Fatou's lemma

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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 of Fatou's lemma[edit]

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

Fatou's lemma. Consider a measure space with a countably additive measure . Select a measurable subset . 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 ,

  1. the non-negative sequence pointwise non-decreases, i.e. , for every ;
  2. , 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, and the integrals and exist (i.e. differ from ).

  • 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[edit]

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[edit]

Lebesgue integral as measure[edit]

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

.

Then is a measure on .

Proof[edit]

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"[edit]

Lemma 2. Let be a measure, and , where

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

.
Proof[edit]

Let . The sequence , where , consists of pairwise disjoint sets, , and . By countable additivity,

as required.

Proof of theorem[edit]

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[edit]

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

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

The role of non-negativity[edit]

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 fn(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 fn has integral −1, hence the inequality in Fatou's lemma fails.

Reverse Fatou lemma[edit]

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 fn ≤ g for all n, then

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

Proof[edit]

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

Extensions and variations of Fatou's lemma[edit]

Integrable lower bound[edit]

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 fn ≥ −g for all n, then

Proof[edit]

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

Pointwise convergence[edit]

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

Proof[edit]

Note that f has to agree with the limit inferior of the functions fn 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[edit]

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

Proof[edit]

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[edit]

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 μn is a sequence of measures on the measurable space (S,Σ) such that (see Convergence of measures)

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

Fatou's lemma for conditional expectations[edit]

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[edit]

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[edit]

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 Xn. 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 Yk ≤ Xn, 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 Yk, 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[edit]

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[edit]

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.

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