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

In what follows, denotes the \sigma-algebra of Borel sets on [0,+\infty]. Fatou's lemma remains true if its assumptions hold \mu-almost everywhere. In other words, it is enough that there is a null set N such that the values {f_n(x)} 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 N.

Proof

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.

Via the Monotone Convergence Theorem

let. Then: Since and infima and suprema of measurable functions are measurable we see that f is measurable. By the Monotone Convergence Theorem and property (1), the sup and integral may be interchanged: where the last step used property (2).

From "first principles"

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 f and g_n are measurable. Denote by the set of simple -measurable functions such that on X. Now we turn to the main theorem The proof is complete.

Examples for strict inequality

Equip the space S with the Borel σ-algebra and the Lebesgue measure. These sequences converge on S pointwise (respectively uniformly) to the zero function (with zero integral), but every f_n 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 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 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 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 g - f_n. Since this sequence is defined \mu-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 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 (S,Σ) such that (see Convergence of measures) Then, with fn non-negative integrable functions and f being their pointwise limit inferior, we have !Proof Let Then μ(E-K)=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(A) is infinite since where M is the (necessarily finite) 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(A) 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 (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 k\geq N 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 Y<Sub>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 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 Since we have hence This implies the assertion.