Fatou's lemma


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
provided that is -measurable. .
For use in the proof, define a sequence of functions by.
Remark 3. For every,
Remark 4. The proof below does not use any properties of Lebesgue integral except those established here.
Remark 5. In the proof below, we apply the monotonic property of Lebesgue integral to non-negative functions only. Specifically, let the functions be -measurable.
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, 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,, so we have,, and furthermore the sequence is non-decreasing. Recall that, and 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 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",
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.
These sequences converge on pointwise to the zero function, 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 and for every x ≥ 0 we even have fn = 0 for all n > x. However, every function fn 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. 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.

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. If there exists an integrable function g on S such that fn ≥ −g for all n, then

Proof

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

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 μn is a sequence of measures on the measurable space such that
Then, with fn non-negative integrable functions and f being their pointwise limit inferior, we have
Let
Then μ=0 and
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
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 is infinite since
where M is the maximum value of that φ attains.
Next, we define
We have that
But An 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 is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define
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,
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
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 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
by the monotonicity of conditional expectation, hence
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

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
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 denotes the positive part of a real x, monotonicity of conditional expectation and the standard version of Fatou's lemma for conditional expectations imply
Since
we have
hence
This implies the assertion.