[proofplan] We exploit the absolute [continuity](/page/Continuity) of the [Lebesgue integral](/page/Lebesgue%20Integral) to convert the $L^1$ integrability of $X$ into an $\varepsilon$-$\delta$ condition on [sets](/page/Set) of small measure. We then use Markov's inequality applied to $|\mathbb{E}[X \mid \mathcal{G}]|$ — whose $L^1$ norm is controlled by that of $X$ via the conditional triangle inequality — to show that the set $\{|\mathbb{E}[X \mid \mathcal{G}]| \geq \lambda\}$ has uniformly small measure for $\lambda$ large, regardless of the choice of sub-$\sigma$-algebra $\mathcal{G}$. The defining property of conditional expectation then transfers the tail [integral](/page/Integral) of $|\mathbb{E}[X \mid \mathcal{G}]|$ back to an integral of $|X|$ over a set of small measure, which is small by absolute continuity. [/proofplan] [step:Extract the $\varepsilon$-$\delta$ condition from absolute continuity of the integral of $|X|$] Fix $\varepsilon > 0$. Since $X \in L^1(\Omega, \mathcal{F}, \mathbb{P})$, the set [function](/page/Function) \begin{align*} \nu \colon \mathcal{F} &\to [0, \infty) \\ A &\mapsto \int_A |X| \, d\mathbb{P} \end{align*} is a finite measure on $(\Omega, \mathcal{F})$ that is absolutely continuous with respect to $\mathbb{P}$. By absolute continuity of the Lebesgue integral, there exists $\delta > 0$ such that for every $A \in \mathcal{F}$, \begin{align*} \mathbb{P}(A) < \delta \implies \mathbb{E}[|X| \mathbb{1}_A] < \varepsilon. \end{align*} [guided] Fix $\varepsilon > 0$. The proof requires an $\varepsilon$-$\delta$ link between the $\mathbb{P}$-measure of a set and the integral of $|X|$ over that set. Where does this come from? Since $X \in L^1(\Omega, \mathcal{F}, \mathbb{P})$, the set function \begin{align*} \nu \colon \mathcal{F} &\to [0, \infty) \\ A &\mapsto \int_A |X| \, d\mathbb{P} = \mathbb{E}[|X| \mathbb{1}_A] \end{align*} defines a finite measure on $(\Omega, \mathcal{F})$ with $\nu(\Omega) = \mathbb{E}[|X|] < \infty$. This measure is absolutely continuous with respect to $\mathbb{P}$: if $\mathbb{P}(A) = 0$, then $|X| \mathbb{1}_A = 0$ $\mathbb{P}$-a.s., so $\nu(A) = 0$. The standard characterisation of absolute continuity for finite measures gives the $\varepsilon$-$\delta$ form: there exists $\delta > 0$ such that for every $A \in \mathcal{F}$, \begin{align*} \mathbb{P}(A) < \delta \implies \mathbb{E}[|X| \mathbb{1}_A] < \varepsilon. \end{align*} This $\delta$ depends on $\varepsilon$ and on the function $X$, but not on the choice of $A$ or of any sub-$\sigma$-algebra. It is this uniformity that will drive the entire argument. [/guided] [/step] [step:Choose $\lambda$ so that every conditional expectation has tail measure less than $\delta$] Let $\mathcal{G} \subset \mathcal{F}$ be an arbitrary sub-$\sigma$-algebra, and write $Y := \mathbb{E}[X \mid \mathcal{G}]$, which exists by the [Existence and Uniqueness of Conditional Expectation](/theorems/1147) since $X \in L^1(\Omega, \mathcal{F}, \mathbb{P})$. By the [Conditional Triangle Inequality](/theorems/1148) (part (vi)), $|Y| \leq \mathbb{E}[|X| \mid \mathcal{G}]$ $\mathbb{P}$-a.s. Integrating both sides and applying the [Averaging Property](/theorems/1148) (part (i)) to $|X|$ yields \begin{align*} \mathbb{E}[|Y|] \leq \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}]\bigr] = \mathbb{E}[|X|]. \end{align*} We apply [Markov's Inequality](/theorems/514) to the non-negative random variable $|Y|$ at level $\lambda > 0$. Since $|Y| \colon \Omega \to [0, \infty)$ is $\mathcal{G}$-measurable (hence $\mathcal{F}$-measurable) and non-negative, the hypotheses are satisfied: \begin{align*} \mathbb{P}(|Y| \geq \lambda) \leq \frac{\mathbb{E}[|Y|]}{\lambda} \leq \frac{\mathbb{E}[|X|]}{\lambda}. \end{align*} Choose $\lambda := \mathbb{E}[|X|] / \delta + 1$, so that \begin{align*} \frac{\mathbb{E}[|X|]}{\lambda} = \frac{\mathbb{E}[|X|]}{\mathbb{E}[|X|]/\delta + 1} < \frac{\mathbb{E}[|X|]}{\mathbb{E}[|X|]/\delta} = \delta. \end{align*} Hence $\mathbb{P}(|Y| \geq \lambda) < \delta$. This bound holds for every sub-$\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$ simultaneously, since $\lambda$ depends only on $\mathbb{E}[|X|]$ and $\delta$ — neither of which involves $\mathcal{G}$. [guided] The goal is to find a single threshold $\lambda > 0$, independent of $\mathcal{G}$, such that the set where $|\mathbb{E}[X \mid \mathcal{G}]|$ exceeds $\lambda$ has $\mathbb{P}$-measure strictly less than $\delta$. Write $Y := \mathbb{E}[X \mid \mathcal{G}]$, which exists by the [Existence and Uniqueness of Conditional Expectation](/theorems/1147) since $X \in L^1(\Omega, \mathcal{F}, \mathbb{P})$. We first need an $L^1$ bound on $Y$ that is uniform over all sub-$\sigma$-algebras. The [Conditional Triangle Inequality](/theorems/1148) (part (vi)) gives $|Y| = |\mathbb{E}[X \mid \mathcal{G}]| \leq \mathbb{E}[|X| \mid \mathcal{G}]$ $\mathbb{P}$-a.s. Since conditional expectation preserves non-negativity by the [Positivity Property](/theorems/1148) (part (iv)), $\mathbb{E}[|X| \mid \mathcal{G}] \geq 0$ a.s., so $|Y|$ is bounded above by a non-negative conditional expectation. Integrating and applying the [Averaging Property](/theorems/1148) (part (i)) to $|X|$: \begin{align*} \mathbb{E}[|Y|] \leq \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}]\bigr] = \mathbb{E}[|X|]. \end{align*} This is the $L^1$ contraction property of conditional expectation. The right-hand side $\mathbb{E}[|X|]$ does not depend on $\mathcal{G}$ — this uniformity is the essential point. Now apply [Markov's Inequality](/theorems/514) to the non-negative [measurable function](/page/Measurable%20Functions) $|Y|$ at level $\lambda > 0$. Since $|Y| \colon \Omega \to [0, \infty)$ is $\mathcal{G}$-measurable (hence $\mathcal{F}$-measurable) and non-negative, the hypotheses of Markov's inequality are satisfied: \begin{align*} \mathbb{P}(|Y| \geq \lambda) \leq \frac{\mathbb{E}[|Y|]}{\lambda} \leq \frac{\mathbb{E}[|X|]}{\lambda}. \end{align*} To make this strictly less than $\delta$, we need $\lambda > \mathbb{E}[|X|] / \delta$. Choose $\lambda := \mathbb{E}[|X|] / \delta + 1$. Then \begin{align*} \frac{\mathbb{E}[|X|]}{\lambda} = \frac{\mathbb{E}[|X|]}{\mathbb{E}[|X|]/\delta + 1} < \frac{\mathbb{E}[|X|]}{\mathbb{E}[|X|]/\delta} = \delta, \end{align*} so $\mathbb{P}(|Y| \geq \lambda) < \delta$. The key observation is that $\lambda$ depends only on $\mathbb{E}[|X|]$ and $\delta$, both of which are determined by $X$ and $\varepsilon$ alone. The same $\lambda$ works for every sub-$\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$. [/guided] [/step] [step:Transfer the tail integral of $|Y|$ to an integral of $|X|$ via the integral-matching property] Fix an arbitrary sub-$\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$ and set $Y := \mathbb{E}[X \mid \mathcal{G}]$ as before. The set $A := \{|Y| \geq \lambda\}$ belongs to $\mathcal{G}$ since $Y$ is $\mathcal{G}$-measurable. By the [Conditional Triangle Inequality](/theorems/1148) (part (vi)), $|Y| \leq \mathbb{E}[|X| \mid \mathcal{G}]$ $\mathbb{P}$-a.s. Since $\mathbb{1}_A \geq 0$, multiplying both sides by $\mathbb{1}_A$ preserves the inequality. Taking expectations (which preserves the ordering by monotonicity of the integral): \begin{align*} \mathbb{E}[|Y| \mathbb{1}_A] \leq \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}] \cdot \mathbb{1}_A\bigr]. \end{align*} Since $A \in \mathcal{G}$, the integral-matching condition in the [Existence and Uniqueness of Conditional Expectation](/theorems/1147) (part (ii)), applied to the integrable random variable $|X|$, gives \begin{align*} \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}] \cdot \mathbb{1}_A\bigr] = \mathbb{E}[|X| \mathbb{1}_A]. \end{align*} Combining: \begin{align*} \mathbb{E}[|Y| \mathbb{1}_A] \leq \mathbb{E}[|X| \mathbb{1}_A]. \end{align*} [guided] We need to bound $\mathbb{E}[|Y| \mathbb{1}_{|Y| \geq \lambda}]$. The difficulty is that both $|Y|$ and $\mathbb{1}_{|Y| \geq \lambda}$ are defined in terms of $Y = \mathbb{E}[X \mid \mathcal{G}]$, so we cannot directly apply bounds on $X$. The strategy is to replace $|Y|$ by $\mathbb{E}[|X| \mid \mathcal{G}]$ (which dominates $|Y|$ pointwise a.s.) and then use the integral-matching property to pass from the conditional expectation back to $|X|$. Set $A := \{|Y| \geq \lambda\}$. This set belongs to $\mathcal{G}$ because $Y$ is $\mathcal{G}$-measurable, so $|Y|$ is $\mathcal{G}$-measurable, and the preimage of $[\lambda, \infty)$ under a measurable function is measurable. By the [Conditional Triangle Inequality](/theorems/1148) (part (vi)), $|Y| \leq \mathbb{E}[|X| \mid \mathcal{G}]$ $\mathbb{P}$-a.s. Since $\mathbb{1}_A \geq 0$, multiplying preserves the inequality: \begin{align*} |Y| \mathbb{1}_A \leq \mathbb{E}[|X| \mid \mathcal{G}] \cdot \mathbb{1}_A \quad \mathbb{P}\text{-a.s.} \end{align*} Taking expectations (which preserves the inequality by monotonicity of the integral): \begin{align*} \mathbb{E}[|Y| \mathbb{1}_A] \leq \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}] \cdot \mathbb{1}_A\bigr]. \end{align*} Now comes the key step. Since $A \in \mathcal{G}$ and $\mathbb{1}_A$ is a bounded $\mathcal{G}$-measurable function, the integral-matching condition from the [Existence and Uniqueness of Conditional Expectation](/theorems/1147) (part (ii)) gives \begin{align*} \mathbb{E}\bigl[\mathbb{E}[|X| \mid \mathcal{G}] \cdot \mathbb{1}_A\bigr] = \mathbb{E}[|X| \mathbb{1}_A]. \end{align*} This is the defining identity of conditional expectation: integrating a conditional expectation against a $\mathcal{G}$-measurable indicator recovers the unconditional integral over the same set. Combining: \begin{align*} \mathbb{E}[|Y| \mathbb{1}_A] \leq \mathbb{E}[|X| \mathbb{1}_A]. \end{align*} The right-hand side is now an integral of the original function $|X|$ over a set $A$ whose $\mathbb{P}$-measure we have already controlled. [/guided] [/step] [step:Conclude uniform integrability by combining absolute continuity with the uniform tail bound] From the previous steps, for every sub-$\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$ with $Y = \mathbb{E}[X \mid \mathcal{G}]$ and $A = \{|Y| \geq \lambda\}$: 1. $\mathbb{P}(A) < \delta$, with $\lambda$ independent of $\mathcal{G}$. 2. $\mathbb{E}[|Y| \mathbb{1}_A] \leq \mathbb{E}[|X| \mathbb{1}_A]$. Since $A \in \mathcal{F}$ satisfies $\mathbb{P}(A) < \delta$, the absolute continuity condition from the first step gives $\mathbb{E}[|X| \mathbb{1}_A] < \varepsilon$. Therefore \begin{align*} \mathbb{E}\bigl[|\mathbb{E}[X \mid \mathcal{G}]| \cdot \mathbb{1}_{|\mathbb{E}[X \mid \mathcal{G}]| \geq \lambda}\bigr] \leq \mathbb{E}[|X| \mathbb{1}_{|Y| \geq \lambda}] < \varepsilon. \end{align*} This holds for every sub-$\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$ with the same $\lambda$. Hence \begin{align*} \sup_{\mathcal{G} \subset \mathcal{F}} \mathbb{E}\bigl[|\mathbb{E}[X \mid \mathcal{G}]| \cdot \mathbb{1}_{|\mathbb{E}[X \mid \mathcal{G}]| \geq \lambda}\bigr] \leq \varepsilon, \end{align*} which is the definition of uniform integrability for the family $\{\mathbb{E}[X \mid \mathcal{G}] : \mathcal{G} \subset \mathcal{F}\}$. [guided] We assemble the three ingredients: - **Absolute continuity** (first step): there exists $\delta > 0$ such that $\mathbb{P}(A) < \delta$ implies $\mathbb{E}[|X| \mathbb{1}_A] < \varepsilon$. - **Uniform tail bound** (second step): there exists $\lambda > 0$ (depending only on $\mathbb{E}[|X|]$ and $\delta$) such that $\mathbb{P}(|Y| \geq \lambda) < \delta$ for every $Y = \mathbb{E}[X \mid \mathcal{G}]$. - **Integral transfer** (third step): $\mathbb{E}[|Y| \mathbb{1}_{|Y| \geq \lambda}] \leq \mathbb{E}[|X| \mathbb{1}_{|Y| \geq \lambda}]$. Chaining these: the set $A = \{|Y| \geq \lambda\}$ has $\mathbb{P}(A) < \delta$ by the second step, so $\mathbb{E}[|X| \mathbb{1}_A] < \varepsilon$ by the first step. The integral transfer from the third step then gives \begin{align*} \mathbb{E}[|Y| \mathbb{1}_{|Y| \geq \lambda}] \leq \mathbb{E}[|X| \mathbb{1}_{|Y| \geq \lambda}] < \varepsilon. \end{align*} Since neither $\lambda$ nor $\varepsilon$ depends on the choice of $\mathcal{G}$, the bound $\mathbb{E}[|Y| \mathbb{1}_{|Y| \geq \lambda}] < \varepsilon$ holds uniformly over all sub-$\sigma$-algebras $\mathcal{G} \subset \mathcal{F}$. Taking the supremum: \begin{align*} \sup_{\mathcal{G} \subset \mathcal{F}} \mathbb{E}\bigl[|\mathbb{E}[X \mid \mathcal{G}]| \cdot \mathbb{1}_{|\mathbb{E}[X \mid \mathcal{G}]| \geq \lambda}\bigr] \leq \varepsilon. \end{align*} As $\varepsilon > 0$ was arbitrary (with $\lambda$ depending on $\varepsilon$), this confirms that the family $\{\mathbb{E}[X \mid \mathcal{G}] : \mathcal{G} \text{ a sub-}\sigma\text{-algebra of } \mathcal{F}\}$ is uniformly integrable. [/guided] [/step]