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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, ${\displaystyle \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}}$ denotes the ${\displaystyle \sigma }$-algebra of Borel sets on ${\displaystyle }$.

Theorem — Fatou's lemma. Given a measure space ${\displaystyle (\Omega ,{\mathcal {F}},\mu )}$ and a set ${\displaystyle X\in {\mathcal {F}},}$ let ${\displaystyle \{f_{n}\}}$ be a sequence of ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable non-negative functions ${\displaystyle f_{n}:X\to }$. Define the function ${\displaystyle f:X\to }$ by ${\displaystyle f(x)=\liminf _{n\to \infty }f_{n}(x),}$ for every ${\displaystyle x\in X}$. Then ${\displaystyle f}$ is ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable, and

${\displaystyle \int _{X}f\,d\mu \leq \liminf _{n\to \infty }\int _{X}f_{n}\,d\mu ,}$

where the integrals and the Limit inferior may be infinite.

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

Proof

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

Via the Monotone Convergence Theorem

let ${\displaystyle \textstyle g_{n}(x)=\inf _{k\geq n}f_{k}(x)}$. Then:

1. the sequence ${\displaystyle \{g_{n}(x)\}_{n}}$ is pointwise non-decreasing at any x and
2. ${\displaystyle g_{n}\leq f_{n}}$, ${\displaystyle \forall n\in \mathbb {N} }$.

Since

${\displaystyle f(x)=\liminf _{n\to \infty }f_{n}(x)=\sup _{n}\inf _{k\geq n}f_{k}(x)=\sup _{n}g_{n}(x)}$,

and infima and suprema of measurable functions are measurable we see that ${\displaystyle f}$ is measurable.

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

${\displaystyle {\begin{aligned}\int _{X}f\,d\mu &=\int _{X}\sup _{n}g_{n}\,d\mu \\&=\sup _{n}\int _{X}g_{n}\,d\mu \\&=\liminf _{n\to \infty }\int _{X}g_{n}\,d\mu \\&\leq \liminf _{n\to \infty }\int _{X}f_{n}\,d\mu ,\end{aligned}}}$

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 and the fact that the functions ${\displaystyle f}$ and ${\displaystyle g_{n}}$ are measurable.

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

Monotonicity — 

• If ${\displaystyle f\leq g}$ everywhere on ${\displaystyle X,}$ then

${\displaystyle \int _{X}f\,d\mu \leq \int _{X}g\,d\mu .}$

• If ${\displaystyle X_{1},X_{2}\in {\mathcal {F}}}$ and ${\displaystyle X_{1}\subseteq X_{2},}$ then

${\displaystyle \int _{X_{1}}f\,d\mu \leq \int _{X_{2}}f\,d\mu .}$

• If f is nonnegative and ${\displaystyle S=\cup _{i=1}^{\infty }S_{i}}$, where ${\displaystyle S_{1}\subseteq \ldots \subseteq S_{i}\subseteq \ldots \subseteq S}$ is a non-decreasing chain of ${\displaystyle \mu }$-measurable sets, then

${\displaystyle \int _{S}{f\,d\mu }=\lim _{n\to \infty }{\int _{S_{n}}{f\,d\mu }}}$

1. Since ${\displaystyle f\leq g,}$ we have

${\displaystyle \operatorname {SF} (f)\subseteq \operatorname {SF} (g).}$

By definition of Lebesgue integral and the properties of supremum,

${\displaystyle \int _{X}f\,d\mu =\sup _{s\in {\rm {SF}}(f)}\int _{X}s\,d\mu \leq \sup _{s\in {\rm {SF}}(g)}\int _{X}s\,d\mu =\int _{X}g\,d\mu .}$

2. Let ${\displaystyle {\mathbf {1} }_{X_{1}}}$ be the indicator function of the set ${\displaystyle X_{1}.}$ It can be deduced from the definition of Lebesgue integral that

${\displaystyle \int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu =\int _{X_{1}}f\,d\mu }$

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

${\displaystyle \int _{X_{1}}f\,d\mu =\int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu \leq \int _{X_{2}}f\,d\mu .}$

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

${\displaystyle \int _{X}{f\,d\mu }\geq \lim _{n\to \infty }{\int _{S_{n}}{f\,d\mu }}}$.

For the reverse, suppose g ∈ SF(f) with ${\displaystyle \textstyle \int _{X}{f\,d\mu }-\epsilon \leq \int _{X}{g\,d\mu }}$ By the above,

${\displaystyle \int _{X}{f\,d\mu }-\epsilon \leq \int _{X}{g\,d\mu }=\lim _{n\to \infty }{\int _{S_{n}}{g\,d\mu }}\leq \lim _{n\to \infty }{\int _{S_{n}}{f\,d\mu }}}$

Now we turn to the main theorem

Step 1 — ${\displaystyle g_{n}=g_{n}(x)}$ is ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable, for every ${\displaystyle n\geq 1}$, as is ${\displaystyle f}$.

Recall the closed intervals generate the Borel σ-algebra. Thus it suffices to show, for every ${\displaystyle t\in }$, that ${\displaystyle g_{n}^{-1}()\in {\mathcal {F}}}$. Now observe that

${\displaystyle {\begin{aligned}g_{n}^{-1}()&=\left\{x\in X\mid g_{n}(x)\geq t\right\}\\&=\left\{x\in X\;\left|\;\inf _{k\,\geq \,n}f_{k}(x)\geq t\right.\right\}\\&=\bigcap _{k\,\geq \,n}\left\{x\in X\mid f_{k}(x)\geq t\right\}\\&=\bigcap _{k\,\geq \,n}f_{k}^{-1}()\end{aligned}}}$

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

Similarly, it is enough to verify that ${\displaystyle f^{-1}()\in {\mathcal {F}}}$, for every ${\displaystyle t\in }$. Since the sequence ${\displaystyle \{g_{n}(x)\}}$ pointwise non-decreases,

${\displaystyle f^{-1}()=\bigcap _{n}g_{n}^{-1}()\in {\mathcal {F}}}$.

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

${\displaystyle B_{k}^{s,t}=\{x\in X\mid t\cdot s(x)\leq g_{k}(x)\}\subseteq X.}$

Then ${\displaystyle B_{k}^{s,t}\in {\mathcal {F}}}$, ${\displaystyle B_{k}^{s,t}\subseteq B_{k+1}^{s,t}}$, and ${\displaystyle \textstyle X=\bigcup _{k}B_{k}^{s,t}}$.

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

${\displaystyle s=\sum _{i=1}^{m}c_{i}\cdot \mathbf {1} _{A_{i}}}$.

Then

${\displaystyle B_{k}^{s,t}=\bigcup _{i=1}^{m}{\Bigl (}g_{k}^{-1}{\Bigl (}{\Bigr )}\cap A_{i}{\Bigr )}}$.

Since the pre-image ${\displaystyle g_{k}^{-1}{\Bigl (}{\Bigr )}}$ of the Borel set ${\displaystyle }$ under the measurable function ${\displaystyle g_{k}}$ is measurable, and ${\displaystyle \sigma }$-algebras are closed under finite intersection and unions, the first claim follows.

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

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

${\displaystyle x_{0}\in X\setminus \bigcup _{k}B_{k}^{s,t}=\bigcap _{k}(X\setminus B_{k}^{s,t})}$

Then ${\displaystyle g_{k}(x_{0})<t\cdot s(x_{0})}$, for every ${\displaystyle k}$. Taking the limit as ${\displaystyle k\to \infty }$,

${\displaystyle f(x_{0})\leq t\cdot s(x_{0})<s(x_{0}).}$

This contradicts our initial assumption that ${\displaystyle s\leq f}$.

Step 3 — From step 2 and monotonicity,

${\displaystyle \lim _{n}\int _{B_{n}^{s,t}}s\,d\mu =\int _{X}s\,d\mu .}$

Step 4 — For every ${\displaystyle s\in \operatorname {SF} (f)}$,

${\displaystyle \int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$.

Indeed, using the definition of ${\displaystyle B_{k}^{s,t}}$, the non-negativity of ${\displaystyle g_{k}}$, and the monotonicity of Lebesgue integral, we have

${\displaystyle \forall k\geq 1\qquad \int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \int _{B_{k}^{s,t}}g_{k}\,d\mu \leq \int _{X}g_{k}\,d\mu }$.

In accordance with Step 4, as ${\displaystyle k\to \infty }$ the inequality becomes

${\displaystyle t\int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$.

Taking the limit as ${\displaystyle t\uparrow 1}$ yields

${\displaystyle \int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$,

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 ${\displaystyle g_{n}\leq f_{n}}$:

${\displaystyle {\begin{aligned}\int _{X}f\,d\mu &=\sup _{s\in \operatorname {SF} (f)}\int _{X}s\,d\mu \\&\leq \lim _{k}\int _{X}g_{k}\,d\mu \\&=\liminf _{k}\int _{X}g_{k}\,d\mu \\&\leq \liminf _{k}\int _{X}f_{k}\,d\mu \end{aligned}}}$

The proof is complete.

Examples for strict inequality

Equip the space ${\displaystyle S}$ with the Borel σ-algebra and the Lebesgue measure.

• Example for a probability space: Let ${\displaystyle S=}$ denote the unit interval. For every natural number ${\displaystyle n}$ define

${\displaystyle f_{n}(x)={\begin{cases}n&{\text{for }}x\in (0,1/n),\\0&{\text{otherwise.}}\end{cases}}}$

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

${\displaystyle f_{n}(x)={\begin{cases}{\frac {1}{n}}&{\text{for }}x\in ,\\0&{\text{otherwise.}}\end{cases}}}$

These sequences ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ converge on ${\displaystyle S}$ pointwise (respectively uniformly) to the zero function (with zero integral), but every ${\displaystyle f_{n}}$ 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

${\displaystyle f_{n}(x)={\begin{cases}-{\frac {1}{n}}&{\text{for }}x\in ,\\0&{\text{otherwise.}}\end{cases}}}$

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

${\displaystyle \limsup _{n\to \infty }\int _{S}f_{n}\,d\mu \leq \int _{S}\limsup _{n\to \infty }f_{n}\,d\mu .}$

Note: Here g integrable means that g is measurable and that ${\displaystyle \textstyle \int _{S}g\,d\mu <\infty }$.

Sketch of proof

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence ${\displaystyle g-f_{n}.}$ Since ${\displaystyle \textstyle \int _{S}g\,d\mu <+\infty ,}$ this sequence is defined ${\displaystyle \mu }$-almost everywhere and non-negative.

Extensions and variations of Fatou's lemma

Integrable lower bound

Let ${\displaystyle f_{1},f_{2},\ldots }$ be a sequence of extended real-valued measurable functions defined on a measure space ${\displaystyle (S,\Sigma ,\mu )}$. If there exists an integrable function ${\displaystyle g}$ on ${\displaystyle S}$ such that ${\displaystyle f_{n}\geq -g}$ for all ${\displaystyle n}$, then

${\displaystyle \int _{S}\liminf _{n\to \infty }f_{n}\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .}$

Proof

Apply Fatou's lemma to the non-negative sequence given by ${\displaystyle f_{n}+g}$.

Pointwise convergence

If in the previous setting the sequence ${\displaystyle f_{1},f_{2},\ldots }$ converges pointwise to a function ${\displaystyle f}$ ${\displaystyle \mu }$-almost everywhere on ${\displaystyle S}$, then

${\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu \,.}$

Proof

Note that ${\displaystyle f}$ has to agree with the limit inferior of the functions ${\displaystyle f_{n}}$ 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 ${\displaystyle f_{1},f_{2},\ldots }$ converges in measure to a function ${\displaystyle f}$.

Proof

There exists a subsequence such that

${\displaystyle \lim _{k\to \infty }\int _{S}f_{n_{k}}\,d\mu =\liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .}$

Since this subsequence also converges in measure to ${\displaystyle f}$, there exists a further subsequence, which converges pointwise to ${\displaystyle f}$ 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 ${\displaystyle \mu }$. Suppose that ${\displaystyle \mu _{n}}$ is a sequence of measures on the measurable space ${\displaystyle (M,\Sigma )}$ such that (see Convergence of measures)

${\displaystyle \forall E\in {\mathcal {F}}\colon \;\mu _{n}(E)\to \mu (E)}$.

Then, with ${\displaystyle f_{n}}$ non-negative integrable functions and ${\displaystyle f}$ being their pointwise limit inferior, we have

${\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu _{n}.}$

Proof

We will prove something a bit stronger here. Namely, we will allow ${\displaystyle f_{n}}$ to converge ${\displaystyle \mu }$-almost everywhere on a subset ${\displaystyle E}$ of ${\displaystyle S}$. We seek to show that ${\displaystyle \int _{E}f\,d\mu \leq \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu _{n}\,.}$ Let ${\displaystyle K=\{x\in E|f_{n}(x)\rightarrow f(x)\}}$. Then μ(E-K)=0 and ${\displaystyle \int _{E}f\,d\mu =\int _{E-K}f\,d\mu ,~~~\int _{E}f_{n}\,d\mu =\int _{E-K}f_{n}\,d\mu ~\forall n\in \mathbb {N} .}$ Thus, replacing ${\displaystyle E}$ by ${\displaystyle E-K}$ we may assume that ${\displaystyle f_{n}}$ converge to ${\displaystyle f}$ pointwise on ${\displaystyle E}$. Next, note that for any simple function ${\displaystyle \phi }$ we have ${\displaystyle \int _{E}\phi \,d\mu =\lim _{n\to \infty }\int _{E}\phi \,d\mu _{n}.}$ Hence, by the definition of the Lebesgue Integral, it is enough to show that if ${\displaystyle \phi }$ is any non-negative simple function less than or equal to ${\displaystyle f}$, then ${\displaystyle \int _{E}\phi \,d\mu \leq \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{n}}$ Let a be the minimum non-negative value of φ. Define ${\displaystyle A=\{x\in E|\phi (x)>a\}}$ We first consider the case when ${\displaystyle \int _{E}\phi \,d\mu =\infty }$. We must have that ${\displaystyle \mu (A)}$ is infinite since ${\displaystyle \int _{E}\phi \,d\mu \leq M\mu (A),}$ where M is the (necessarily finite) maximum value of that ${\displaystyle \phi }$ attains. Next, we define ${\displaystyle A_{n}=\{x\in E|f_{k}(x)>a~\forall k\geq n\}.}$ We have that ${\displaystyle A\subseteq \bigcup _{n}A_{n}\Rightarrow \mu (\bigcup _{n}A_{n})=\infty .}$ But ${\displaystyle A_{n}}$ is a nested increasing sequence of functions and hence, by the continuity from below ${\displaystyle \mu }$, ${\displaystyle \lim _{n\rightarrow \infty }\mu (A_{n})=\infty .}$. Thus, ${\displaystyle \lim _{n\to \infty }\mu _{n}(A_{n})=\mu (A_{n})=\infty .}$. At the same time, ${\displaystyle \int _{E}f_{n}\,d\mu _{n}\geq a\mu _{n}(A_{n})\Rightarrow \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu _{n}=\infty =\int _{E}\phi \,d\mu ,}$ proving the claim in this case. The remaining case is when ${\displaystyle \int _{E}\phi \,d\mu <\infty }$. We must have that ${\displaystyle \mu (A)}$ is finite. Denote, as above, by ${\displaystyle M}$ the maximum value of ${\displaystyle \phi }$ and fix ${\displaystyle \epsilon >0}$ Define ${\displaystyle A_{n}=\{x\in E|f_{k}(x)>(1-\epsilon )\phi (x)~\forall k\geq n\}.}$ Then An is a nested increasing sequence of sets whose union contains ${\displaystyle A}$. Thus, ${\displaystyle A-A_{n}}$ is a decreasing sequence of sets with empty intersection. Since ${\displaystyle A}$ has finite measure (this is why we needed to consider the two separate cases), ${\displaystyle \lim _{n\rightarrow \infty }\mu (A-A_{n})=0.}$ Thus, there exists n such that ${\displaystyle \mu (A-A_{k})<\epsilon ,~\forall k\geq n.}$ Therefore, since ${\displaystyle \lim _{n\to \infty }\mu _{n}(A-A_{k})=\mu (A-A_{k}),}$ there exists N such that ${\displaystyle \mu _{k}(A-A_{k})<\epsilon ,~\forall k\geq N.}$ Hence, for ${\displaystyle k\geq N}$ ${\displaystyle \int _{E}f_{k}\,d\mu _{k}\geq \int _{A_{k}}f_{k}\,d\mu _{k}\geq (1-\epsilon )\int _{A_{k}}\phi \,d\mu _{k}.}$ At the same time, ${\displaystyle \int _{E}\phi \,d\mu _{k}=\int _{A}\phi \,d\mu _{k}=\int _{A_{k}}\phi \,d\mu _{k}+\int _{A-A_{k}}\phi \,d\mu _{k}.}$ Hence, ${\displaystyle (1-\epsilon )\int _{A_{k}}\phi \,d\mu _{k}\geq (1-\epsilon )\int _{E}\phi \,d\mu _{k}-\int _{A-A_{k}}\phi \,d\mu _{k}.}$ Combining these inequalities gives that ${\displaystyle \int _{E}f_{k}\,d\mu _{k}\geq (1-\epsilon )\int _{E}\phi \,d\mu _{k}-\int _{A-A_{k}}\phi \,d\mu _{k}\geq \int _{E}\phi \,d\mu _{k}-\epsilon \left(\int _{E}\phi \,d\mu _{k}+M\right).}$ Hence, sending ${\displaystyle \epsilon }$ to 0 and taking the liminf in ${\displaystyle n}$, we get that ${\displaystyle \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{k}\geq \int _{E}\phi \,d\mu ,}$ 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 ${\displaystyle \scriptstyle (\Omega ,\,{\mathcal {F}},\,\mathbb {P} )}$; 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 ${\displaystyle \scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )}$ and let ${\displaystyle \scriptstyle {\mathcal {G}}\,\subset \,{\mathcal {F}}}$ be a sub-σ-algebra. Then

${\displaystyle \mathbb {E} {\Bigl }\leq \liminf _{n\to \infty }\,\mathbb {E} }$   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

${\displaystyle Y_{k}=\inf _{n\geq k}X_{n}.}$

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

${\displaystyle \mathbb {E} \leq \mathbb {E} }$   almost surely

by the monotonicity of conditional expectation, hence

${\displaystyle \mathbb {E} \leq \inf _{n\geq k}\mathbb {E} }$   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

${\displaystyle {\begin{aligned}\mathbb {E} {\Bigl }&=\mathbb {E} =\mathbb {E} {\Bigl }=\lim _{k\to \infty }\mathbb {E} \\&\leq \lim _{k\to \infty }\inf _{n\geq k}\mathbb {E} =\liminf _{n\to \infty }\,\mathbb {E} .\end{aligned}}}$

Extension to uniformly integrable negative parts

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

${\displaystyle X_{n}^{-}:=\max\{-X_{n},0\},\qquad n\in {\mathbb {N} },}$

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

${\displaystyle \mathbb {E} {\bigl }<\varepsilon ,\qquad {\text{for all }}n\in \mathbb {N} ,\,{\text{almost surely}}}$,

then

${\displaystyle \mathbb {E} {\Bigl }\leq \liminf _{n\to \infty }\,\mathbb {E} }$   almost surely.

Note: On the set where

${\displaystyle X:=\liminf _{n\to \infty }X_{n}}$

satisfies

${\displaystyle \mathbb {E} =\infty ,}$

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

${\displaystyle \mathbb {E} {\bigl }<\varepsilon \qquad {\text{for all }}n\in \mathbb {N} ,\,{\text{almost surely}}.}$

Since

${\displaystyle X+c\leq \liminf _{n\to \infty }(X_{n}+c)^{+},}$

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

${\displaystyle \mathbb {E} +c\leq \mathbb {E} {\Bigl }\leq \liminf _{n\to \infty }\mathbb {E} }$   almost surely.

Since

${\displaystyle (X_{n}+c)^{+}=(X_{n}+c)+(X_{n}+c)^{-}\leq X_{n}+c+X_{n}^{-}1_{\{X_{n}^{-}>c\}},}$

we have

${\displaystyle \mathbb {E} \leq \mathbb {E} +c+\varepsilon }$   almost surely,

hence

${\displaystyle \mathbb {E} \leq \liminf _{n\to \infty }\mathbb {E} +\varepsilon }$   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.
• 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.

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