta(M_k)\le 2^{-k}\) , and define

\[ \Phi(x)=\sum_{k=1}^\infty(x-M_k)^{+}. \]

For each \(x\) only finitely many terms are nonzero, so \(\Phi\) is finite; it is nondecreasing and convex because each term is. If \(x\ge 2M_k\) , then \((x-M_j)^{+}\ge x/2\) for each \(j\le k\) , so \(\Phi(x)\ge kx/2\) ; hence \(\Phi(x)/x\to\infty\) . Finally \((\lvert X_i\rvert-M_k)^{+}\le\lvert X_i\rvert\mathbf 1_{\{\lvert X_i\rvert>M_k\}}\) , so by the Monotone Convergence Theorem

\[ \mathbb E\,\Phi(\lvert X_i\rvert)=\sum_k\mathbb E(\lvert X_i\rvert-M_k)^{+}\le\sum_k\eta(M_k)\le\sum_k2^{-k}=1 \]

for every \(i\) .

The criterion places uniform integrability precisely: it is boundedness in a class slightly smaller than \(L^1\) , where “slightly” depends on the family. No single \(\Phi\) serves for all uniformly integrable families (Exercise 16).

Proposition 5.5 (stability). If \((X_i)_{i\in I}\) and \((Y_i)_{i\in I}\) are uniformly integrable families on one probability space, so is \((X_i+Y_i)_{i\in I}\) . If \((X_n)_{n\ge 1}\) is uniformly integrable, so is the sequence of averages \(\bar X_n=\tfrac1n(X_1+\dots+X_n)\) .

Proof. Use the two faces of Proposition 5.2. For sums, \(\sup_i\mathbb E\lvert X_i+Y_i\rvert\le\sup_i\mathbb E\lvert X_i\rvert+\sup_i\mathbb E\lvert Y_i\rvert\) , and given \(\varepsilon\) , if \(\delta\) serves both families for \(\varepsilon/2\) then \(\mathbb P(A)<\delta\) gives \(\mathbb E[\lvert X_i+Y_i\rvert;A]\le\mathbb E[\lvert X_i\rvert;A]+\mathbb E[\lvert Y_i\rvert;A]<\varepsilon\) . For averages, \(\mathbb E[\lvert\bar X_n\rvert;A]\le\tfrac1n\sum_{k\le n}\mathbb E[\lvert X_k\rvert;A]\le\sup_k\mathbb E[\lvert X_k\rvert;A]\) for every event \(A\) , including \(A=\Omega\) , so both faces pass from \((X_n)\) to \((\bar X_n)\) .

Trying to prove the statement about sums from Definition 5.1 directly is instructive: the event \(\{\lvert X_i+Y_i\rvert>M\}\) is not controlled by the events \(\{\lvert X_i\rvert>M\}\) and \(\{\lvert Y_i\rvert>M\}\) in any convenient way, whereas face (b) quantifies over all events and so does not care which event is at hand.

Example 5.2: A uniformly integrable family of exponential laws

Let \(X_a\) be exponential with rate \(a\) , with density \(ae^{-ax}\) on \((0,\infty)\) . Integrating by parts,

\[ \mathbb E[X_a;X_a>M]=\int_M^\infty x\,a\,e^{-ax}\,dx=\Bigl(M+\frac1a\Bigr)e^{-aM}. \]

For the family \(\{X_a:a\ge 1\}\) this is largest at \(a=1\) , since both factors decrease in \(a\) , so

\[ \eta(M)=(M+1)e^{-M}\longrightarrow 0, \]

and the family is uniformly integrable; for instance \(\eta(5)=6e^{-5}\approx 0.0404\) . The family \(\{X_a:0<a\le 1\}\) is not: \(\mathbb E X_a=1/a\) is unbounded, so face (a) fails. Proposition 5.3 gives the first conclusion without integration, since \(\mathbb E X_a^2=2/a^2\le 2\) for \(a\ge 1\) .

Example 5.3: Integrable, but in no smaller power class

Uniform integrability can hold when no power \(p>1\) is available. Let \(X\ge e\) have the tail

\[ \mathbb P(X>x)=\frac{e}{x(\log x)^2},\qquad x\ge e . \]

By the tail formula (Proposition 1.12) and the substitution \(u=\log x\) ,

\[ \mathbb E X=e+\int_e^\infty\frac{e}{x(\log x)^2}\,dx=e+e\int_1^\infty\frac{du}{u^2}=2e\approx 5.437 . \]

So \(X\) is integrable, and the one-member family \(\{X\}\) is uniformly integrable. But for every \(p>1\) ,

\[ \mathbb E X^p\ge\int_e^\infty p\,x^{p-1}\,\frac{e}{x(\log x)^2}\,dx=p\,e\int_1^\infty\frac{e^{(p-1)u}}{u^2}\,du=\infty , \]

so Proposition 5.3 never applies. Theorem 5.4 promises a gauge all the same, and here one can be written down: \(\Phi(x)=x\sqrt{\log x}\) for \(x\ge e\) , continued by \(\Phi(x)=x\) on \([0,e]\) . It is convex and nondecreasing, \(\Phi(x)/x=\sqrt{\log x}\to\infty\) , and since \(\Phi'(x)=\sqrt{\log x}+\tfrac{1}{2\sqrt{\log x}}\) for \(x>e\) ,

\[ \mathbb E\,\Phi(X)=\Phi(e)+\int_e^\infty\Phi'(x)\,\mathbb P(X>x)\,dx=e+e\int_1^\infty\Bigl(u^{-3/2}+\tfrac12u^{-5/2}\Bigr)du=e+\frac{7e}{3}=\frac{10e}{3}\approx 9.061 . \]

The gauge \(x\log x\) would already be too strong: the same computation gives \(\int_1^\infty u^{-1}\,du=\infty\) . The room between \(L^1\) and the nearest usable gauge can be arbitrarily narrow.

Example 5.4: Uniformly integrable with no dominating variable

Return to the opening example: \(A_1,A_2,\dots\) independent with \(\mathbb P(A_n)=1/n\) and \(X_n=\sqrt n\,\mathbf 1_{A_n}\) . We have \(\mathbb E X_n^2=n\cdot\tfrac1n=1\) for all \(n\) , so the sequence is uniformly integrable by Proposition 5.3. Directly, \(\mathbb E[X_n;X_n>M]\) equals \(n^{-1/2}\) when \(n>M^2\) and \(0\) otherwise, so \(\eta(M)<1/M\) .

No integrable variable dominates the sequence. Any \(Y\) with \(Y\ge X_n\) for all \(n\) satisfies \(Y\ge\sup_nX_n\) , and for \(t>0\) and integers \(N>n>t^2\) ,

\[ \mathbb P\Bigl(\max_{n\le k\le N}X_k\le t\Bigr)=\mathbb P\Bigl(\bigcap_{k=n}^{N}A_k^{c}\Bigr)=\prod_{k=n}^{N}\Bigl(1-\frac1k\Bigr)=\frac{n-1}{N}\longrightarrow 0\quad(N\to\infty), \]

so \(\sup_nX_n>t\) almost surely for every \(t\) ; that is, \(\sup_nX_n=\infty\) almost surely. For the same reason the sequence does not converge almost surely: \(X_n\ge 1\) infinitely often. It does converge to \(0\) in probability, since \(\mathbb P(X_n\ne 0)=1/n\) .

So here is a sequence that converges in probability and not almost surely, has no dominating variable, and nevertheless satisfies \(\mathbb E X_n=n^{-1/2}\to 0\) . Neither the Dominated Convergence Theorem nor its extension to convergence in probability (Corollary 4.6) can account for that limit. The theorem of the next section does.

5.3 The Vitali convergence theorem

Theorem 5.6 (Vitali convergence theorem). Let \(X_1,X_2,\dots\) be integrable random variables with \(X_n\to X\) in probability. The following are equivalent.

(i) The sequence \((X_n)\) is uniformly integrable.

(ii) \(X\) is integrable and \(\mathbb E\lvert X_n-X\rvert\to 0\) .

(iii) \(X\) is integrable and \(\mathbb E\lvert X_n\rvert\to\mathbb E\lvert X\rvert\) .

Proof. (i) implies (ii). By Theorem 4.4 some subsequence converges to \(X\) almost surely, and Fatou’s lemma of Measure Theory, valid for nonnegative measurable functions, gives \(\mathbb E\lvert X\rvert\le\liminf_k\mathbb E\lvert X_{n_k}\rvert\le\sup_n\mathbb E\lvert X_n\rvert\) , which is finite by Proposition 5.2. So \(X\) is integrable. The constant sequence \((X)\) is then uniformly integrable, and by Proposition 5.5 so is \((X_n-X)\) . Fix \(\varepsilon>0\) and let \(\delta\) be as in face (b) of Proposition 5.2 for the family \((X_n-X)\) . With \(A_n=\{\lvert X_n-X\rvert>\varepsilon\}\) ,

\[ \mathbb E\lvert X_n-X\rvert\le\varepsilon+\mathbb E\bigl[\lvert X_n-X\rvert;A_n\bigr]. \]

Convergence in probability gives \(\mathbb P(A_n)<\delta\) for all large \(n\) , and then the last term is less than \(\varepsilon\) . Hence \(\limsup_n\mathbb E\lvert X_n-X\rvert\le 2\varepsilon\) .

(ii) implies (iii). \(\bigl\lvert\mathbb E\lvert X_n\rvert-\mathbb E\lvert X\rvert\bigr\rvert\le\mathbb E\lvert X_n-X\rvert\) .

(iii) implies (i). For \(M\ge 1\) let \(\psi_M:[0,\infty)\to[0,\infty)\) be the continuous function equal to \(x\) on \([0,M-1]\) , equal to \(0\) on \([M,\infty)\) , and linear on \([M-1,M]\) . It is bounded by \(M\) and satisfies

\[ x\,\mathbf 1_{\{x>M\}}\ \le\ x-\psi_M(x)\ \le\ x\,\mathbf 1_{\{x>M-1\}}\qquad(x\ge 0). \]

By Corollary 4.6, \(\psi_M(\lvert X_n\rvert)\to\psi_M(\lvert X\rvert)\) in probability, and since these variables are bounded by \(M\) , their expectations converge. Using the left inequality, hypothesis (iii), and then the right inequality,

\[ \limsup_n\mathbb E\bigl[\lvert X_n\rvert;\lvert X_n\rvert>M\bigr]\le\lim_n\Bigl(\mathbb E\lvert X_n\rvert-\mathbb E\,\psi_M(\lvert X_n\rvert)\Bigr)=\mathbb E\lvert X\rvert-\mathbb E\,\psi_M(\lvert X\rvert)\le\mathbb E\bigl[\lvert X\rvert;\lvert X\rvert>M-1\bigr]. \]

Given \(\varepsilon>0\) , integrability of \(X\) provides \(M_0\) with \(\mathbb E[\lvert X\rvert;\lvert X\rvert>M_0-1]<\varepsilon\) , and then there is \(n_0\) with \(\mathbb E[\lvert X_n\rvert;\lvert X_n\rvert>M_0]<\varepsilon\) for all \(n\ge n_0\) . The finitely many integrable variables \(X_1,\dots,X_{n_0-1}\) admit a level \(M_1\ge M_0\) with \(\mathbb E[\lvert X_n\rvert;\lvert X_n\rvert>M_1]<\varepsilon\) for \(n<n_0\) . Since tail contributions decrease as the level rises, \(\sup_n\mathbb E[\lvert X_n\rvert;\lvert X_n\rvert>M]\le\varepsilon\) for all \(M\ge M_1\) .

Corollary 5.7 (limits of moments). Let \(X_n\to X\) in probability. If \((X_n)\) is uniformly integrable, then \(\mathbb E X_n\to\mathbb E X\) . More generally, for \(p\ge 1\) and \(X_n\in L^p\) , \(X_n\to X\) in \(L^p\) if and only if the family \((\lvert X_n\rvert^p)\) is uniformly integrable.

Proof. The first statement follows from (ii), since \(\lvert\mathbb E X_n-\mathbb E X\rvert\le\mathbb E\lvert X_n-X\rvert\) . For the second, note that \(\lvert X_n\rvert^p\to\lvert X\rvert^p\) in probability by Corollary 4.6. If \(X_n\to X\) in \(L^p\) , then \(\lVert X_n\rVert_p\to\lVert X\rVert_p\) by Minkowski’s inequality, so \(\mathbb E\lvert X_n\rvert^p\to\mathbb E\lvert X\rvert^p<\infty\) , and the implication from (iii) to (i), applied to the variables \(\lvert X_n\rvert^p\) , gives uniform integrability. Conversely, if \((\lvert X_n\rvert^p)\) is uniformly integrable then \(\lvert X\rvert^p\) is integrable by the first step of the theorem, and \(\lvert X_n-X\rvert^p\le 2^{p-1}(\lvert X_n\rvert^p+\lvert X\rvert^p)\) shows that \((\lvert X_n-X\rvert^p)\) is uniformly integrable; it tends to \(0\) in probability, so its expectation tends to \(0\) by the implication from (i) to (ii).

Vitali’s theorem contains the Dominated Convergence Theorem for convergence in probability, since a dominated family is uniformly integrable, and it cannot be improved, since it is an equivalence. Condition (iii) deserves notice for its economy: to upgrade convergence in probability to convergence in \(L^1\) it is enough that one sequence of numbers, \(\mathbb E\lvert X_n\rvert\) , converge to the right limit. For nonnegative variables this says that convergence of the means to the mean of the limit already forces convergence in \(L^1\) , a fact known as Scheffé’s lemma.

In Example 5.1 the three conditions can be compared directly. The sequence \(X_n=n^{a}\mathbf 1_{(0,1/n)}\) tends to \(0\) in probability for every \(a\) ; it is uniformly integrable exactly for \(a<1\) ; and \(\mathbb E\lvert X_n\rvert=n^{a-1}\) tends to \(\mathbb E\lvert 0\rvert=0\) exactly for \(a<1\) . For the sequence of Example 5.4 the theorem gives \(\mathbb E\lvert X_n\rvert\to 0\) from uniform integrability, with no domination available.

Example 5.5: Densities that converge in total variation

Let \(f_n(x)=(1+\tfrac1n)x^{1/n}\) on \((0,1)\) . Each \(f_n\) is a probability density, and \(f_n(x)\to 1\) for every \(x\in(0,1)\) . Regard \(f_n\) as a random variable on \(((0,1),\mathcal B,\lambda)\) . It converges almost surely to the constant \(1\) , and \(\mathbb E f_n=1=\mathbb E\,1\) for every \(n\) . Condition (iii) of Theorem 5.6 holds, so (ii) holds: \(\int_0^1\lvert f_n-1\rvert\,dx\to 0\) .

The integral can be computed to confirm this. The density \(f_n\) crosses the level \(1\) at \(x_n=(\tfrac{n}{n+1})^{n}\) , and since \(\int_0^1(f_n-1)\,dx=0\) the areas on either side of the crossing are equal:

\[ \int_0^1\lvert f_n-1\rvert\,dx=2\int_{x_n}^{1}(f_n-1)\,dx=2\bigl(x_n-x_n^{1+1/n}\bigr)=\frac{2x_n}{n+1}, \]

using \(x_n^{1/n}=\tfrac{n}{n+1}\) . For \(n=1\) this is \(\int_0^1\lvert 2x-1\rvert\,dx=\tfrac12\) ; for \(n=9\) it is \(\tfrac15(0.9)^9\approx 0.0775\) ; and as \(n\to\infty\) , \(x_n\to e^{-1}\) and the integral behaves like \(2/(en)\) .

Contrast the densities \(g_n=n\mathbf 1_{(0,1/n)}\) . They converge to \(0\) at every point, their integrals are all \(1\) , and the integral of the limit is \(0\) . Condition (iii) fails, the limit is not a density, and \(\int\lvert g_n-0\rvert\,dx=1\) for all \(n\) . The mass of \(g_n\) has escaped; the mass of \(f_n\) has not.

Common mistake. Taking boundedness in \(L^1\) for uniform integrability. The argument “\(\mathbb E\lvert X_n\rvert\le 1\) for all \(n\) and \(X_n\to X\) almost surely, so \(\mathbb E X_n\to\mathbb E X\) ” is false, and the spike \(n\mathbf 1_{(0,1/n)}\) refutes it. From a bound on \(\mathbb E\lvert X_n\rvert\) one gets only Fatou’s inequality \(\mathbb E\lvert X\rvert\le\liminf_n\mathbb E\lvert X_n\rvert\) , which can be strict. The question to ask is the one in face (b) of Proposition 5.2: can a member of the family carry a fixed amount of expectation on an event of arbitrarily small probability? If it can, the family is not uniformly integrable, however small the expectations are. A bound on \(\mathbb E\lvert X_n\rvert^p\) for some \(p>1\) , or on \(\mathbb E\,\Phi(\lvert X_n\rvert)\) for a gauge \(\Phi\) , answers the question; a bound on \(\mathbb E\lvert X_n\rvert\) does not.

5.4 Where the condition returns

Uniform integrability is a tool that later chapters reach for whenever a limit must be taken inside an expectation without a dominating variable in sight. Three such places are worth announcing, so that the definition is recognised when it reappears.

In Chapter 6, a single integrable variable \(X\) generates a large family: its conditional expectations given all possible sub-sigma-algebras. That family is uniformly integrable, although its members are in general not dominated by any integrable variable that is simple to name. In Chapter 7 a martingale that is bounded in \(L^1\) converges almost surely, but its limit may have the wrong expectation; the martingales for which nothing is lost in the limit are exactly the uniformly integrable ones, and for them the limit recovers the whole martingale. In Chapter 8 the averages of an i.i.d. integrable sequence are uniformly integrable by Proposition 5.5, so the law of large numbers holds in \(L^1\) as well as almost surely.

All three uses involve conditioning on information. The next chapter makes sense of conditioning when the conditioning event has probability zero, which is the usual case and the one that the elementary definition cannot handle.

Exercises

Prove directly from Definition 5.1 that a finite family of integrable random variables is uniformly integrable and that the union of two uniformly integrable families is uniformly integrable.

Hint

For finitely many integrable variables, each \(\mathbb E[\lvert X_i\rvert;\lvert X_i\rvert>M]\) tends to \(0\) by dominated convergence, and a maximum of finitely many null sequences is null. For a union, \(\eta(M)\) of the union is the larger of the two values of \(\eta(M)\) .
On \(((0,1),\mathcal B,\lambda)\) let \(X_n=n^{a}\mathbf 1_{(0,n^{-b})}\) with \(a,b>0\) . For which pairs \((a,b)\) is the sequence bounded in \(L^1\) , and for which is it uniformly integrable?

Answer

\(\mathbb E X_n=n^{a-b}\) , so the sequence is bounded in \(L^1\) exactly when \(a\le b\) . It is uniformly integrable exactly when \(a<b\) : for \(a<b\) , \(\eta(M)\le M^{(a-b)/a}\to 0\) , while for \(a=b\) every term with \(n^{a}>M\) contributes \(1\) .
Prove the statement about sums in Proposition 5.5 in full, and explain why a direct argument from Definition 5.1 is awkward.

Solution

Let \((X_i)\) and \((Y_i)\) be uniformly integrable, indexed by the same set. By Proposition 5.2 each family is bounded in \(L^1\) and uniformly absolutely continuous, and it suffices to verify these two properties for the sums.

Boundedness. \(\sup_i\mathbb E\lvert X_i+Y_i\rvert\le\sup_i\mathbb E\lvert X_i\rvert+\sup_i\mathbb E\lvert Y_i\rvert<\infty\) .

Uniform absolute continuity. Let \(\varepsilon>0\) . There are \(\delta_1,\delta_2>0\) such that \(\mathbb P(A)<\delta_1\) implies \(\mathbb E[\lvert X_i\rvert;A]<\varepsilon/2\) for all \(i\) , and \(\mathbb P(A)<\delta_2\) implies \(\mathbb E[\lvert Y_i\rvert;A]<\varepsilon/2\) for all \(i\) . With \(\delta=\min(\delta_1,\delta_2)\) and \(\mathbb P(A)<\delta\) ,

\[ \mathbb E[\lvert X_i+Y_i\rvert;A]\le\mathbb E[\lvert X_i\rvert;A]+\mathbb E[\lvert Y_i\rvert;A]<\varepsilon\qquad\text{for all }i . \]

So the sums are uniformly integrable. A direct argument from Definition 5.1 has to bound \(\mathbb E[\lvert X_i+Y_i\rvert;\lvert X_i+Y_i\rvert>M]\) , and the event in question is not contained in \(\{\lvert X_i\rvert>M\}\cup\{\lvert Y_i\rvert>M\}\) in a way that splits the integrand; one is led to compare with \(\{\lvert X_i\rvert>M/2\}\cup\{\lvert Y_i\rvert>M/2\}\) and then still has the cross terms \(\mathbb E[\lvert X_i\rvert;\lvert Y_i\rvert>M/2]\) to control, which is exactly a statement of type (b).
Let \((X_n)\) be uniformly integrable. Prove that \(\tfrac1n\,\mathbb E\max_{k\le n}\lvert X_k\rvert\to 0\) .

Solution

Let \(\eta(M)=\sup_k\mathbb E[\lvert X_k\rvert;\lvert X_k\rvert>M]\) . For every \(k\) and every level \(M\) , \(\lvert X_k\rvert\le M+\lvert X_k\rvert\mathbf 1_{\{\lvert X_k\rvert>M\}}\) , so

\[ \max_{k\le n}\lvert X_k\rvert\le M+\sum_{k\le n}\lvert X_k\rvert\,\mathbf 1_{\{\lvert X_k\rvert>M\}} . \]

The maximum has been bounded by a sum only on the part where the variables are large, which is where the sum is affordable. Taking expectations and dividing by \(n\) ,

\[ \frac1n\,\mathbb E\max_{k\le n}\lvert X_k\rvert\le\frac Mn+\eta(M). \]

Let \(n\to\infty\) with \(M\) fixed to get \(\limsup_n\tfrac1n\mathbb E\max_{k\le n}\lvert X_k\rvert\le\eta(M)\) , and then let \(M\to\infty\) ; the right side tends to \(0\) by uniform integrability. Without uniform integrability the claim can fail even for bounded expectations: for independent copies of the spikes the maximum can be of order \(n\) .
Let \(X_n\ge 0\) with \(\mathbb E X_n=1\) for all \(n\) and \(X_n\to 0\) in probability. Prove that \((X_n)\) is not uniformly integrable, and say which of the three conditions of Theorem 5.6 your argument uses.

Answer

If the sequence were uniformly integrable, Theorem 5.6 would give \(\mathbb E\lvert X_n\rvert\to\mathbb E\lvert 0\rvert=0\) , contradicting \(\mathbb E X_n=1\) . The argument uses the implication from (i) to (iii).
Show that the family of all random variables \(X\) with \(\mathbb E[\lvert X\rvert\log^{+}\lvert X\rvert]\le 1\) is uniformly integrable, where \(\log^{+}x=\max(\log x,0)\) , and that it is not bounded in \(L^p\) for any \(p>1\) .

Answer

Apply Theorem 5.4 with \(\Phi(x)=x\log^{+}x\) , which is nondecreasing and convex with \(\Phi(x)/x\to\infty\) . The variable of Example 5.3 modified to have tail \(e/(x(\log x)^3)\) , suitably scaled, lies in the family and in no \(L^p\) with \(p>1\) .
Let \(X_\alpha\) have density \(\alpha x^{-\alpha-1}\) on \([1,\infty)\) , for \(\tfrac32\le\alpha\le 2\) . Compute \(\mathbb E[X_\alpha;X_\alpha>M]\) , find \(\eta(M)\) for this family, and decide whether the family is uniformly integrable and whether it is bounded in \(L^2\) .

Solution

For \(M\ge 1\) ,

\[ \mathbb E[X_\alpha;X_\alpha>M]=\int_M^\infty x\cdot\alpha x^{-\alpha-1}\,dx=\frac{\alpha}{\alpha-1}\,M^{1-\alpha}. \]

On \(\tfrac32\le\alpha\le 2\) the factor \(\alpha/(\alpha-1)\) decreases from \(3\) to \(2\) , and for \(M\ge 1\) the factor \(M^{1-\alpha}\) also decreases in \(\alpha\) . Both are largest at \(\alpha=\tfrac32\) , so

\[ \eta(M)=\sup_{3/2\le\alpha\le 2}\frac{\alpha}{\alpha-1}M^{1-\alpha}=3M^{-1/2}\longrightarrow 0 . \]

The family is uniformly integrable. It is not bounded in \(L^2\) ; in fact no member has a finite second moment, because \(\mathbb E X_\alpha^2=\int_1^\infty\alpha x^{1-\alpha}\,dx=\infty\) for \(\alpha\le 2\) . Proposition 5.3 does apply with a smaller power: for \(p=\tfrac54\) , \(\mathbb E X_\alpha^{5/4}=\alpha/(\alpha-\tfrac54)\le\tfrac32/\tfrac14=6\) on the whole range. The example shows that the proposition should be tried with \(p\) close to \(1\) before concluding that it does not apply.
A student argues: “The variables satisfy \(\mathbb E\lvert X_n\rvert\le 1\) and \(X_n\to X\) almost surely, so by dominated convergence \(\mathbb E X_n\to\mathbb E X\) .” Identify the hypothesis of the Dominated Convergence Theorem that has been replaced by a weaker one, show by example that the conclusion fails, and state the conclusion that the given hypotheses do support.

Solution

The Dominated Convergence Theorem requires a single integrable variable \(Y\) with \(\lvert X_n\rvert\le Y\) for all \(n\) , which amounts to \(\mathbb E\sup_n\lvert X_n\rvert<\infty\) . The student has only \(\sup_n\mathbb E\lvert X_n\rvert\le 1\) : the supremum has been moved outside the expectation, and that is a strictly weaker statement. The spike \(X_n=n\mathbf 1_{(0,1/n)}\) on \((0,1)\) shows the conclusion can fail: \(\mathbb E\lvert X_n\rvert=1\) , \(X_n\to 0\) at every point, and \(\mathbb E X_n=1\) does not tend to \(0\) . Here \(\sup_nX_n(\omega)\) is the largest integer below \(1/\omega\) , which is not integrable, so no dominating variable exists.

What the hypotheses do support is Fatou’s lemma: \(\mathbb E\lvert X\rvert\le\liminf_n\mathbb E\lvert X_n\rvert\le 1\) , so the limit is integrable, with an inequality that may be strict. To conclude \(\mathbb E X_n\to\mathbb E X\) one needs uniform integrability (Theorem 5.6), for which a bound on \(\mathbb E\lvert X_n\rvert^p\) with some \(p>1\) would suffice.
Prove directly, without Theorem 5.6, that if \(X_n\to X\) in \(L^1\) then \((X_n)\) is uniformly integrable, by handling finitely many indices separately from the rest.

Hint

Given \(\varepsilon\) , choose \(n_0\) with \(\mathbb E\lvert X_n-X\rvert<\varepsilon\) for \(n\ge n_0\) . The finite family \(X,X_1,\dots,X_{n_0}\) is uniformly integrable; let \(\delta\) serve it in face (b) of Proposition 5.2. For \(n\ge n_0\) and \(\mathbb P(A)<\delta\) , \(\mathbb E[\lvert X_n\rvert;A]\le\mathbb E\lvert X_n-X\rvert+\mathbb E[\lvert X\rvert;A]<2\varepsilon\) .
Let \(f_n\) and \(f\) be probability densities on \(\mathbb R\) with \(f_n\to f\) almost everywhere. Prove that \(\int\lvert f_n-f\rvert\,dx\to 0\) by applying the Dominated Convergence Theorem to \((f-f_n)^{+}\) .

Solution

Since \(f_n\) and \(f\) both integrate to \(1\) , \(\int(f-f_n)\,dx=0\) , which says that the positive and negative parts of \(f-f_n\) have equal integrals. Hence

\[ \int\lvert f_n-f\rvert\,dx=\int(f-f_n)^{+}\,dx+\int(f-f_n)^{-}\,dx=2\int(f-f_n)^{+}\,dx . \]

Now \(0\le(f-f_n)^{+}\le f\) , because \(f_n\ge 0\) , and \(f\) is integrable. Also \((f-f_n)^{+}\to 0\) almost everywhere, since \(f_n\to f\) . The Dominated Convergence Theorem, with dominating function \(f\) , gives \(\int(f-f_n)^{+}\,dx\to 0\) , and therefore \(\int\lvert f_n-f\rvert\,dx\to 0\) .

The proof needs no domination of the \(f_n\) themselves, which may be unbounded and have no common integrable majorant; only the part of \(f_n\) lying below \(f\) is controlled, and the part above is then forced by the equality of total masses. This is the implication from (iii) to (ii) of Theorem 5.6 in the special case of nonnegative functions, proved by hand.
Suppose \(X_n\to X\) in distribution and \((X_n)\) is uniformly integrable. Prove that \(X\) is integrable and \(\mathbb E X_n\to\mathbb E X\) , using only the definition of convergence in distribution through bounded continuous functions.

Solution

Let \(C=\sup_n\mathbb E\lvert X_n\rvert\) , finite by Proposition 5.2, and \(\eta(M)=\sup_n\mathbb E[\lvert X_n\rvert;\lvert X_n\rvert>M]\) .

Integrability of \(X\) . For each \(M\) the function \(x\mapsto\lvert x\rvert\wedge M\) is bounded and continuous, so \(\mathbb E[\lvert X\rvert\wedge M]=\lim_n\mathbb E[\lvert X_n\rvert\wedge M]\le C\) . Letting \(M\uparrow\infty\) , the Monotone Convergence Theorem gives \(\mathbb E\lvert X\rvert\le C\) .

Convergence of the means. Let \(\varphi_M(x)=\max(-M,\min(x,M))\) , a bounded continuous function with \(\lvert x-\varphi_M(x)\rvert\le\lvert x\rvert\mathbf 1_{\{\lvert x\rvert>M\}}\) . Then

\[ \lvert\mathbb E X_n-\mathbb E X\rvert\le\mathbb E\lvert X_n-\varphi_M(X_n)\rvert+\lvert\mathbb E\varphi_M(X_n)-\mathbb E\varphi_M(X)\rvert+\mathbb E\lvert\varphi_M(X)-X\rvert . \]

The first term is at most \(\eta(M)\) for every \(n\) . The second tends to \(0\) as \(n\to\infty\) by convergence in distribution. The third is at most \(\mathbb E[\lvert X\rvert;\lvert X\rvert>M]\) . Hence \(\limsup_n\lvert\mathbb E X_n-\mathbb E X\rvert\le\eta(M)+\mathbb E[\lvert X\rvert;\lvert X\rvert>M]\) for every \(M\) , and both terms tend to \(0\) as \(M\to\infty\) , the first by uniform integrability and the second by integrability of \(X\) . Only laws were used, as must be the case for a statement about convergence in distribution.
Let \(X_1,X_2,\dots\) be identically distributed and integrable, with no independence assumed. Prove that the averages \(\tfrac1n(X_1+\dots+X_n)\) form a uniformly integrable sequence.

Hint

A family of identically distributed integrable variables has the same \(\eta(M)\) as a single one of them, so it is uniformly integrable. Then apply the statement about averages in Proposition 5.5.
Let \(X_n\to X\) in probability with \(\sup_n\mathbb E X_n^2<\infty\) . Prove that \(\mathbb E\lvert X_n\rvert^p\to\mathbb E\lvert X\rvert^p\) for every \(1\le p<2\) , and give an example in which \(\mathbb E X_n^2\) does not converge to \(\mathbb E X^2\) .

Solution

Let \(C=\sup_n\mathbb E X_n^2\) and fix \(1\le p<2\) . The variables \(\lvert X_n\rvert^p\) satisfy \(\mathbb E(\lvert X_n\rvert^p)^{2/p}=\mathbb E X_n^2\le C\) with exponent \(2/p>1\) , so the family \((\lvert X_n\rvert^p)\) is uniformly integrable by Proposition 5.3. By Corollary 4.6, \(\lvert X_n\rvert^p\to\lvert X\rvert^p\) in probability. Theorem 5.6, from (i) to (iii), applied to these variables gives that \(\lvert X\rvert^p\) is integrable and \(\mathbb E\lvert X_n\rvert^p\to\mathbb E\lvert X\rvert^p\) . With \(p=1\) and Corollary 5.7, also \(\mathbb E X_n\to\mathbb E X\) .

At the endpoint \(p=2\) the argument breaks, because a bound on \(\mathbb E X_n^2\) gives no higher moment for the family \((X_n^2)\) . An example: \(X_n=\sqrt n\,\mathbf 1_{(0,1/n)}\) on \((0,1)\) . It tends to \(0\) at every point, \(\mathbb E X_n^2=1\) for all \(n\) , and \(\mathbb E X^2=0\) . For \(p<2\) , in agreement with the first part, \(\mathbb E X_n^p=n^{p/2-1}\to 0\) .
Prove that \((X_i)\) is uniformly integrable if and only if \(\sup_i\mathbb E(\lvert X_i\rvert-M)^{+}\to 0\) as \(M\to\infty\) .

Hint

For \(x\ge 0\) , \((x-M)^{+}\le x\mathbf 1_{\{x>M\}}\le 2(x-M/2)^{+}\) ; the second inequality holds because \(x>M\) implies \(x-M/2>x/2\) .
Let \(X\) be integrable. Prove that the family \(\{X\mathbf 1_A:A\in\mathcal F\}\) is uniformly integrable, and deduce that \(\mathbb E[X;A_n]\to 0\) whenever \(\mathbb P(A_n)\to 0\) .

Answer

\(\lvert X\mathbf 1_A\rvert\le\lvert X\rvert\) , so the family is dominated by the integrable variable \(\lvert X\rvert\) and is uniformly integrable. Face (b) of Proposition 5.2 for the single variable \(X\) gives \(\mathbb E[\lvert X\rvert;A_n]\to 0\) when \(\mathbb P(A_n)\to 0\) .
Let \(\Phi:[0,\infty)\to[0,\infty)\) satisfy \(\Phi(x)/x\to\infty\) . Construct an integrable random variable \(X\) with \(\mathbb E\,\Phi(\lvert X\rvert)=\infty\) , showing that no single gauge in Theorem 5.4 works for every uniformly integrable family.

Hint

Choose \(x_1<x_2<\cdots\) with \(\Phi(x_k)\ge k\,2^k\,x_k\) and let \(\mathbb P(X=x_k)=c\,2^{-k}/x_k\) , with \(c\) a normalizing constant and the remaining mass at \(0\) . Then \(\mathbb E X=c\sum_k2^{-k}<\infty\) and \(\mathbb E\,\Phi(X)\ge c\sum_kk=\infty\) .
Use the tail formula of Proposition 1.12 to show that \(\mathbb E[Y;Y>M]=M\,\mathbb P(Y>M)+\int_M^\infty\mathbb P(Y>y)\,dy\) for \(Y\ge 0\) , and restate uniform integrability of a family as a condition on its tail probabilities.

Answer

Apply the tail formula to \((Y-M)^{+}\) : \(\mathbb E(Y-M)^{+}=\int_M^\infty\mathbb P(Y>y)\,dy\) , and \(\mathbb E[Y;Y>M]=M\,\mathbb P(Y>M)+\mathbb E(Y-M)^{+}\) . A family is uniformly integrable if and only if \(\sup_i\bigl(M\,\mathbb P(\lvert X_i\rvert>M)+\int_M^\infty\mathbb P(\lvert X_i\rvert>y)\,dy\bigr)\to 0\) .
Modify the typewriter sequence of Example 4.1 by giving the \(n\) -th term the height \(2^{k/2}\) in generation \(k\) . Show that the modified sequence is uniformly integrable, converges to \(0\) in \(L^1\) , converges at no sample point, and has \(\sup_nX_n=\infty\) everywhere.

Answer

With \(X_n=2^{k/2}\mathbf 1_{I_n}\) in generation \(k\) , \(\mathbb E X_n^2=2^k\cdot 2^{-k}=1\) , so Proposition 5.3 gives uniform integrability; \(\mathbb E X_n=2^{-k/2}\to 0\) , which is convergence to \(0\) in \(L^1\) , as Theorem 5.6 predicts. Each sample point is hit once per generation with height \(2^{k/2}\to\infty\) and misses the other terms, so the sequence converges nowhere and its supremum is infinite everywhere.