Uniform Integrability and the Vitali Theorem

Convergence almost everywhere does not imply convergence in the mean unless mass is prevented from escaping, and uniform integrability is the precise condition that prevents it. We define uniform integrability, characterise it on a finite measure space as boundedness together with uniform absolute continuity, and prove the Vitali convergence theorem that a sequence converging in measure converges in L-one if and only if it is uniformly integrable. This sharpens the dominated convergence theorem and is the hypothesis behind the convergence of martingales in the mean.

2 equations

5 results

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The dominated convergence theorem trades pointwise convergence for convergence in the mean by assuming a single integrable dominator, but the dominator is more than is needed. The exact requirement is that the integral over the region where the functions are large is uniformly small, a condition called uniform integrability, and under it convergence in measure upgrades to convergence in L-one. This post proves the Vitali convergence theorem, the sharp form of dominated convergence, on a finite measure space $(\Omega,\mathcal A,\mu)$ with $\mu(\Omega)<\infty$ [1], [2].

#Uniform integrability

Definition1

A family $\mathcal F\subseteq L^1(\mu)$ is uniformly integrable when

\lim_{M\to\infty}\sup_{f\in\mathcal F}\int_{\{\abs f>M\}}\abs f\,d\mu=0. \tag{1}

A single integrable function is uniformly integrable, since $\int_{\{\abs f>M\}}\abs f\to 0$ as $M\to\infty$ by dominated convergence, and so is any family dominated by one integrable $g$ , because $\int_{\{\abs f>M\}} \abs f\le\int_{\{g>M\}}g$ . Uniform integrability is exactly the strengthening of this to a uniform bound across the family, and on a finite measure space it has a clean characterisation.

Proposition2

On a finite measure space, $\mathcal F$ is uniformly integrable if and only if it is bounded in $L^1$ and uniformly absolutely continuous, meaning for every $\varepsilon>0$ there is a $\delta>0$ with $\int_A\abs f\,d\mu<\varepsilon$ for all $f\in\mathcal F$ whenever $\mu(A)<\delta$ .

Proof

Suppose $\mathcal F$ is uniformly integrable and fix $\varepsilon$ . Take $M$ with $\sup_f\int_{\{\abs f>M\}} \abs f<\varepsilon/2$ . Then $\int\abs f\le M\mu(\Omega)+\varepsilon/2$ gives the $L^1$ bound, and for any set $A$ , $\int_A\abs f\le M\mu(A)+\int_{\{\abs f>M\}}\abs f<M\mu(A)+\varepsilon/2$ , which is below $\varepsilon$ once $\mu(A)<\delta:=\varepsilon/(2M)$ , the uniform absolute continuity. Conversely, suppose the family is $L^1$ -bounded by $C$ and uniformly absolutely continuous, and fix $\varepsilon$ with its $\delta$ . By Markov's inequality, $\mu(\{\abs f>M\})\le C/M<\delta$ once $M>C/ \delta$ , so $\int_{\{\abs f>M\}}\abs f<\varepsilon$ for all $f$ by uniform absolute continuity, which is uniform integrability.

#The Vitali convergence theorem

Theorem3

On a finite measure space, let $f_n$ converge to $f$ in measure. Then $f_n\to f$ in $L^1$ if and only if $(f_n)$ is uniformly integrable, and in that case $f\in L^1$ .

Proof

Suppose $(f_n)$ is uniformly integrable. It is then bounded in $L^1$ , and since $f_n\to f$ in measure, the Riesz theorem yields a subsequence $f_{n_k}\to f$ almost everywhere. Applying Fatou's lemma along it gives $\int\abs f\le\liminf_k\int\abs{f_{n_k}}\le \sup_n\norm{f_n}_1<\infty$ , so $f\in L^1$ . Let $\varphi_M(x)=\max(-M,\min(M,x))$ be the truncation at level $M$ , which is $1$ -Lipschitz and satisfies $\abs{x-\varphi_M(x)}=(\abs x-M)^+\le\abs x\,\mathbf 1_{\{\abs x>M\}}$ . Split

\int\abs{f_n-f}\le\int\abs{f_n-\varphi_M(f_n)}+\int\abs{\varphi_M(f_n)-\varphi_M(f)}+\int\abs{\varphi_M(f)- f}. \tag{2}

Given $\varepsilon>0$ , uniform integrability and $f\in L^1$ provide an $M$ with $\sup_n\int_{\{\abs{f_n}>M\}} \abs{f_n}<\varepsilon$ and $\int_{\{\abs f>M\}}\abs f<\varepsilon$ , so the first and third terms of Equation (2) are each below $\varepsilon$ for every $n$ . The middle term has integrand bounded by $2M$ and, because $\varphi_M$ is Lipschitz and $f_n\to f$ in measure, $\varphi_M(f_n)\to\varphi_M (f)$ in measure. Write $g_n:=\abs{\varphi_M(f_n)-\varphi_M(f)}$ , so $g_n\to 0$ in measure with $0\le g_n\le 2M$ and the constant $2M$ integrable on the finite measure space. The ordinary bounded convergence theorem needs almost-everywhere convergence, so argue through subsequences. Every subsequence of $(g_n)$ has, by the Riesz theorem, a further subsequence converging to $0$ almost everywhere, and the ordinary bounded convergence theorem sends $\int g_n\to 0$ along it; since every subsequence of the nonnegative sequence $(\int g_n)$ thus has a sub-subsequence tending to $0$ , the full sequence $\int\abs{\varphi_M(f_n)-\varphi_M( f)}\to 0$ . Hence $\limsup_n\int\abs{f_n-f}\le 2\varepsilon$ , and $\varepsilon$ being arbitrary, $f_n\to f$ in $L^1$ .

Conversely, suppose $f_n\to f$ in $L^1$ . Convergence in measure follows from Markov's inequality, $\mu(\{\abs{f_n-f}>\varepsilon\})\le\norm{f_n-f }_1/\varepsilon\to 0$ . For uniform integrability, $\int_{\{\abs{f_n}>M\}}\abs{f_n}\le\int_{\{\abs{f_n}>M\}} \abs f+\norm{f_n-f}_1$ , and the single function $f$ is uniformly integrable while $\mu(\{\abs{f_n}>M\})\le (\sup_n\norm{f_n}_1)/M\to 0$ uniformly in $n$ . The first term is uniformly small for large $M$ by the absolute continuity of $\int\abs f$ , and the second is small once $n$ is large. The finitely many remaining $f_n$ are each uniformly integrable on their own, so the whole family is.

The Vitali theorem contains the dominated convergence theorem, since a dominated sequence is uniformly integrable and convergence almost everywhere implies convergence in measure on a finite measure space, but it is strictly stronger, because it replaces the dominator with the weaker tail condition and gives an exact characterisation rather than a mere sufficient condition. The cost is the finiteness of the measure, which the truncation argument uses through the bounded convergence theorem.

#Uniform integrability in probability

On a probability space, where $\mu$ is a probability and integrals are expectations, uniform integrability is the hinge of the convergence theory of martingales. A martingale converges in $L^1$ if and only if it is uniformly integrable, the condition that lets the almost-sure limit also be the mean limit. The de la Vallee Poussin criterion makes the condition checkable, since a family with $\sup_f\int\Phi(\abs f)\,d\mu<\infty$ for some nonnegative measurable $\Phi$ with $\Phi(t)/t\to\infty$ as $t\to\infty$ , such as $\Phi(t)=t^2$ , is uniformly integrable. On $\{\abs f>M\}$ one has $\abs f\le\Phi(\abs f)/\inf_{t>M}(\Phi(t)/t)$ , so the tail integral is bounded by $\sup_f\int\Phi(\abs f)\,d\mu$ divided by a quantity tending to infinity with $M$ , hence controlled by the superlinear moment. Uniform integrability is therefore the precise dividing line between the sequences whose limits keep their mass and those that leak it to infinity. It is what turns a limit in distribution into a limit of expectations and what closes a martingale at its terminal value.

[1]

G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed. Wiley, 1999.

[2]

P. Billingsley, Probability and Measure, 3rd ed. Wiley, 1995.

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cite

@misc{uniform-integrability,
  author = {Zac Kienzle},
  title  = {Uniform Integrability and the Vitali Theorem},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/uniform-integrability}
}

Uniform Integrability and the Vitali Theorem

2 equations

5 results

11 connections

#Uniform integrability

Definition1

A family $\mathcal F\subseteq L^1(\mu)$ is uniformly integrable when

\lim_{M\to\infty}\sup_{f\in\mathcal F}\int_{\{\abs f>M\}}\abs f\,d\mu=0. \tag{1}

A single integrable function is uniformly integrable, since

\int_{\{\abs f>M\}}\abs f\to 0

M\to\infty

by dominated convergence, and so is any family dominated by one integrable

g

, because

\int_{\{\abs f>M\}} \abs f\le\int_{\{g>M\}}g

. Uniform integrability is exactly the strengthening of this to a uniform bound across the family, and on a finite measure space it has a clean characterisation.

Proposition2

Proof

#The Vitali convergence theorem

Theorem3

On a finite measure space, let $f_n$ converge to $f$ in measure. Then $f_n\to f$ in $L^1$ if and only if $(f_n)$ is uniformly integrable, and in that case $f\in L^1$ .

Proof

\int\abs{f_n-f}\le\int\abs{f_n-\varphi_M(f_n)}+\int\abs{\varphi_M(f_n)-\varphi_M(f)}+\int\abs{\varphi_M(f)- f}. \tag{2}

#Uniform integrability in probability

On a probability space, where

\mu

is a probability and integrals are expectations, uniform integrability is the hinge of the convergence theory of martingales. A martingale converges in

L^1

if and only if it is uniformly integrable, the condition that lets the almost-sure limit also be the mean limit. The de la Vallee Poussin criterion makes the condition checkable, since a family with

\sup_f\int\Phi(\abs f)\,d\mu<\infty

for some nonnegative measurable

\Phi

with

\Phi(t)/t\to\infty

t\to\infty

, such as

\Phi(t)=t^2

, is uniformly integrable. On

\{\abs f>M\}

one has

\abs f\le\Phi(\abs f)/\inf_{t>M}(\Phi(t)/t)

, so the tail integral is bounded by

\sup_f\int\Phi(\abs f)\,d\mu

divided by a quantity tending to infinity with

M

, hence controlled by the superlinear moment. Uniform integrability is therefore the precise dividing line between the sequences whose limits keep their mass and those that leak it to infinity. It is what turns a limit in distribution into a limit of expectations and what closes a martingale at its terminal value.

[1]

G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed. Wiley, 1999.

[2]

P. Billingsley, Probability and Measure, 3rd ed. Wiley, 1995.