Fatou's lemma - AbsoluteAstronomy.com

Mathematics

Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...

, Fatou's lemma establishes an inequality relating the integral

Integral

Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...

(in the sense of Lebesgue

Lebesgue integration

In mathematics, Lebesgue integration, named after French mathematician Henri Lebesgue , refers to both the general theory of integration of a function with respect to a general measure, and to the specific case of integration of a function defined on a subset of the real line or a higher...

) of the limit inferior

Limit superior and limit inferior

In mathematics, the limit inferior and limit superior of a sequence can be thought of as limiting bounds on the sequence...

of a sequence

Sequence

In mathematics, a sequence is an ordered list of objects . Like a set, it contains members , and the number of terms is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence...

of function

Function (mathematics)

In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...

s to the limit inferior of integrals of these functions. The lemma

Lemma (mathematics)

In mathematics, a lemma is a proven proposition which is used as a stepping stone to a larger result rather than as a statement in-and-of itself...

is named after Pierre Fatou

Pierre Fatou

Pierre Joseph Louis Fatou was a French mathematician working in the field of complex analytic dynamics. He entered the École Normale Supérieure in Paris in 1898 to study mathematics and graduated in 1901 when he was appointed an astronomy post in the Paris Observatory...

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

Fatou–Lebesgue theorem

In mathematics, the Fatou–Lebesgue theorem establishes a chain of inequalities relating the integrals of the limit inferior and the limit superior of a sequence of functions to the limit inferior and the limit superior of integrals of these functions...

and Lebesgue's dominated convergence theorem

Dominated convergence theorem

In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which two limit processes commute, namely Lebesgue integration and almost everywhere convergence of a sequence of functions...

Standard statement of Fatou's lemma

Let f₁, f₂, f₃, . . . be a sequence of non-negative measurable

Measurable function

In mathematics, particularly in measure theory, measurable functions are structure-preserving functions between measurable spaces; as such, they form a natural context for the theory of integration...

functions on a measure space (S,Σ,μ). Define the function f : S → [0, ∞] pointwise by

f(s) =\liminf_{n\to\infty} f_n(s),\qquad s\in S.

Then f is measurable and

\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.

Note: The functions are allowed to attain the value +∞

Extended real number line

In mathematics, the affinely extended real number system is obtained from the real number system R by adding two elements: +∞ and −∞ . The projective extended real number system adds a single object, ∞ and makes no distinction between "positive" or "negative" infinity...

and the integrals may also be infinite.

Proof

Fatou's lemma may be proved directly as in the first proof presented below, which is an elaboration on the one that can be found in Royden (see the references). The second proof is shorter but uses the monotone convergence theorem.

Examples for strict inequality

Equip the space

S

with the Borel σ-algebra

Borel algebra

In mathematics, a Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement...

and the Lebesgue measure

Lebesgue measure

In measure theory, the Lebesgue measure, named after French mathematician Henri Lebesgue, is the standard way of assigning a measure to subsets of n-dimensional Euclidean space. For n = 1, 2, or 3, it coincides with the standard measure of length, area, or volume. In general, it is also called...

.NEWLINE

Example for a probability space
Probability space
In probability theory, a probability space or a probability triple is a mathematical construct that models a real-world process consisting of states that occur randomly. A probability space is constructed with a specific kind of situation or experiment in mind...

: Let $S=[0,1]$ denote the unit interval
Unit interval
In mathematics, the unit interval is the closed interval , that is, the set of all real numbers that are greater than or equal to 0 and less than or equal to 1...

. For every natural number
Natural number
In mathematics, the natural numbers are the ordinary whole numbers used for counting and ordering . These purposes are related to the linguistic notions of cardinal and ordinal numbers, respectively...

$n$ define

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NEWLINE

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

Example with uniform convergence: Let $S$ denote the set of all real number
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...

s. Define

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NEWLINE

NEWLINE f_n(x)=\begin{cases}\frac1n&\text{for }x\in [0,n],\\ 0&\text{otherwise.} \end{cases} These sequences

(f_n)_{n\in\N}

converge on

S

pointwise (respectively uniformly) to the zero function (with zero integral), but every

f_n

has integral one.

A counterexample

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

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

This sequence converges uniformly on S to the zero function (with zero integral) and for every x ≥ 0 we even have f_n(x) = 0 for all n > x (so for every point x the limit 0 is reached in a finite number of steps). However, every function f_n has integral −1, hence the inequality in Fatou's lemma fails.

Reverse Fatou lemma

Let f₁, f₂, . . . be a sequence of extended real

Extended real number line

-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

\int_S\limsup_{n\to\infty}f_n\,d\mu
\le\int_Sg\,d\mu.

Note: Here g integrable means that g is measurable and that

\textstyle\int_S g\,d\mu<\infty

Proof

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

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 a non-negative integrable function g on S such that f_n ≥ −g for all n, then

\int_S \liminf_{n\to\infty} f_n\,d\mu
 \le \liminf_{n\to\infty} \int_S f_n\,d\mu.\

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

Pointwise convergence

In mathematics, pointwise convergence is one of various senses in which a sequence of functions can converge to a particular function.-Definition:...

to a function f μ-almost everywhere

Almost everywhere

In measure theory , a property holds almost everywhere if the set of elements for which the property does not hold is a null set, that is, a set of measure zero . In cases where the measure is not complete, it is sufficient that the set is contained within a set of measure zero...

on S, then

\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.

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

Convergence in measure

Convergence in measure can refer to two distinct mathematical concepts which both generalizethe concept of convergence in probability.-Definitions:...

to a function f.

Proof

There exists a subsequence such that

\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 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

Convergence of measures

In mathematics, more specifically measure theory, there are various notions of the convergence of measures. Three of the most common notions of convergence are described below.-Total variation convergence of measures:...

)

\mu_n(E)\to \mu(E),~\forall E\in \Sigma.

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

\int_S f\,d\mu \leq \liminf_{n\to \infty} \int_S f_n\, d\mu_n.

NEWLINENEWLINENEWLINENEWLINENEWLINE

Proof

f_n to converge μ-almost everywhere

Almost everywhere

on a subset E of S. We seek to show that

\int_E f\,d\mu \le \liminf_{n\to\infty} \int_E f_n\,d\mu_n\,.

Let

K=\{x\in E>f_n(x)\rightarrow f(x)\}

. Then μ(E-K)=0 and

\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.

Thus, replacing E by E-K we may assume that f_n converge to f pointwise

Pointwise convergence

In mathematics, pointwise convergence is one of various senses in which a sequence of functions can converge to a particular function.-Definition:...

on E. Next, note that for any simple function φ we have

\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 φ is any non-negative simple function less than or equal to f, then

\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

A=\{x\in E |\phi(x)>a\}

We first consider the case when

\int_{E}\phi\, d\mu=\infty

. We must have that μ(A) is infinite since

\int_{E}\phi\, d\mu \leq M\mu(A),

where M is the (necessarily finite) maximum value of that φ attains. Next, we define

A_n=\{x\in E |f_k(x)>a~\forall k\geq n \}.

We have that

A\subseteq \bigcup_n A_n \Rightarrow \mu(\bigcup_n A_n)=\infty.

But A_n is a nested increasing sequence of functions and hence, by the continuity from below μ,

\lim_{n\rightarrow \infty} \mu(A_n)=\infty.

. Thus,

\lim_{n\to\infty}\mu_n(A_n)=\mu(A_n)=\infty.

. At the same time,

\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

\int_{E}\phi\, d\mu<\infty

. We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define

A_n=\{x\in E|f_k(x)>(1-\epsilon)\phi(x)~\forall k\geq n\}.

Then A_n is a nested increasing sequence of sets whose union contains A. Thus, A-A_n 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),

\lim_{n\rightarrow \infty} \mu(A-A_n)=0.

Thus, there exists n such that

\mu(A-A_k)<\epsilon ,~\forall k\geq n.

Therefore, since

\lim_{n\to \infty} \mu_n(A-A_k)=\mu(A-A_k),

there exists N such that

\mu_k(A-A_k)<\epsilon,~\forall k\geq N.

Hence, for

k\geq N

\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,

\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,

(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

\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 ε to 0 and taking the liminf in n, we get that

\liminf_{n\rightarrow \infty} \int_{E} f_n \, d\mu_k \geq \int_E \phi \, d\mu,

completing the proof.

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Fatou's lemma for conditional expectations

In probability theory

Probability theory

Probability theory is the branch of mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either be single...

, 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

Probability space

In probability theory, a probability space or a probability triple is a mathematical construct that models a real-world process consisting of states that occur randomly. A probability space is constructed with a specific kind of situation or experiment in mind...

\scriptstyle(\Omega,\,\mathcal F,\,\mathbb P)

; the integrals turn into expectation

Expected value

In probability theory, the expected value of a random variable is the weighted average of all possible values that this random variable can take on...

s. In addition, there is also a version for conditional expectation

Conditional expectation

In probability theory, a conditional expectation is the expected value of a real random variable with respect to a conditional probability distribution....

Standard version

Let X₁, X₂, . . . be a sequence of non-negative random variables on a probability space

\scriptstyle(\Omega,\mathcal F,\mathbb P)

and let

\scriptstyle \mathcal G\,\subset\,\mathcal F

be a sub-σ-algebra. Then

\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]

almost surely

Almost surely

In probability theory, one says that an event happens almost surely if it happens with probability one. The concept is analogous to the concept of "almost everywhere" in measure theory...

. 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

Y_k=\inf_{n\ge 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

\mathbb{E}[Y_k|\mathcal G]\le\mathbb{E}[X_n|\mathcal G]

almost surely by the monotonicity of conditional expectation, hence

\mathbb{E}[Y_k|\mathcal G]\le\inf_{n\ge k}\mathbb{E}[X_n|\mathcal G]

almost surely, because the countable union of the exceptional sets of probability zero is again a null set

Null set

In mathematics, a null set is a set that is negligible in some sense. For different applications, the meaning of "negligible" varies. In measure theory, any set of measure 0 is called 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

\begin{align}
\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]
&=\mathbb{E}[X|\mathcal G]\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]

Extension to uniformly integrable negative parts

Let X₁, X₂, . . . be a sequence of random variables on a probability space

\scriptstyle(\Omega,\mathcal F,\mathbb P)

and let

\scriptstyle \mathcal G\,\subset\,\mathcal F

be a sub-σ-algebra. If the negative parts

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

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

, then

\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]

almost surely. Note: On the set where

X:=\liminf_{n\to\infty}X_n

satisfies

\mathbb{E}[\max\{X,0\}\,|\,\mathcal G]=\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

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

Since

X+c\le\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

\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]

almost surely. Since

(X_n+c)^+=(X_n+c)+(X_n+c)^-\le X_n+c+X_n^-1_{\{X_n^->c\}},

we have

\mathbb{E}[(X_n+c)^+\,|\,\mathcal G]
\le\mathbb{E}[X_n\,|\,\mathcal G]+c+\varepsilon

almost surely, hence

\mathbb{E}[X\,|\,\mathcal G]\le
\liminf_{n\to\infty}\mathbb{E}[X_n\,|\,\mathcal G]+\varepsilon

almost surely. This implies the assertion.

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