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 for every . Then is -measurable, and
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 .
Fatou's lemma does not require the monotone convergence theorem, but the latter can be used to provide a quick and natural proof. A proof directly from the definitions of integrals is given further below.
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 and the fact that the functions and are measurable.
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 Borelσ-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
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 :
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.
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 .
Let be a sequence of extended real-valued measurable functions defined on a measure space . If there exists an integrable function on such that for all , then
Note that has to agree with the limit inferior of the functions almost everywhere, and that the values of the integrand on a set of measure zero have no influence on the value of the integral.
Since this subsequence also converges in measure to , there exists a further subsequence, which converges pointwise to almost everywhere, hence the previous variation of Fatou's lemma is applicable to this subsubsequence.
In all of the above statements of Fatou's Lemma, the integration was carried out with respect to a single fixed measure . Suppose that is a sequence of measures on the measurable space such that (see Convergence of measures)
.
Then, with non-negative integrable functions and being their pointwise limit inferior, we have
Proof
We will prove something a bit stronger here. Namely, we will allow to converge -almost everywhere on a subset of . We seek to show that
Let
.
Then μ(E-K)=0 and
Thus, replacing by we may assume that converge to pointwise on . Next, note that for any simple function we have
Hence, by the definition of the Lebesgue Integral, it is enough to show that if is any non-negative simple function less than or equal to , then
Let a be the minimum non-negative value of φ. Define
We first consider the case when .
We must have that is infinite since
where M is the (necessarily finite) maximum value of that attains.
Next, we define
We have that
But is a nested increasing sequence of functions and hence, by the continuity from below ,
.
Thus,
.
At the same time,
proving the claim in this case.
The remaining case is when . We must have that is finite. Denote, as above, by the maximum value of and fix Define
Then An is a nested increasing sequence of sets whose union contains . Thus, is a decreasing sequence of sets with empty intersection. Since 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 to 0 and taking the liminf in , we get that
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
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.
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
Royden, H. L. (1988). Real Analysis (3rd ed.). London: Collier Macmillan. ISBN0-02-404151-3.
Weir, Alan J. (1973). "The Convergence Theorems". Lebesgue Integration and Measure. Cambridge: Cambridge University Press. pp. 93–118. ISBN0-521-08728-7.