Uniform integrability begins with a failure of ordinary integrability to control a family. A single integrable function has small tails: if $f \in L^1(E, \mathcal E, \mu)$, then the integral of $|f|$ over the set where $|f|$ is large tends to $0$. But when a whole family of functions is moving at once, each member may have a small tail while the location or height of the tail changes from member to member. Uniform integrability asks for one tail estimate that works for the entire family. [example: A Bounded $L^1$ Family with Moving Spikes] Let $E=(0,1)$ with [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$, and define \begin{align*} f_n(x)=n\,\mathbb{1}_{(0,1/n)}(x). \end{align*} For each $n$, the function $f_n$ is measurable, nonnegative, and supported on $(0,1/n)$. Its $L^1$ norm is \begin{align*} \int_E |f_n|\,d\mathcal L^1=\int_0^1 n\,\mathbb{1}_{(0,1/n)}(x)\,d\mathcal L^1(x)=\int_0^{1/n} n\,d\mathcal L^1(x)=n\mathcal L^1((0,1/n))=n\cdot\frac{1}{n}=1. \end{align*} Thus $\sup_n\|f_n\|_{L^1(E)}=1$, so the family is bounded in $L^1(E)$. Now fix $M>0$ and choose an integer $n>M$. Since $f_n(x)=n$ for every $x\in(0,1/n)$ and $f_n(x)=0$ for $x\in(1/n,1)$, the level set is \begin{align*} \{|f_n|>M\}=(0,1/n). \end{align*} Therefore \begin{align*} \int_{\{|f_n|>M\}} |f_n|\,d\mathcal L^1=\int_0^{1/n} n\,d\mathcal L^1(x)=n\mathcal L^1((0,1/n))=1. \end{align*} So no matter how large the threshold $M$ is, some member of the family still has total mass $1$ above that threshold. The mass has not become large in total; instead it has concentrated on smaller and smaller sets, showing that $L^1$ boundedness alone does not prevent concentration. [/example] The example explains why uniform integrability is a child of integrability, not a replacement for it. Integrability is a property of one function. Uniform integrability is a stability condition for a family: it says that the integrable mass of all members can be controlled uniformly on the parts where the functions are large. ## Definition The most robust definition should not depend on a special description of the bad set. A family may fail by putting large values on tiny sets, but in applications the tiny sets often come from [convergence in measure](/page/Convergence%20in%20Measure), stopping times, or approximation errors rather than from explicit level sets. Uniform integrability therefore asks for two things at once: bounded total mass, and a uniform version of the absolute continuity of the integral. [definition: Uniform Integrability] Let $(E, \mathcal E, \mu)$ be a [measure space](/page/Measure%20Space). A family $\mathcal F \subset L^1(E, \mathcal E, \mu)$ is uniformly integrable if \begin{align*} \sup_{f\in\mathcal F}\int_E |f|\,d\mu &< \infty \end{align*} and for every $\varepsilon>0$ there exists $\delta>0$ such that for every $A\in\mathcal E$ with $\mu(A)<\delta$, \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu &< \varepsilon. \end{align*} [/definition] This is a family version of the absolute continuity of the integral. For one integrable function $f$, small measure of $A$ forces $\int_A |f|\,d\mu$ to be small. Uniform integrability demands that the same $\delta$ work for every function in the family, while also excluding unbounded total $L^1$ mass. The definition is written in terms of arbitrary small sets, but in practice one often detects failure by looking at large values. The spike example failed because the set $\{|f_n|>M\}$ still carried order-one mass no matter how large $M$ was chosen. This raises the first structural question: on the measure spaces where most probability and finite-volume analysis takes place, is it enough to control the large-value tails uniformly? The next theorem answers exactly that question. [quotetheorem:9925] For random variables, the same statement is written with expectation. Thus a family $\mathcal X\subset L^1(\Omega,\mathcal F,\mathbb P)$ is uniformly integrable exactly when \begin{align*} \lim_{K\to\infty}\sup_{X\in\mathcal X}\mathbb E\big[|X|\,\mathbb{1}_{\{|X|>K\}}\big] &= 0. \end{align*} This form is usually the quickest way to check the condition on probability spaces, while the small-set formulation is the definition that remains stable on general measure spaces. There is one more convention to record before moving on. On $L^1(\mathbb R^n)$, many analysis texts use "uniform integrability" for a two-part compactness condition: no concentration on small sets and no escape of mass to infinity. Other texts reserve "uniform integrability" for the small-set or large-value condition alone and call the second condition tightness. This page follows the two-part Euclidean convention only when it explicitly says "uniform integrability in $L^1(\mathbb R^n)$" or "Euclidean uniform integrability"; otherwise, uniform integrability means the general-measure definition above. ## Tails, Concentration, and Boundedness Uniform integrability lives between pointwise integrability and stronger $L^p$ control. It is weaker than having a common $L^p$ bound for some $p>1$, but stronger than boundedness in $L^1$. The difference is exactly whether the family can hide mass in rare large values. ### Higher Integrability The most common sufficient condition comes from comparing the $L^1$ tail to a higher moment. If the family is uniformly bounded in $L^p$ for some $p>1$, then placing mass far out in the tail becomes increasingly expensive. [quotetheorem:9926] The finite-measure hypothesis is doing real work. It turns the estimate on small sets into \begin{align*} \int_A |f|\,d\mu &\le \|f\|_{L^p(E)}\mu(A)^{1-1/p}, \end{align*} and it also gives uniform $L^1$ boundedness from the uniform $L^p$ bound. On an infinite measure space, a single function may belong to $L^p$ but not to $L^1$, so a uniform $L^p$ bound alone cannot imply uniform integrability in the $L^1$ sense. The theorem explains why uniform integrability is often invisible in finite-measure $L^p$ theory for $p>1$: it is automatically supplied by the stronger norm. At the endpoint $p=1$, the mechanism disappears, and concentration becomes possible. [example: Higher Moments Rule Out the Spike] On $(0,1)$ with Lebesgue measure, let $f_n(x)=n\mathbb{1}_{(0,1/n)}(x)$. Since $f_n(x)=n$ on $(0,1/n)$ and $f_n(x)=0$ on $(1/n,1)$, its $L^1$ norm is \begin{align*} \|f_n\|_{L^1(0,1)}=\int_0^1 |f_n|\,d\mathcal L^1=\int_0^{1/n} n\,d\mathcal L^1. \end{align*} Because $n$ is constant on $(0,1/n)$, \begin{align*} \int_0^{1/n} n\,d\mathcal L^1=n\,\mathcal L^1((0,1/n))=n\cdot \frac{1}{n}=1. \end{align*} Thus $\|f_n\|_{L^1(0,1)}=1$ for every $n$. For $p>1$, the $p$th power of the $L^p$ norm is \begin{align*} \|f_n\|_{L^p(0,1)}^p=\int_0^1 |f_n|^p\,d\mathcal L^1. \end{align*} Using again that $f_n=n$ on $(0,1/n)$ and $f_n=0$ outside that interval, \begin{align*} \int_0^1 |f_n|^p\,d\mathcal L^1=\int_0^{1/n} n^p\,d\mathcal L^1. \end{align*} Since $n^p$ is constant on $(0,1/n)$, \begin{align*} \int_0^{1/n} n^p\,d\mathcal L^1=n^p\,\mathcal L^1((0,1/n))=n^p\cdot \frac{1}{n}=n^{p-1}. \end{align*} Therefore \begin{align*} \|f_n\|_{L^p(0,1)}=(n^{p-1})^{1/p}=n^{(p-1)/p}. \end{align*} Because $p>1$, the exponent $(p-1)/p$ is positive, so $n^{(p-1)/p}\to\infty$. The spike family is bounded in $L^1$ but unbounded in every $L^p$ with $p>1$, so higher integrability detects exactly the growing height that keeps the mass concentrated on shrinking intervals. [/example] ### Superlinear Gauges The $L^p$ theorem is useful but too rigid: many uniformly integrable families have no common $L^p$ bound for any fixed $p>1$. To capture the exact amount of extra control beyond $L^1$, one uses a convex gauge that grows faster than linearly but may grow much slower than $t^p$. [definition: De la Vallee Poussin Function] A finite-valued function $\Phi:[0,\infty)\to[0,\infty)$ is a de la Vallee Poussin function if $\Phi$ is convex, nondecreasing, satisfies $\Phi(0)=0$, and \begin{align*} \lim_{t\to\infty}\frac{\Phi(t)}{t} &= \infty. \end{align*} [/definition] The point of introducing such a function is to measure growth just beyond linear. If a family has uniformly bounded $\Phi(|f|)$-integrals for a superlinear convex gauge $\Phi$, then large values have become too expensive to carry significant mass. The next criterion says that this is not merely sufficient; on finite measure spaces it characterizes uniform integrability. [quotetheorem:951] This criterion is a bridge from the qualitative definition to Orlicz-space methods. It also explains the slogan that uniform integrability is slightly more than $L^1$ boundedness, but not necessarily as much as any fixed $L^p$ bound. ## Convergence in $L^1$ The strongest reason to isolate uniform integrability is that it identifies when weak-looking convergence becomes convergence in mean. Almost sure convergence or convergence in probability controls where values go pointwise; uniform integrability controls how much mass can be lost in exceptional sets. ### Vitali's Theorem Dominated convergence proves $L^1$ convergence by comparing all functions to one integrable majorant. Many natural approximations do not admit such a majorant. [Vitali's theorem](/theorems/1228) replaces domination by the exact family condition needed to make tails harmless. [quotetheorem:950] Vitali's theorem should be read as the precise repair for the failure of convergence in probability to control averages. The convergence-in-probability hypothesis says that typical values approach the limit, while uniform integrability rules out a small exceptional region carrying a persistent amount of expectation. The conclusion is therefore not a purely pointwise phenomenon: it depends on excluding concentration of mass in the tails. [example: Convergence in Probability Without $L^1$ Convergence] Let $X_n=n\mathbb{1}_{(0,1/n)}$ on the [probability space](/page/Probability%20Space) $((0,1),\mathcal B((0,1)),\mathcal L^1)$. We first show that $X_n\xrightarrow{\mathbb P}0$. Fix $\varepsilon>0$, and choose $N\in\mathbb N$ with $N>\varepsilon$. If $n\ge N$, then $n>\varepsilon$, so $X_n(x)=n>\varepsilon$ exactly on $(0,1/n)$ and $X_n(x)=0\le \varepsilon$ on $(0,1)\setminus(0,1/n)$. Hence \begin{align*} \{|X_n|>\varepsilon\}=(0,1/n). \end{align*} Therefore, for every $n\ge N$, \begin{align*} \mathcal L^1(\{|X_n|>\varepsilon\})=\mathcal L^1((0,1/n))=\frac{1}{n}. \end{align*} Since $1/n\to0$, it follows that $\mathcal L^1(\{|X_n|>\varepsilon\})\to0$ for every $\varepsilon>0$, which is exactly convergence in probability to $0$. However, the $L^1$ distance from $X_n$ to $0$ does not go to $0$. Since $X_n\ge0$, \begin{align*} \mathbb E[|X_n-0|]=\int_0^1 |X_n|\,d\mathcal L^1=\int_0^1 X_n\,d\mathcal L^1. \end{align*} Using $X_n=n$ on $(0,1/n)$ and $X_n=0$ outside that interval, \begin{align*} \int_0^1 X_n\,d\mathcal L^1=\int_0^{1/n} n\,d\mathcal L^1. \end{align*} Since $n$ is constant on $(0,1/n)$, \begin{align*} \int_0^{1/n} n\,d\mathcal L^1=n\,\mathcal L^1((0,1/n))=n\cdot\frac{1}{n}=1. \end{align*} Thus $\mathbb E[|X_n-0|]=1$ for every $n$, so $X_n$ cannot converge to $0$ in $L^1$. The failure is not pointwise probabilistic convergence; it is the persistent unit mass carried by values of height $n$ on intervals of length $1/n$, which is exactly the lack of uniform integrability needed in Vitali's theorem. [/example] ### Necessity for $L^1$ Limits Vitali's theorem would be less convincing if uniform integrability were only a convenient sufficient hypothesis. The converse stability result shows that any sequence that really converges in mean already has uniform tail control, so the condition belongs to the structure of $L^1$ convergence. [quotetheorem:9927] This result often supplies uniform integrability after the fact: once convergence in mean has been proved by another method, the sequence automatically has the tail control needed for later limiting arguments. ## Weak Compactness in $L^1$ In reflexive spaces, bounded sequences have weakly compact subsequences. The space $L^1$ is not reflexive, so boundedness alone cannot give weak compactness. Uniform integrability is the replacement compactness condition for concentration; on non-compact spaces, one must also prevent mass from escaping to infinity. ### Relative Weak Compactness Compactness in applications rarely means that a whole set has a strong $L^1$ limit. More often, one wants weak limits that survive after passing to subsequences or taking closures. This motivates the weak compactness language used in the $L^1$ [compactness theorem](/theorems/2748). [definition: Relative Weak Compactness] Let $X$ be a [Banach space](/page/Banach%20Space), and let $\sigma(X,X^*)$ denote the [weak topology](/page/Weak%20Topology) on $X$ induced by the [dual space](/page/Dual%20Space) $X^*$. A subset $K\subset X$ is relatively weakly compact if its closure in the topology $\sigma(X,X^*)$ is compact in the topology $\sigma(X,X^*)$. [/definition] Relative weak compactness is the [sequential compactness](/page/Sequential%20Compactness) substitute one wants for variational limits and probability limits. In $L^1$, the key question is whether the analytic tail condition is exactly the same as this topological compactness condition. The [Dunford-Pettis theorem](/theorems/990) answers that question on finite measure spaces. [quotetheorem:9928] This theorem gives uniform integrability a structural meaning. It is not only a tail estimate; it is the condition that restores weak compactness at the non-reflexive endpoint $L^1$. This is the $L^1$ compactness theorem usually called Dunford-Pettis in measure theory. It should not be confused with the broader Dunford-Pettis property of Banach spaces, which concerns weakly compact operators and is a different functional-analytic notion. ### Escape to Infinity The finite-measure theorem cannot by itself handle $\mathbb R^n$, because a uniformly integrable family can still move all its mass away from every compact set. To recover weak compactness on non-compact domains, uniform integrability must be paired with tightness in space. [example: Translation Prevents Compactness on $\mathbb R$] Let $g\in L^1(\mathbb R)$ be nonzero, and define $f_n(x)=g(x-n)$. By the [translation invariance of Lebesgue measure](/theorems/4911), with the change of variables $y=x-n$, \begin{align*} \|f_n\|_{L^1(\mathbb R)}=\int_{\mathbb R}|g(x-n)|\,d\mathcal L^1(x)=\int_{\mathbb R}|g(y)|\,d\mathcal L^1(y)=\|g\|_{L^1(\mathbb R)}. \end{align*} Thus the total $L^1$ mass is constant along the translated family. If $g$ is bounded, say $|g|\le B$ almost everywhere, then $|f_n|\le B$ almost everywhere for every $n$. Hence for every $M\ge B$, \begin{align*} \{|f_n|>M\}=\varnothing \end{align*} up to a null set, so the family has no large-value tail. The obstruction is spatial. Fix $R>0$. The mass of $f_n$ inside $(-R,R)$ is \begin{align*} \int_{(-R,R)}|f_n|\,d\mathcal L^1=\int_{-R}^{R}|g(x-n)|\,d\mathcal L^1(x)=\int_{-R-n}^{R-n}|g(y)|\,d\mathcal L^1(y). \end{align*} Since $g\in L^1(\mathbb R)$, for every $\varepsilon>0$ there is $A>0$ such that \begin{align*} \int_{(-\infty,-A)}|g|\,d\mathcal L^1<\varepsilon. \end{align*} If $n>R+A$, then $R-n<-A$, and therefore \begin{align*} (-R-n,R-n)\subset(-\infty,-A). \end{align*} For such $n$, \begin{align*} \int_{(-R,R)}|f_n|\,d\mathcal L^1=\int_{-R-n}^{R-n}|g|\,d\mathcal L^1\le\int_{(-\infty,-A)}|g|\,d\mathcal L^1<\varepsilon. \end{align*} Consequently, \begin{align*} \int_{\mathbb R\setminus(-R,R)}|f_n|\,d\mathcal L^1=\|f_n\|_{L^1(\mathbb R)}-\int_{(-R,R)}|f_n|\,d\mathcal L^1>\|g\|_{L^1(\mathbb R)}-\varepsilon. \end{align*} Because $R>0$ and $\varepsilon>0$ were arbitrary, \begin{align*} \sup_{n\in\mathbb N}\int_{\mathbb R\setminus(-R,R)}|f_n|\,d\mathcal L^1=\|g\|_{L^1(\mathbb R)} \end{align*} for every $R>0$. Since $g$ is nonzero, $\|g\|_{L^1(\mathbb R)}>0$, so the spatial tail cannot tend to $0$ as $R\to\infty$. The family avoids concentration in height when $g$ is bounded, but its mass drifts away from every fixed bounded interval, so uniform integrability in the Euclidean sense fails. [/example] The translation example identifies the missing hypothesis, but the compactness question is still open: after ruling out concentration and drifting bumps, do weak limits exist? The following definition records the standard Euclidean convention for the same term. Throughout this subsection, the phrase "uniform integrability in $L^1(\mathbb R^n)$" means this two-part Euclidean condition, not merely the general small-set condition. It is not a different pathology; it is the form uniform integrability takes when the underlying space itself has a direction in which mass can escape. [definition: Uniform Integrability in $L^1(\mathbb R^n)$] Let $\mathcal F \subset L^1(\mathbb R^n)$, with integration taken with respect to $\mathcal L^n$. The family $\mathcal F$ is uniformly integrable in $L^1(\mathbb R^n)$ if it is uniformly integrable as a family on the measure space $(\mathbb R^n,\mathcal B(\mathbb R^n),\mathcal L^n)$ and \begin{align*} \lim_{R\to\infty}\sup_{f\in\mathcal F}\int_{\mathbb R^n\setminus B(0,R)} |f|\,d\mathcal L^n &= 0. \end{align*} [/definition] This Euclidean convention combines the two independent controls that are both needed on $\mathbb R^n$. The small-set condition handles concentration and high values, while the spatial tail condition says that most of the $L^1$ mass lies in one large ball, uniformly over the family. The terminology is conventional rather than universal. In sources that use the phrase "uniform integrability" only for the small-set condition, the definition above would be described as uniform integrability together with tightness. These two estimates are useful only if they lead to compactness. In PDE, probability, and the calculus of variations, one often has a sequence of densities on $\mathbb R^n$ and needs a weakly convergent subsequence rather than just separate tail bounds. The next theorem is the reason the Euclidean convention is worth naming: it turns "no concentration and no escape to infinity" into the weak compactness conclusion that replaces reflexive boundedness at the $L^1$ endpoint. [quotetheorem:9929] This is the form that appears in analysis of PDE and calculus of variations when mass distributions are allowed to move through space. A compactness argument must rule out both spike formation and loss of mass at infinity. ## Martingales and Conditional Expectation Uniform integrability is also the condition that makes limiting random variables retain their expectations. For martingales, ordinary $L^1$ boundedness controls each time marginal, but it may not control the terminal limit strongly enough. Uniform integrability is the hypothesis that makes optional limiting operations behave well. ### Conditional Expectation [Conditional expectation](/page/Conditional%20Expectation) is an averaging operation, and averaging should not create new tails. This principle gives one of the most important sources of uniformly integrable families: all conditional expectations of a single integrable [random variable](/page/Random%20Variable) share one tail bound. [quotetheorem:1161] This theorem explains why uniformly integrable martingales arise naturally from terminal variables: if $M_n=\mathbb E[X\mid\mathcal F_n]$, then the whole martingale inherits uniform integrability from $X$. ### Martingale Limits A martingale may converge almost surely while still losing expectation through rare large events. To prevent that loss, the martingale definition must include the same uniform tail control used in Vitali's theorem. [definition: Uniformly Integrable Martingale] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space with a filtration $(\mathcal F_n)_{n\in\mathbb N}$, meaning $\mathcal F_n\subset\mathcal F_{n+1}\subset\mathcal F$ for every $n\in\mathbb N$. A martingale $(M_n)_{n\in\mathbb N}$ adapted to $(\mathcal F_n)_{n\in\mathbb N}$, where each $M_n:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is integrable and $\mathcal F_n$-measurable, is uniformly integrable if the family $\{M_n:n\in\mathbb N\}$ is uniformly integrable in $L^1(\Omega,\mathcal F,\mathbb P)$. [/definition] The definition packages a tail condition into a martingale property, but the reason to impose it is a convergence question: when does a martingale limit still represent the earlier martingale by conditional expectation? The next theorem says that uniform integrability is the hypothesis that gives both $L^1$ convergence and terminal representation. [quotetheorem:1163] The theorem shows the probabilistic role of uniform integrability: it is the condition that prevents expectation from disappearing into rare events as time tends to infinity. ## Testing Uniform Integrability In applications, the definition is rarely checked directly from the limit over $K$. One usually proves a stronger estimate, reduces to a known criterion, or finds a counterexample by locating a bad family of sets. ### Small-Set Obstructions A useful negative test is to search for small sets on which a fixed amount of mass persists. Such a test directly contradicts uniform absolute continuity of integrals, so it detects concentration even when the tail sets are hard to compute. [quotetheorem:9930] The spike example is exactly this obstruction with $A_n=(0,1/n)$ and $\varepsilon_0=1$. This diagnostic is often easier than estimating all tail sets. [example: Indicator Functions of Moving Small Sets] Let $(E,\mathcal E,\mu)$ be a finite measure space, and let $A_n\in\mathcal E$ satisfy $\mu(A_n)\to0$. We show that the family $\{\mathbb{1}_{A_n}:n\in\mathbb N\}$ is uniformly integrable. First, each $\mathbb{1}_{A_n}$ is integrable because $0\le \mathbb{1}_{A_n}\le 1$ and $\mu(E)<\infty$. For every $n$, \begin{align*} \int_E |\mathbb{1}_{A_n}|\,d\mu=\int_E \mathbb{1}_{A_n}\,d\mu=\mu(A_n)\le \mu(E). \end{align*} Hence \begin{align*} \sup_{n\in\mathbb N}\int_E |\mathbb{1}_{A_n}|\,d\mu\le \mu(E)<\infty. \end{align*} Now let $\varepsilon>0$ and choose $\delta=\varepsilon$. If $A\in\mathcal E$ and $\mu(A)<\delta$, then for every $n\in\mathbb N$, \begin{align*} \int_A |\mathbb{1}_{A_n}|\,d\mu=\int_A \mathbb{1}_{A_n}\,d\mu=\mu(A\cap A_n). \end{align*} Since $A\cap A_n\subset A$, monotonicity of measure gives \begin{align*} \mu(A\cap A_n)\le \mu(A)<\delta=\varepsilon. \end{align*} Therefore \begin{align*} \sup_{n\in\mathbb N}\int_A |\mathbb{1}_{A_n}|\,d\mu<\varepsilon. \end{align*} So the same $\delta$ works for every indicator in the family, and the family is uniformly integrable. The supports may move and shrink, but the height stays equal to $1$, so no fixed amount of mass can remain trapped on sets whose measure tends to $0$. [/example] ### Stability Operations Limiting arguments often replace a family by linear combinations, dominated subfamilies, positive and negative parts, or truncations. A useful compactness condition must survive these operations; otherwise it would be too fragile for analysis. [quotetheorem:9931] These stability properties let the hypothesis pass through truncations, decompositions into positive and negative parts, and domination arguments. ## Beyond and Connected Topics Uniform integrability is the endpoint compactness condition for $L^1$. In [Lebesgue integration](/page/Integral), it refines the absolute continuity of the integral from a single function to a family of functions. The natural next step is to study how this interacts with convergence theorems such as dominated convergence, Fatou's lemma, and convergence in measure. In functional analysis, the Dunford-Pettis theorem places uniform integrability inside the weak compactness theory of Banach spaces. This connects the topic to [weak convergence](/page/Weak%20Convergence), dual spaces, and the special non-reflexive behaviour of $L^1$. In probability, uniform integrability is central to martingale convergence, optional stopping, and convergence of expectations. It is the condition that lets convergence in probability combine with tail control to produce $L^1$ convergence, and it is one of the standard hypotheses under which limiting random variables preserve expected values. In PDE and the [calculus of variations](/page/Calculus%20of%20Variations), uniform integrability is one part of compactness for densities and measures. On unbounded domains, it must be paired with tightness to prevent mass from leaving every compact set. ## References Androma, [Integral](/page/Integral). Androma, [Weak Convergence](/page/Weak%20Convergence). Androma, [Calculus of Variations](/page/Calculus%20of%20Variations). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Billingsley, *Probability and Measure* (1995). Bogachev, *Measure Theory* (2007). Dunford and Schwartz, *Linear Operators, Part I* (1958). Kallenberg, *Foundations of Modern Probability* (2002).