# Fatou's lemma

In mathematics, **Fatou's lemma** establishes an inequality relating the integral (in the sense of Lebesgue) 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

Let *f*_{1}, *f*_{2}, *f*_{3}, . . . be a sequence of non-negative measurable functions on a measure space (*S*,*Σ*,*μ*). Define the function *f* : *S* → [0, ∞] a.e. pointwise limit by

Then *f * is measurable and

**Note:** The functions are allowed to attain the value +∞ and the integrals may also be infinite.

### Proof

Fatou's lemma may be proved directly as in the first proof presented below, which is an elaboration on the one that can be found in Royden (see the references). The second proof is shorter but uses the monotone convergence theorem - which is usually proved using Fatou's lemma and thus creating a circular argument.

## Examples for strict inequality

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

- Example for a probability space: Let denote the unit interval. For every natural number define

- Example with uniform convergence: Let denote the set of all real numbers. Define

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

## The role of non-negativity

A suitable assumption concerning the negative parts of the sequence *f*_{1}, *f*_{2}, . . . 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 *f _{n}*(

*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

*f*has integral −1, hence the inequality in Fatou's lemma fails.

_{n}## Reverse Fatou lemma

Let *f*_{1}, *f*_{2}, . . . 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 *f*_{n} ≤ *g* for all *n*, then

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

### Proof

Apply Fatou's lemma to the non-negative sequence given by *g* – *f*_{n}.

## Extensions and variations of Fatou's lemma

### Integrable lower bound

Let *f*_{1}, *f*_{2}, . . . 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 *f*_{n} ≥ −*g* for all *n*, then

#### Proof

Apply Fatou's lemma to the non-negative sequence given by *f*_{n} + *g*.

### Pointwise convergence

If in the previous setting the sequence *f*_{1}, *f*_{2}, . . . 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 *f*_{n} 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 *f*_{1}, *f*_{2}, . . . 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 μ_{n} is a sequence of measures on the measurable space (*S*,*Σ*) such that (see Convergence of measures)

Then, with *f _{n}* non-negative integrable functions and

*f*being their pointwise limit inferior, we have

Proof We will prove something a bit stronger here. Namely, we will allow *f*_{n}to converge μ-almost everywhere on a subset E of S. We seek to show thatLet

Then

*μ(E-K)=0*andThus, replacing

*E*by*E-K*we may assume that*f*_{n}converge to*f*pointwise on E. Next, note that for any simple function*φ*we haveHence, 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,*thenLet

*a*be the minimum non-negative value of*φ.*DefineWe first consider the case when . We must have that

*μ(A)*is infinite sincewhere

*M*is the (necessarily finite) maximum value of that*φ*attains.Next, we define

We have that

But

*A*is a nested increasing sequence of functions and hence, by the continuity from below_{n}*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.*DefineThen

*A*is a nested increasing sequence of sets whose union contains_{n}*A.*Thus,*A-A*is a decreasing sequence of sets with empty intersection. Since_{n}*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

At the same time,

Hence,

Combining these inequalities gives that

Hence, sending

*ε*to 0 and taking the liminf in n, we get thatcompleting 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 *X*_{1}, *X*_{2}, . . . defined on a probability space ; the integrals turn into expectations. In addition, there is also a version for conditional expectations.

### Standard version

Let *X*_{1}, *X*_{2}, . . . 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 *X*_{n}. For every natural number *k* define pointwise the random variable

Then the sequence *Y*_{1}, *Y*_{2}, . . . is increasing and converges pointwise to *X*.
For *k* ≤ *n*, we have *Y*_{k} ≤ *X*_{n}, 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*_{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 *X*_{1}, *X*_{2}, . . . 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.

## References

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