Fatou's lemma

Short description: Lemma in measure theory

In mathematics, Fatou's lemma establishes an inequality relating the Lebesgue integral of the limit inferior of a sequence of functions to the limit inferior of integrals of these functions. The lemma is named after Pierre Fatou.

Fatou's lemma can be used to prove the Fatou–Lebesgue theorem and Lebesgue's dominated convergence theorem.

Standard statement

In what follows, [math]\displaystyle{ \operatorname{\mathcal B}_{\R_{\geq 0}} }[/math] denotes the [math]\displaystyle{ \sigma }[/math]-algebra of Borel sets on [math]\displaystyle{ [0,+\infty] }[/math].

Theorem — Fatou's lemma. Given a measure space [math]\displaystyle{ (\Omega,\mathcal{F},\mu) }[/math] and a set [math]\displaystyle{ X \in \mathcal{F}, }[/math] let [math]\displaystyle{ \{f_n\} }[/math] be a sequence of [math]\displaystyle{ (\mathcal{F}, \operatorname{\mathcal B}_{\R_{\geq 0}}) }[/math]-measurable non-negative functions [math]\displaystyle{ f_n: X\to [0,+\infty] }[/math]. Define the function [math]\displaystyle{ f: X\to [0,+\infty] }[/math] by setting [math]\displaystyle{ f(x) =\liminf_{n\to\infty} f_n(x), }[/math] for every [math]\displaystyle{ x\in X }[/math].

Then [math]\displaystyle{ f }[/math] is [math]\displaystyle{ (\mathcal{F}, \operatorname{\mathcal B}_{\R_{\geq 0}}) }[/math]-measurable, and also [math]\displaystyle{ \int_X f\,d\mu \le \liminf_{n\to\infty} \int_X f_n\,d\mu }[/math], where the integrals may be infinite.

Fatou's lemma remains true if its assumptions hold [math]\displaystyle{ \mu }[/math]-almost everywhere. In other words, it is enough that there is a null set [math]\displaystyle{ N }[/math] such that the values [math]\displaystyle{ \{f_n(x)\} }[/math] are non-negative for every [math]\displaystyle{ {x\in X\setminus N}. }[/math] To see this, note that the integrals appearing in Fatou's lemma are unchanged if we change each function on [math]\displaystyle{ N }[/math].

Proof

Fatou's lemma does not require the monotone convergence theorem, but the latter can be used to provide a quick proof. A proof directly from the definitions of integrals is given further below.

In each case, the proof begins by analyzing the properties of [math]\displaystyle{ \textstyle g_n(x)=\inf_{k\geq n}f_k(x) }[/math]. These satisfy:

1. the sequence [math]\displaystyle{ \{g_n(x)\}_n }[/math] is pointwise non-decreasing at any x and
2. [math]\displaystyle{ g_n\leq f_n }[/math], [math]\displaystyle{ \forall n \in \N }[/math].

Since [math]\displaystyle{ f(x) =\liminf_{n\to\infty} f_n(x) = \lim_{n\to \infty} \inf_{k\geq n} f_k(x)=\lim_{n\to\infty}{g_n(x)} }[/math], we immediately see that f is measurable.

Via the Monotone Convergence Theorem

Moreover,

[math]\displaystyle{ \int_X f\,d\mu=\int_X\lim_{n\to \infty}g_n\,d\mu }[/math]

By the Monotone Convergence Theorem and property (1), the limit and integral may be interchanged:

[math]\displaystyle{ \begin{align} \int_X f\,d\mu&=\lim_{n\to \infty}\int_X g_n\,d\mu\\ &=\liminf_{n\to \infty}\int_X g_n\,d\mu\\ &\leq \liminf_{n\to \infty}\int_X f_n\,d\mu, \end{align} }[/math]

where the last step used property (2).

From "first principles"

To demonstrate that the monotone convergence theorem is not "hidden", the proof below does not use any properties of Lebesgue integral except those established here.

Denote by [math]\displaystyle{ \operatorname{SF}(f) }[/math] the set of simple [math]\displaystyle{ (\mathcal{F}, \operatorname{\mathcal B}_{\R_{\geq 0}}) }[/math]-measurable functions [math]\displaystyle{ s:X\to [0,\infty) }[/math] such that [math]\displaystyle{ 0\leq s\leq f }[/math] on [math]\displaystyle{ X }[/math].

Monotonicity — 

• If [math]\displaystyle{ f \leq g }[/math] everywhere on [math]\displaystyle{ X, }[/math] then

[math]\displaystyle{ \int_X f\,d\mu \leq \int_X g\,d\mu. }[/math]

• If [math]\displaystyle{ X_1,X_2 \in \mathcal{F} }[/math] and [math]\displaystyle{ X_1 \subseteq X_2, }[/math] then

[math]\displaystyle{ \int_{X_1} f\,d\mu \leq \int_{X_2} f\,d\mu. }[/math]

• If f is nonnegative and [math]\displaystyle{ S= \cup^\infty_{i=1}S_i }[/math], where [math]\displaystyle{ S_1\subseteq\ldots\subseteq S_i\subseteq\ldots\subseteq S }[/math] is a non-decreasing chain of [math]\displaystyle{ \mu }[/math]-measurable sets, then

[math]\displaystyle{ \int_S{f\,d\mu}=\lim_{n\to\infty}{\int_{S_n}{f\,d\mu}} }[/math]

Proof

1. Since [math]\displaystyle{ f \leq g, }[/math] we have

[math]\displaystyle{ \operatorname{SF}(f) \subseteq \operatorname{SF}(g). }[/math]

By definition of Lebesgue integral and the properties of supremum,

[math]\displaystyle{ \int_X f\,d\mu = \sup_{s\in {\rm SF}(f)}\int_X s\,d\mu \leq \sup_{s\in {\rm SF}(g)}\int_X s\,d\mu = \int_X g\,d\mu. }[/math]

2. Let [math]\displaystyle{ {\mathbf 1}_{X_1} }[/math] be the indicator function of the set [math]\displaystyle{ X_1. }[/math] It can be deduced from the definition of Lebesgue integral that

[math]\displaystyle{ \int_{X_2} f\cdot {\mathbf 1}_{X_1} \,d\mu = \int_{X_1} f \,d\mu }[/math]

if we notice that, for every [math]\displaystyle{ s \in {\rm SF}(f\cdot {\mathbf 1}_{X_1}), }[/math] [math]\displaystyle{ s=0 }[/math] outside of [math]\displaystyle{ X_1. }[/math] Combined with the previous property, the inequality [math]\displaystyle{ f\cdot {\mathbf 1}_{X_1} \leq f }[/math] implies

[math]\displaystyle{ \int_{X_1} f \,d\mu = \int_{X_2} f\cdot {\mathbf 1}_{X_1} \,d\mu \leq \int_{X_2} f \,d\mu. }[/math]

3. First note that the claim holds if f is the indicator function of a set, by monotonicity of measures. By linearity, this also immediately implies the claim for simple functions.

Since any simple function supported on S_n is simple and supported on X, we must have

[math]\displaystyle{ \int_X{f\,d\mu}\geq\lim_{n\to\infty}{\int_{S_n}{f\,d\mu}} }[/math].

For the reverse, suppose g ∈ SF(f) with [math]\displaystyle{ \textstyle \int_X{f\,d\mu}-\epsilon\leq\int_X{g\,d\mu} }[/math] By the above,

[math]\displaystyle{ \int_X{f\,d\mu}-\epsilon\leq\int_X{g\,d\mu}=\lim_{n\to\infty}{\int_{S_n}{g\,d\mu}}\leq\lim_{n\to\infty}{\int_{S_n}{f\,d\mu}} }[/math]

Now we turn to the main theorem

Step 1 — [math]\displaystyle{ g_n=g_n(x) }[/math] is [math]\displaystyle{ (\mathcal{F}, \operatorname{\mathcal B}_{\R_{\geq 0}}) }[/math]-measurable, for every [math]\displaystyle{ n\geq 1 }[/math], as is [math]\displaystyle{ f }[/math].

Proof

Recall the closed intervals generate the Borel σ-algebra. Thus it suffices to show, for every [math]\displaystyle{ t\in [-\infty,+\infty] }[/math], that [math]\displaystyle{ g^{-1}_n([t,+\infty])\in\mathcal{F} }[/math]. Now observe that

[math]\displaystyle{ \begin{align} g^{-1}_n([t,+\infty])&=\left\{x\in X\mid g_n(x)\geq t\right\}\\[3pt] &=\left\{x\in X \; \left| \; \inf_{k \, \geq \, n} f_k(x) \geq t \right. \right\}\\[3pt] &=\bigcap_{k \, \geq \, n}\left\{x\in X\mid f_k(x)\geq t\right\}\\[3pt] &=\bigcap_{k \, \geq \, n} f^{-1}_k([t,+\infty]) \end{align} }[/math]

Every set on the right-hand side is from [math]\displaystyle{ \mathcal{F} }[/math], which is closed under countable intersections. Thus the left-hand side is also a member of [math]\displaystyle{ \mathcal{F} }[/math].

Similarly, it is enough to verify that [math]\displaystyle{ f^{-1}([0,t])\in\mathcal{F} }[/math], for every [math]\displaystyle{ t\in [-\infty,+\infty] }[/math]. Since the sequence [math]\displaystyle{ \{g_n(x)\} }[/math] pointwise non-decreases,

[math]\displaystyle{ f^{-1}([0,t])=\bigcap_{n}g^{-1}_n([0,t])\in\mathcal{F} }[/math].

Step 2 — Given a simple function [math]\displaystyle{ s\in\operatorname{SF}(f) }[/math] and a real number [math]\displaystyle{ t\in (0,1) }[/math], define

[math]\displaystyle{ B^{s,t}_k=\{x\in X\mid t\cdot s(x)\leq g_k(x)\}\subseteq X. }[/math]

Then [math]\displaystyle{ B^{s,t}_k\in\mathcal{F} }[/math], [math]\displaystyle{ B^{s,t}_k\subseteq B^{s,t}_{k+1} }[/math], and [math]\displaystyle{ \textstyle X=\bigcup_k B^{s,t}_k }[/math].

Proof

Step 2a. To prove the first claim, write s as a weighted sum of indicator functions of disjoint sets:

[math]\displaystyle{ s=\sum^m_{i=1}c_i\cdot\mathbf{1}_{A_i} }[/math].

Then

[math]\displaystyle{ B^{s,t}_k=\bigcup^m_{i=1}\Bigl(g^{-1}_k\Bigl([t\cdot c_i,+\infty]\Bigr)\cap A_i\Bigr) }[/math].

Since the pre-image [math]\displaystyle{ g^{-1}_k\Bigl([t\cdot c_i,+\infty]\Bigr) }[/math] of the Borel set [math]\displaystyle{ [t\cdot c_i,+\infty] }[/math] under the measurable function [math]\displaystyle{ g_k }[/math] is measurable, and [math]\displaystyle{ \sigma }[/math]-algebras are closed under finite intersection and unions, the first claim follows.

Step 2b. To prove the second claim, note that, for each [math]\displaystyle{ k }[/math] and every [math]\displaystyle{ x\in X }[/math], [math]\displaystyle{ g_k(x) \leq g_{k+1}(x). }[/math]

Step 2c. To prove the third claim, suppose for contradiction there exists

[math]\displaystyle{ x_0\in X\setminus\bigcup_k B^{s,t}_k=\bigcap_k(X\setminus B^{s,t}_k) }[/math]

Then [math]\displaystyle{ g_k(x_0)\lt t\cdot s(x_0) }[/math], for every [math]\displaystyle{ k }[/math]. Taking the limit as [math]\displaystyle{ k\to\infty }[/math],

[math]\displaystyle{ f(x_0)\leq t\cdot s(x_0)\lt s(x_0). }[/math]

This contradicts our initial assumption that [math]\displaystyle{ s\leq f }[/math].

Step 3 — From step 2 and monotonicity,

[math]\displaystyle{ \lim_n\int_{B^{s,t}_n}s\,d\mu=\int_Xs\,d\mu. }[/math]

Step 4 — For every [math]\displaystyle{ s\in\operatorname{SF}(f) }[/math],

[math]\displaystyle{ \int_X s\,d\mu\leq \lim_k\int_X g_k\,d\mu }[/math].

Proof

Indeed, using the definition of [math]\displaystyle{ B^{s,t}_k }[/math], the non-negativity of [math]\displaystyle{ g_k }[/math], and the monotonicity of Lebesgue integral, we have

[math]\displaystyle{ \forall k \geq 1 \qquad \int_{B^{s,t}_k}t\cdot s\,d\mu\leq \int_{B^{s,t}_k} g_k\,d\mu\leq \int_X g_k\,d\mu }[/math].

In accordance with Step 4, as [math]\displaystyle{ k\to\infty }[/math] the inequality becomes

[math]\displaystyle{ t\int_X s\,d\mu\leq\lim_k\int_X g_k\,d\mu }[/math].

Taking the limit as [math]\displaystyle{ t\uparrow 1 }[/math] yields

[math]\displaystyle{ \int_X s\,d\mu\leq\lim_k\int_X g_k\,d\mu }[/math],

as required.

Step 5 — To complete the proof, we apply the definition of Lebesgue integral to the inequality established in Step 4 and take into account that [math]\displaystyle{ g_n\leq f_n }[/math]:

[math]\displaystyle{ \begin{align} \int_X f \,d\mu&=\sup_{s\in\operatorname{SF}(f)}\int_X s\,d\mu\\ &\leq\lim_k\int_X g_k\,d\mu\\ &=\liminf_k\int_X g_k\,d\mu\\ &\leq\liminf_k\int_X f_k\,d\mu \end{align} }[/math]

The proof is complete.

Examples for strict inequality

Equip the space [math]\displaystyle{ S }[/math] with the Borel σ-algebra and the Lebesgue measure.

• Example for a probability space: Let [math]\displaystyle{ S=[0,1] }[/math] denote the unit interval. For every natural number [math]\displaystyle{ n }[/math] define

[math]\displaystyle{ f_n(x)=\begin{cases}n&\text{for }x\in (0,1/n),\\ 0&\text{otherwise.} \end{cases} }[/math]

• Example with uniform convergence: Let [math]\displaystyle{ S }[/math] denote the set of all real numbers. Define

[math]\displaystyle{ f_n(x)=\begin{cases}\frac1n&\text{for }x\in [0,n],\\ 0&\text{otherwise.} \end{cases} }[/math]

These sequences [math]\displaystyle{ (f_n)_{n\in\N} }[/math] converge on [math]\displaystyle{ S }[/math] pointwise (respectively uniformly) to the zero function (with zero integral), but every [math]\displaystyle{ f_n }[/math] has integral one.

The role of non-negativity

A suitable assumption concerning the negative parts of the sequence f₁, f₂, . . . of functions is necessary for Fatou's lemma, as the following example shows. Let S denote the half line [0,∞) with the Borel σ-algebra and the Lebesgue measure. For every natural number n define

[math]\displaystyle{ f_n(x)=\begin{cases}-\frac1n&\text{for }x\in [n,2n],\\ 0&\text{otherwise.} \end{cases} }[/math]

This sequence converges uniformly on S to the zero function and the limit, 0, is reached in a finite number of steps: for every x ≥ 0, if n > x, then f_n(x) = 0. However, every function f_n has integral −1. Contrary to Fatou's lemma, this value is strictly less than the integral of the limit (0).

As discussed in § Extensions and variations of Fatou's lemma below, the problem is that there is no uniform integrable bound on the sequence from below, while 0 is the uniform bound from above.

Reverse Fatou lemma

Let f₁, f₂, . . . be a sequence of extended real-valued measurable functions defined on a measure space (S,Σ,μ). If there exists a non-negative integrable function g on S such that f_n ≤ g for all n, then

[math]\displaystyle{ \limsup_{n\to\infty}\int_S f_n\,d\mu\leq\int_S\limsup_{n\to\infty}f_n\,d\mu. }[/math]

Note: Here g integrable means that g is measurable and that [math]\displaystyle{ \textstyle\int_S g\,d\mu\lt \infty }[/math].

Sketch of proof

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence [math]\displaystyle{ g - f_n. }[/math] Since [math]\displaystyle{ \textstyle\int_Sg\,d\mu \lt +\infty, }[/math] this sequence is defined [math]\displaystyle{ \mu }[/math]-almost everywhere and non-negative.

Extensions and variations of Fatou's lemma

Integrable lower bound

Let f₁, f₂, . . . be a sequence of extended real-valued measurable functions defined on a measure space (S,Σ,μ). If there exists an integrable function g on S such that f_n ≥ −g for all n, then

[math]\displaystyle{ \int_S \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu. }[/math]

Proof

Apply Fatou's lemma to the non-negative sequence given by f_n + g.

Pointwise convergence

If in the previous setting the sequence f₁, f₂, . . . converges pointwise to a function f μ-almost everywhere on S, then

[math]\displaystyle{ \int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,. }[/math]

Proof

Note that f has to agree with the limit inferior of the functions f_n almost everywhere, and that the values of the integrand on a set of measure zero have no influence on the value of the integral.

Convergence in measure

The last assertion also holds, if the sequence f₁, f₂, . . . converges in measure to a function f.

Proof

There exists a subsequence such that

[math]\displaystyle{ \lim_{k\to\infty} \int_S f_{n_k}\,d\mu=\liminf_{n\to\infty} \int_S f_n\,d\mu. }[/math]

Since this subsequence also converges in measure to f, there exists a further subsequence, which converges pointwise to f almost everywhere, hence the previous variation of Fatou's lemma is applicable to this subsubsequence.

Fatou's Lemma with Varying Measures

In all of the above statements of Fatou's Lemma, the integration was carried out with respect to a single fixed measure μ. Suppose that μ_n is a sequence of measures on the measurable space (S,Σ) such that (see Convergence of measures)

[math]\displaystyle{ \forall E\in \mathcal{F} \colon\; \mu_n(E)\to \mu(E) }[/math].

Then, with f_n non-negative integrable functions and f being their pointwise limit inferior, we have

[math]\displaystyle{ \int_S f\,d\mu \leq \liminf_{n\to \infty} \int_S f_n\, d\mu_n. }[/math]

Proof

We will prove something a bit stronger here. Namely, we will allow fn to converge μ-almost everywhere on a subset E of S. We seek to show that [math]\displaystyle{ \int_E f\,d\mu \le \liminf_{n\to\infty} \int_E f_n\,d\mu_n\,. }[/math] Let [math]\displaystyle{ K=\{x\in E|f_n(x)\rightarrow f(x)\} }[/math]. Then μ(E-K)=0 and [math]\displaystyle{ \int_{E}f\,d\mu=\int_{E-K}f\,d\mu,~~~\int_{E}f_n\,d\mu=\int_{E-K}f_n\,d\mu ~\forall n\in \N. }[/math] Thus, replacing E by E-K we may assume that fn converge to f pointwise on E. Next, note that for any simple function φ we have [math]\displaystyle{ \int_{E}\phi\, d\mu=\lim_{n\to \infty} \int_{E} \phi\, d\mu_n. }[/math] Hence, by the definition of the Lebesgue Integral, it is enough to show that if φ is any non-negative simple function less than or equal to f, then [math]\displaystyle{ \int_{E}\phi \,d\mu\leq \liminf_{n\rightarrow \infty} \int_{E}f_n\,d\mu_n }[/math] Let a be the minimum non-negative value of φ. Define [math]\displaystyle{ A=\{x\in E |\phi(x)\gt a\} }[/math] We first consider the case when [math]\displaystyle{ \int_{E}\phi\, d\mu=\infty }[/math]. We must have that μ(A) is infinite since [math]\displaystyle{ \int_{E}\phi\, d\mu \leq M\mu(A), }[/math] where M is the (necessarily finite) maximum value of that φ attains. Next, we define [math]\displaystyle{ A_n=\{x\in E |f_k(x)\gt a~\forall k\geq n \}. }[/math] We have that [math]\displaystyle{ A\subseteq \bigcup_n A_n \Rightarrow \mu(\bigcup_n A_n)=\infty. }[/math] But An is a nested increasing sequence of functions and hence, by the continuity from below μ, [math]\displaystyle{ \lim_{n\rightarrow \infty} \mu(A_n)=\infty. }[/math]. Thus, [math]\displaystyle{ \lim_{n\to\infty}\mu_n(A_n)=\mu(A_n)=\infty. }[/math]. At the same time, [math]\displaystyle{ \int_E f_n\, d\mu_n \geq a \mu_n(A_n) \Rightarrow \liminf_{n\to \infty}\int_E f_n \, d\mu_n = \infty = \int_E \phi\, d\mu, }[/math] proving the claim in this case. The remaining case is when [math]\displaystyle{ \int_{E}\phi\, d\mu\lt \infty }[/math]. We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define [math]\displaystyle{ A_n=\{x\in E|f_k(x)\gt (1-\epsilon)\phi(x)~\forall k\geq n\}. }[/math] Then An is a nested increasing sequence of sets whose union contains A. Thus, A-An is a decreasing sequence of sets with empty intersection. Since A has finite measure (this is why we needed to consider the two separate cases), [math]\displaystyle{ \lim_{n\rightarrow \infty} \mu(A-A_n)=0. }[/math] Thus, there exists n such that [math]\displaystyle{ \mu(A-A_k)\lt \epsilon ,~\forall k\geq n. }[/math] Therefore, since [math]\displaystyle{ \lim_{n\to \infty} \mu_n(A-A_k)=\mu(A-A_k), }[/math] there exists N such that [math]\displaystyle{ \mu_k(A-A_k)\lt \epsilon,~\forall k\geq N. }[/math] Hence, for [math]\displaystyle{ k\geq N }[/math] [math]\displaystyle{ \int_E f_k \, d\mu_k \geq \int_{A_k}f_k \, d\mu_k \geq (1-\epsilon)\int_{A_k}\phi\, d\mu_k. }[/math] At the same time, [math]\displaystyle{ \int_E \phi \, d\mu_k = \int_A \phi \, d\mu_k = \int_{A_k} \phi \, d\mu_k + \int_{A-A_k} \phi \, d\mu_k. }[/math] Hence, [math]\displaystyle{ (1-\epsilon)\int_{A_k} \phi \, d\mu_k \geq (1-\epsilon)\int_E \phi \, d\mu_k - \int_{A-A_k} \phi \, d\mu_k. }[/math] Combining these inequalities gives that [math]\displaystyle{ \int_{E} f_k \, d\mu_k \geq (1-\epsilon)\int_E \phi \, d\mu_k - \int_{A-A_k} \phi \, d\mu_k \geq \int_E \phi \, d\mu_k - \epsilon\left(\int_{E} \phi \, d\mu_k+M\right). }[/math] Hence, sending ε to 0 and taking the liminf in n, we get that [math]\displaystyle{ \liminf_{n\rightarrow \infty} \int_{E} f_n \, d\mu_k \geq \int_E \phi \, d\mu, }[/math] completing the proof.

Fatou's lemma for conditional expectations

In probability theory, by a change of notation, the above versions of Fatou's lemma are applicable to sequences of random variables X₁, X₂, . . . defined on a probability space [math]\displaystyle{ \scriptstyle(\Omega,\,\mathcal F,\,\mathbb P) }[/math]; the integrals turn into expectations. In addition, there is also a version for conditional expectations.

Standard version

Let X₁, X₂, . . . be a sequence of non-negative random variables on a probability space [math]\displaystyle{ \scriptstyle(\Omega,\mathcal F,\mathbb P) }[/math] and let [math]\displaystyle{ \scriptstyle \mathcal G\,\subset\,\mathcal F }[/math] be a sub-σ-algebra. Then

[math]\displaystyle{ \mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G] }[/math]   almost surely.

Note: Conditional expectation for non-negative random variables is always well defined, finite expectation is not needed.

Proof

Besides a change of notation, the proof is very similar to the one for the standard version of Fatou's lemma above, however the monotone convergence theorem for conditional expectations has to be applied.

Let X denote the limit inferior of the X_n. For every natural number k define pointwise the random variable

[math]\displaystyle{ Y_k=\inf_{n\ge k}X_n. }[/math]

Then the sequence Y₁, Y₂, . . . is increasing and converges pointwise to X. For k ≤ n, we have Y_k ≤ X_n, so that

[math]\displaystyle{ \mathbb{E}[Y_k|\mathcal G]\le\mathbb{E}[X_n|\mathcal G] }[/math]   almost surely

by the monotonicity of conditional expectation, hence

[math]\displaystyle{ \mathbb{E}[Y_k|\mathcal G]\le\inf_{n\ge k}\mathbb{E}[X_n|\mathcal G] }[/math]   almost surely,

because the countable union of the exceptional sets of probability zero is again a null set. Using the definition of X, its representation as pointwise limit of the Y_k, the monotone convergence theorem for conditional expectations, the last inequality, and the definition of the limit inferior, it follows that almost surely

[math]\displaystyle{ \begin{align} \mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr] &=\mathbb{E}[X|\mathcal G] =\mathbb{E}\Bigl[\lim_{k\to\infty}Y_k\,\Big|\,\mathcal G\Bigr] =\lim_{k\to\infty}\mathbb{E}[Y_k|\mathcal G]\\ &\le\lim_{k\to\infty} \inf_{n\ge k}\mathbb{E}[X_n|\mathcal G] =\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]. \end{align} }[/math]

Extension to uniformly integrable negative parts

Let X₁, X₂, . . . be a sequence of random variables on a probability space [math]\displaystyle{ \scriptstyle(\Omega,\mathcal F,\mathbb P) }[/math] and let [math]\displaystyle{ \scriptstyle \mathcal G\,\subset\,\mathcal F }[/math] be a sub-σ-algebra. If the negative parts

[math]\displaystyle{ X_n^-:=\max\{-X_n,0\},\qquad n\in{\mathbb N}, }[/math]

are uniformly integrable with respect to the conditional expectation, in the sense that, for ε > 0 there exists a c > 0 such that

[math]\displaystyle{ \mathbb{E}\bigl[X_n^-1_{\{X_n^-\gt c\}}\,|\,\mathcal G\bigr]\lt \varepsilon, \qquad\text{for all }n\in\mathbb{N},\,\text{almost surely} }[/math],

then

[math]\displaystyle{ \mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G] }[/math]   almost surely.

Note: On the set where

[math]\displaystyle{ X:=\liminf_{n\to\infty}X_n }[/math]

satisfies

[math]\displaystyle{ \mathbb{E}[\max\{X,0\}\,|\,\mathcal G]=\infty, }[/math]

the left-hand side of the inequality is considered to be plus infinity. The conditional expectation of the limit inferior might not be well defined on this set, because the conditional expectation of the negative part might also be plus infinity.

Proof

Let ε > 0. Due to uniform integrability with respect to the conditional expectation, there exists a c > 0 such that

[math]\displaystyle{ \mathbb{E}\bigl[X_n^-1_{\{X_n^-\gt c\}}\,|\,\mathcal G\bigr]\lt \varepsilon \qquad\text{for all }n\in\mathbb{N},\,\text{almost surely}. }[/math]

Since

[math]\displaystyle{ X+c\le\liminf_{n\to\infty}(X_n+c)^+, }[/math]

where x⁺ := max{x,0} denotes the positive part of a real x, monotonicity of conditional expectation (or the above convention) and the standard version of Fatou's lemma for conditional expectations imply

[math]\displaystyle{ \mathbb{E}[X\,|\,\mathcal G]+c \le\mathbb{E}\Bigl[\liminf_{n\to\infty}(X_n+c)^+\,\Big|\,\mathcal G\Bigr] \le\liminf_{n\to\infty}\mathbb{E}[(X_n+c)^+\,|\,\mathcal G] }[/math]   almost surely.

Since

[math]\displaystyle{ (X_n+c)^+=(X_n+c)+(X_n+c)^-\le X_n+c+X_n^-1_{\{X_n^-\gt c\}}, }[/math]

we have

[math]\displaystyle{ \mathbb{E}[(X_n+c)^+\,|\,\mathcal G] \le\mathbb{E}[X_n\,|\,\mathcal G]+c+\varepsilon }[/math]   almost surely,

hence

[math]\displaystyle{ \mathbb{E}[X\,|\,\mathcal G]\le \liminf_{n\to\infty}\mathbb{E}[X_n\,|\,\mathcal G]+\varepsilon }[/math]   almost surely.

This implies the assertion.

References

• Carothers, N. L. (2000). Real Analysis. New York: Cambridge University Press. pp. 321–22. ISBN 0-521-49756-6. https://archive.org/details/realanalysis0000caro. 
• Royden, H. L. (1988). Real Analysis (3rd ed.). London: Collier Macmillan. ISBN 0-02-404151-3. 
• Weir, Alan J. (1973). "The Convergence Theorems". Lebesgue Integration and Measure. Cambridge: Cambridge University Press. pp. 93–118. ISBN 0-521-08728-7.