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[edit]

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

Theorem — 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 also , where the integrals may be infinite.

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 values are non-negative for every To see this, note that the integrals appearing in Fatou's lemma are unchanged if we change each function on .

Proof[edit]

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

In each case, the proof begins by analyzing the properties of . These satisfy:

  1. the sequence is pointwise non-decreasing at any x and
  2. , .

Since , we immediately see that f is measurable.

Via the Monotone Convergence Theorem[edit]

Moreover,

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

where the last step used property (2).

From "first principles"[edit]

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.

Denote by the set of simple -measurable functions such that on .

Monotonicity — 

  • If everywhere on then
  • If and then
  • If f is nonnegative and , where is a non-decreasing chain of -measurable sets, then
Proof

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

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 Sn is simple and supported on X, we must have

.

For the reverse, suppose g ∈ SF(f) with By the above,

Now we turn to the main theorem

Step 1 —  is -measurable, for every , as is .

Proof

Recall the closed intervals generate the Borel σ-algebra. Thus it suffices to show, for every , that . Now observe that

Every set on the right-hand side is from , which is closed under countable intersections. Thus the left-hand side is also a member of .

Similarly, it is enough to verify that , for every . Since the sequence pointwise non-decreases,

.

Step 2 — Given a simple function and a real number , define

Then , , and .

Proof

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

.

Then

.

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

Step 2b. To prove the second claim, note that, for each and every ,

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

Then , for every . Taking the limit as ,

This contradicts our initial assumption that .

Step 3 — From step 2 and monotonicity,

Step 4 — For every ,

.
Proof

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 5 — To complete the proof, we apply the definition of Lebesgue integral to the inequality established in Step 4 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 and the limit, 0, is reached in a finite number of steps: for every x ≥ 0, if n > x, then fn(x) = 0. However, every function fn 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[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 .

Sketch of proof[edit]

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[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.

References[edit]

  • 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.