Chemiluminescence

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 f₁, f₂, f₃, . . . 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

$f(s)={\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{f}_n(s),s\in S.$

Then f  is measurable and

$\underset{S}{\int }fd\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d\mu .$

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.

Direct proof

We will prove something a bit weaker here. Namely, we will allow f_n to converge μ-almost everywhere on a subset E of S. We seek to show that

$\underset{E}{\int }fd\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d\mu .$

Let

$K=\{x\in E|f_n(x)\to f(x)\}$.

Then μ(E-K)=0 and

$\underset{E}{\int }fd\mu =\underset{K}{\int }fd\mu ,\underset{E}{\int }{f}_{n}d\mu =\underset{K}{\int }{f}_{n}d\mu \mathrm{\forall }n\in \mathbb{\mathbb{N}}.$

Thus, replacing E by K we may assume that f_n converge to f pointwise on E. Next, 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

$\underset{E}{\int }\phi d\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d\mu$

We first consider the case when $\underset{E}{\int }\phi =\mathrm{\infty }$. Let a be the minimum non-negative value of φ (it exists since the integral of φ is infinite). Define

$A=\{x\in E|\phi (x)>a\}$

We must have that μ(A) is infinite since

$\underset{E}{\int }\phi \le M\mu (A),$

where M is the (necessarily finite) maximum value of that φ attains.

Next, we define

$A_n=\{x\in E|f_k(x)>a\mathrm{\forall }k\ge n\}.$

We have that

$A\subseteq \bigcup _n{A}_n\Rightarrow \mu (\bigcup _n{A}_n)=\mathrm{\infty }.$

But A_n is a nested increasing sequence of functions and hence, by the continuity from below μ,

$\underset{n\to \mathrm{\infty }}{\lim }\mu (A_n)=\mathrm{\infty }.$.

At the same time,

$\underset{E}{\int }{f}_{n}d\mu \ge a\mu (A_n)\Rightarrow {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d\mu =\mathrm{\infty }=\underset{E}{\int }\phi d\mu ,$

proving the claim in this case.

The remaining case is when $\underset{E}{\int }\phi <\mathrm{\infty }$. We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define

$A_n=\{x\in E|f_k(x)>(1-ϵ)\phi (x)\mathrm{\forall }k\ge n\}.$

Then A_n is a nested increasing sequence of sets whose union contains A. Thus, A-A_n 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),

$\underset{n\to \mathrm{\infty }}{\lim }\mu (A-A_n)=0.$

Thus, there exists n such that

$\mu (A-A_k)<ϵ,\mathrm{\forall }k\ge n.$

Hence, for $k\ge n$

$\underset{E}{\int }{f}_{k}d\mu \ge \underset{A_k}{\int }{f}_{k}d\mu \ge (1-ϵ)\underset{A_k}{\int }\phi d\mu .$

At the same time,

$\underset{E}{\int }\phi d\mu =\underset{A}{\int }\phi d\mu =\underset{A_k}{\int }\phi d\mu +\underset{A-A_k}{\int }\phi d\mu .$

Hence,

$(1-ϵ)\underset{A_k}{\int }\phi d\mu \ge (1-ϵ)\underset{E}{\int }\phi d\mu -\underset{A-A_k}{\int }\phi d\mu .$

Combining these inequalities gives that

$\underset{E}{\int }{f}_{k}d\mu \ge (1-ϵ)\underset{E}{\int }\phi d\mu -\underset{A-A_k}{\int }\phi d\mu \ge \underset{E}{\int }\phi d\mu -ϵ(\underset{E}{\int }\phi d\mu +M).$

Hence, sending ε to 0 and taking the liminf in n, we get that

${\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d\mu \ge \underset{E}{\int }\phi d\mu ,$

completing the proof.

Proof using the monotone convergence theorem

For every natural number k define pointwise the function

$g_k={\inf }_{n\ge k}{f}_n.$

Then the sequence g₁, g₂, . . . of functions is increasing, meaning that g_k ≤ g_k+1 for all k, and converges pointwise to the limit inferior f.

For all k ≤ n we have g_k ≤ f_n, so that by the monotonicity of the integral

$\underset{E}{\int }{g}_{k}d\mu \le \underset{E}{\int }{f}_{n}d\mu ,$

hence

$\underset{E}{\int }{g}_{k}d\mu \le {\inf }_{n\ge k}\underset{E}{\int }{f}_{n}d\mu .$

Using the monotone convergence theorem for the first equality, then the last inequality from above, and finally the definition of the limit inferior, it follows that

$\underset{E}{\int }fd\mu =\underset{k\to \mathrm{\infty }}{\lim }\underset{E}{\int }{g}_{k}d\mu \le \underset{k\to \mathrm{\infty }}{\lim }{\inf }_{n\ge k}\underset{E}{\int }{f}_{n}d\mu ={\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d\mu .$

Examples for strict inequality

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

• Example for a probability space: Let $S=[0,1]$ denote the unit interval. For every natural number $n$ define

$f_n(x)=\{\begin{array}{ll}n& \text{for }x\in (0,1/n),\\ 0& \text{otherwise.}\end{array}$

• Example with uniform convergence: Let $S$ denote the set of all real numbers. Define

$f_n(x)=\{\begin{array}{ll}\frac{1}{n}& \text{for }x\in [0,n],\\ 0& \text{otherwise.}\end{array}$

These sequences $(f_n{)}_{n\in \mathbb{\mathbb{N}}}$ converge on $S$ pointwise (respectively uniformly) to the zero function (with zero integral), but every $f_n$ has integral one.

A counter example

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

$f_n(x)=\{\begin{array}{ll}-\frac{1}{n}& \text{for }x\in [n,2n],\\ 0& \text{otherwise.}\end{array}$

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

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 an integrable function g on S such that f_n ≤ g for all n, then

${\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d\mu \le \underset{S}{\int }{\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{f}_{n}d\mu .$

Note: Here g integrable means that g is measurable and that $\underset{S}{\int }gd\mu <\mathrm{\infty }$.

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 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 f_n ≥ −g for all n, then

$\underset{S}{\int }{\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{f}_{n}d\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d\mu .$

Proof

Apply Fatou's lemma to the non-negative sequence given by f_n + g.

Pointwise convergence

If in the previous setting the sequence f₁, f₂, . . . converges pointwise to a function f μ-almost everywhere on S, then

$\underset{S}{\int }fd\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d\mu .$

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 f₁, f₂, . . . converges in measure to a function f.

Proof

There exists a subsequence such that

$\underset{k\to \mathrm{\infty }}{\lim }\underset{S}{\int }{f}_{n_k}d\mu ={\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d\mu .$

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)

${\mu }_n(E)\to \mu (E),\mathrm{\forall }E\in \mathrm{\Sigma }.$

Then, with f_n non-negative integrable functions and f being their pointwise limit inferior, we have

$\underset{S}{\int }fd\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{S}{\int }{f}_{n}d{\mu }_n.$

Proof

We will prove something a bit stronger here. Namely, we will allow fn to converge μ-almost everywhere on a subset E of S. We seek to show that $\underset{E}{\int }fd\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d{\mu }_n.$ Let $K=\{x\in E|f_n(x)\to f(x)\}$. Then μ(E-K)=0 and $\underset{E}{\int }fd\mu =\underset{E-K}{\int }fd\mu ,\underset{E}{\int }{f}_{n}d\mu =\underset{E-K}{\int }{f}_{n}d\mu \mathrm{\forall }n\in \mathbb{\mathbb{N}}.$ 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 $\underset{E}{\int }\varphi d\mu =\underset{n\to \mathrm{\infty }}{\lim }\underset{E}{\int }\varphi d{\mu }_n.$ 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 $\underset{E}{\int }\varphi d\mu \le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d{\mu }_n$ Let a be the minimum non-negative value of φ. Define $A=\{x\in E|\varphi (x)>a\}$ We first consider the case when $\underset{E}{\int }\varphi d\mu =\mathrm{\infty }$. We must have that μ(A) is infinite since $\underset{E}{\int }\varphi d\mu \le M\mu (A),$ where M is the (necessarily finite) maximum value of that φ attains. Next, we define $A_n=\{x\in E|f_k(x)>a\mathrm{\forall }k\ge n\}.$ We have that $A\subseteq \bigcup _n{A}_n\Rightarrow \mu (\bigcup _n{A}_n)=\mathrm{\infty }.$ But An is a nested increasing sequence of functions and hence, by the continuity from below μ, $\underset{n\to \mathrm{\infty }}{\lim }\mu (A_n)=\mathrm{\infty }.$. Thus, $\underset{n\to \mathrm{\infty }}{\lim }{\mu }_n(A_n)=\mu (A_n)=\mathrm{\infty }.$. At the same time, $\underset{E}{\int }{f}_{n}d{\mu }_n\ge a{\mu }_n(A_n)\Rightarrow {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d{\mu }_n=\mathrm{\infty }=\underset{E}{\int }\varphi d\mu ,$ proving the claim in this case. The remaining case is when $\underset{E}{\int }\varphi d\mu <\mathrm{\infty }$. We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define $A_n=\{x\in E|f_k(x)>(1-ϵ)\varphi (x)\mathrm{\forall }k\ge n\}.$ 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), $\underset{n\to \mathrm{\infty }}{\lim }\mu (A-A_n)=0.$ Thus, there exists n such that $\mu (A-A_k)<ϵ,\mathrm{\forall }k\ge n.$ Therefore, since $\underset{n\to \mathrm{\infty }}{\lim }{\mu }_n(A-A_k)=\mu (A-A_k),$ there exists N such that ${\mu }_k(A-A_k)<ϵ,\mathrm{\forall }k\ge N.$ Hence, for $k\ge N$ $\underset{E}{\int }{f}_{k}d{\mu }_k\ge \underset{A_k}{\int }{f}_{k}d{\mu }_k\ge (1-ϵ)\underset{A_k}{\int }\varphi d{\mu }_k.$ At the same time, $\underset{E}{\int }\varphi d{\mu }_k=\underset{A}{\int }\varphi d{\mu }_k=\underset{A_k}{\int }\varphi d{\mu }_k+\underset{A-A_k}{\int }\varphi d{\mu }_k.$ Hence, $(1-ϵ)\underset{A_k}{\int }\varphi d{\mu }_k\ge (1-ϵ)\underset{E}{\int }\varphi d{\mu }_k-\underset{A-A_k}{\int }\varphi d{\mu }_k.$ Combining these inequalities gives that $\underset{E}{\int }{f}_{k}d{\mu }_k\ge (1-ϵ)\underset{E}{\int }\varphi d{\mu }_k-\underset{A-A_k}{\int }\varphi d{\mu }_k\ge \underset{E}{\int }\varphi d{\mu }_k-ϵ(\underset{E}{\int }\varphi d{\mu }_k+M).$ Hence, sending ε to 0 and taking the liminf in n, we get that ${\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\underset{E}{\int }{f}_{n}d{\mu }_k\ge \underset{E}{\int }\varphi d\mu ,$ 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 $(\mathrm{\Omega },\mathcal{ℱ},\mathbb{\mathbb{P}})$; 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 $(\mathrm{\Omega },\mathcal{ℱ},\mathbb{\mathbb{P}})$ and let $\mathcal{𝒢}\subset \mathcal{ℱ}$ be a sub-σ-algebra. Then

$\mathbb{𝔼}\left[{\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{X}_n\right|\mathcal{𝒢}]\le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\mathbb{𝔼}[X_n|\mathcal{𝒢}]$   almost surely.

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

$Y_k={\inf }_{n\ge k}{X}_n.$

Then the sequence Y₁, Y₂, . . . is increasing and converges pointwise to X. For k ≤ n, we have Y_k ≤ X_n, so that

$\mathbb{𝔼}[Y_k|\mathcal{𝒢}]\le \mathbb{𝔼}[X_n|\mathcal{𝒢}]$   almost surely

by the monotonicity of conditional expectation, hence

$\mathbb{𝔼}[Y_k|\mathcal{𝒢}]\le {\inf }_{n\ge k}\mathbb{𝔼}[X_n|\mathcal{𝒢}]$   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 Y_k, the monotone convergence theorem for conditional expectations, the last inequality, and the definition of the limit inferior, it follows that almost surely

$\begin{array}{rl}\mathbb{𝔼}\left[{\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{X}_n\right|\mathcal{𝒢}]& =\mathbb{𝔼}[X|\mathcal{𝒢}]=\mathbb{𝔼}[\underset{k\to \mathrm{\infty }}{\lim }{Y}_k\left|\mathcal{𝒢}\right]=\underset{k\to \mathrm{\infty }}{\lim }\mathbb{𝔼}[Y_k|\mathcal{𝒢}]\\ & \le \underset{k\to \mathrm{\infty }}{\lim }{\inf }_{n\ge k}\mathbb{𝔼}[X_n|\mathcal{𝒢}]={\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\mathbb{𝔼}[X_n|\mathcal{𝒢}].\end{array}$

Extension to uniformly integrable negative parts

Let X₁, X₂, . . . be a sequence of random variables on a probability space $(\mathrm{\Omega },\mathcal{ℱ},\mathbb{\mathbb{P}})$ and let $\mathcal{𝒢}\subset \mathcal{ℱ}$ be a sub-σ-algebra. If the negative parts

$X_n^{-}:=\max \{-X_n,0\},n\in \mathbb{\mathbb{N}},$

are uniformly integrable with respect to the conditional expectation, in the sense that, for ε > 0 there exists a c > 0 such that

$\mathbb{𝔼}\left[X_n^{-}{1}_{\{X_n^{-}>c\}}\right|\mathcal{𝒢}]<\epsilon ,\text{for all }n\in \mathbb{\mathbb{N}},\text{almost surely}$,

then

$\mathbb{𝔼}\left[{\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{X}_n\right|\mathcal{𝒢}]\le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\mathbb{𝔼}[X_n|\mathcal{𝒢}]$   almost surely.

Note: On the set where

$X:={\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{X}_n$

satisfies

$\mathbb{𝔼}[\max \{X,0\}|\mathcal{𝒢}]=\mathrm{\infty },$

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

$\mathbb{𝔼}\left[X_n^{-}{1}_{\{X_n^{-}>c\}}\right|\mathcal{𝒢}]<\epsilon \text{for all }n\in \mathbb{\mathbb{N}},\text{almost surely}.$

Since

$X+c\le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}(X_n+c{)}^{+},$

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

$\mathbb{𝔼}[X|\mathcal{𝒢}]+c\le \mathbb{𝔼}[{\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}(X_n+c{)}^{+}|\mathcal{𝒢}]\le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\mathbb{𝔼}[(X_n+c{)}^{+}|\mathcal{𝒢}]$   almost surely.

Since

$(X_n+c{)}^{+}=(X_n+c)+(X_n+c{)}^{-}\le X_n+c+X_n^{-}{1}_{\{X_n^{-}>c\}},$

we have

$\mathbb{𝔼}[(X_n+c{)}^{+}|\mathcal{𝒢}]\le \mathbb{𝔼}[X_n|\mathcal{𝒢}]+c+\epsilon$   almost surely,

hence

$\mathbb{𝔼}[X|\mathcal{𝒢}]\le {\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}\mathbb{𝔼}[X_n|\mathcal{𝒢}]+\epsilon$   almost surely.

This implies the assertion.

References

• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534

External links

• Template:Planetmath reference