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Fatou's lemma

释义

Standard statement of Fatou's lemma
Proof Intermediate results Lebesgue integral as measure Proof "Continuity from below" Proof of theorem
Examples for strict inequality
The role of non-negativity
Reverse Fatou lemma
Sketch of proof
Extensions and variations of Fatou's lemma
Integrable lower bound Proof Pointwise convergence Proof Convergence in measure Proof Fatou's Lemma with Varying Measures
Fatou's lemma for conditional expectations
Standard version Proof Extension to uniformly integrable negative parts Proof
References
External links

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 of Fatou's lemma

In what follows, denotes the -algebra of Borel sets on .

Fatou's lemma. Given a measure space and a set let be a sequence of -measurable non-negative functions . Define the function by setting

for every .

Then is -measurable, and

Remark 1. The integrals may be finite or infinite.

Remark 2. Fatou's lemma remains true if its assumptions hold

-almost everywhere. In other words, it is enough that there is a null set

such that the sequence

non-decreases for every

To see why this is true, we start with an observation that allowing the sequence

to pointwise non-decrease almost everywhere causes its pointwise limit

to be undefined on some null set

. On that null set,

may then be defined arbitrarily, e.g. as zero, or in any other way that preserves measurability. To see why this will not affect the outcome, note that since

we have, for every

and

provided that is -measurable. (These equalities follow directly from the definition of Lebesgue integral for a non-negative function).

For the upcoming proof's sake, let .

Remark 3. For every ,

{{ordered list|type=lower-alpha
| the non-negative sequence

pointwise non-decreases, i.e.

, for every ;

, by definition of limit inferior.
}}

Remark 4. The proof below does not use any properties of Lebesgue integral except those established here.

Remark 5 (monotonicity of Lebesgue integral). In the proof below, we apply the monotonic property of Lebesgue integral to non-negative functions only. Specifically (see Remark 4), let the functions be -measurable.

If everywhere on then

If and then

Proof. Denote the set of simple -measurable functions such that

everywhere on

1. Since we have

By definition of Lebesgue integral and the properties of supremum,

2. Let be the indicator function of the set It can be deduced from the definition of Lebesgue integral that

if we notice that, for every outside of Combined with the previous property, the inequality implies

Proof

This proof does not rely on the monotone convergence theorem. However, we do explain how that theorem may be applied.

For those not interested in independent proof, the intermediate results below may be skipped.

Intermediate results

Lebesgue integral as measure

Lemma 1. Let be a measurable space. Consider a simple -measurable non-negative function . For a subset , define

.

Then is a measure on .

Proof

We will only prove countable additivity, leaving the rest up to the reader. Let

, where all the sets are pairwise disjoint. Due to simplicity,

,

for some finite non-negative constants and pairwise disjoint sets such that . By definition of Lebesgue integral,

Since all the sets are pairwise disjoint, the countable additivity of

gives us

Since all the summands are non-negative, the sum of the series, whether this sum is finite or infinite, cannot change if summation order does because the series is either absolutely convergent or diverges to For that reason,

as required.

"Continuity from below"

The following property is a direct consequence of the definition of measure.

Lemma 2. Let be a measure, and , where

is a non-decreasing chain with all its sets -measurable. Then

.

Proof of theorem

Step 1. is -measurable, for every .

Indeed, since the Borel -algebra on is generated by the closed intervals , it suffices to show that, , for every , where denotes the inverse image of under .

Observe that

,

or equivalently,

Note that every set on the right-hand side is from . Since, by definition, is closed under countable intersections, we conclude that the left-hand side is also a member of . The -measurability of follows.

Step 2. Now, we want to show that the function is

-measurable.

If we were to use the monotone convergence theorem, the measurability of would follow easily from Remark 3.

Alternatively, using the technique from Step 1, it is enough to verify that , for every . Since the sequence pointwise non-decreases (see Remark 3), arguing as above, we get

.

Due to the measurability of , the above equivalency implies that

.

End of Step 2.

The proof can proceed in two ways.

Proof using the monotone convergence theorem. By definition, , and the sequence non-decreases for every . Therefore

as required.

Independent proof. To prove the inequality without using the monotone convergence theorem, we need some extra machinery. Denote the set of simple -measurable functions such that

on .

Step 3. Given a simple function and a real number , define

Then , , and .

Step 3a. To prove the first claim, let

for some finite collection of pairwise disjoint measurable sets such that , some (finite) real values , and denoting the indicator function of the set . Then

.

Since the pre-image of the Borel set under the measurable function is measurable, and -algebras, by definition, are closed under finite intersection and unions, the first claim follows.

Step 3b. To prove the second claim, note that, for each

and every

Step 3c. To prove the third claim, we show that .

Indeed, if, to the contrary, , then an element

exists such that , for every . Taking the limit as , get

But by initial assumption, . This is a contradiction.

Step 4. For every simple -measurable non-negative function ,

To prove this, define . By Lemma 1, is a measure on . By "continuity from below" (Lemma 2),

,

as required.

Step 5. We now prove that, for every ,

.

Indeed, using the definition of , the non-negativity of , and the monotonicity of Lebesgue integral, we have

.

In accordance with Step 4, as the inequality becomes

.

Taking the limit as yields

,

as required.

Step 6. To complete the proof, we apply the definition of Lebesgue integral to the inequality established in Step 5 and take into account that :

The proof is complete.

Examples for strict inequality

Equip the space with the Borel σ-algebra and the Lebesgue measure.

Example for a probability space: Let denote the unit interval. For every natural number define

Example with uniform convergence: Let denote the set of all real numbers. Define

These sequences converge on pointwise (respectively uniformly) to the zero function (with zero integral), but every 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

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.

As shown 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

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

Sketch of proof

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence Since this sequence is defined -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

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

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

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)

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

Proof

We will prove something a bit stronger here. Namely, we will allow f_n to converge μ-almost everywhere on a subset E of S. We seek to show that

Let

.

Then μ(E-K)=0 and

Thus, replacing E by E-K we may assume that f_n converge to f pointwise on E. Next, note that for any simple function φ we have

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

Let a be the minimum non-negative value of φ. Define

We first consider the case when .

We must have that μ(A) is infinite since

where M is the (necessarily finite) maximum value of that φ attains.

Next, we define

We have that

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

.

Thus,

.

At the same time,

proving the claim in this case.

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

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),

Thus, there exists n such that

Therefore, since

there exists N such that

Hence, for

At the same time,

Hence,

Combining these inequalities gives that

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

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 ; 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 and let

be a sub-σ-algebra. Then

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

Then the sequence Y₁, Y₂, . . . is increasing and converges pointwise to X.

For k ≤ n, we have Y_k ≤ X_n, so that

almost surely

by the monotonicity of conditional expectation, hence

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

Extension to uniformly integrable negative parts

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

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

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

,

then

almost surely.

Note: On the set where

satisfies

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

Since

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

almost surely.

Since

we have

almost surely,

hence

almost surely.

This implies the assertion.

References

{{cite book |first=N. L. |last=Carothers |title=Real Analysis |location=New York |publisher=Cambridge University Press |year=2000 |isbn=0-521-49756-6 |pages=321–22 |url=https://books.google.com/books?id=4VFDVy1NFiAC&pg=PA321 }}
{{cite book

| last = Royden
| first = H. L.
| title = Real Analysis
| edition = 3rd
| location = London
| publisher = Collier Macmillan
| year = 1988
| isbn = 0-02-404151-3
}}

External links

{{planetmath reference|id=3678|title=Fatou's lemma}}

5 : Inequalities|Lemmas|Theorems in measure theory|Real analysis|Articles containing proofs

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