Fatou's lemma explained

Fatou's lemma should not be confused with Fatou's theorem.

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,

\operatorname{lB}
\bar\R\geq
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

\{fn(x)\}

are non-negative for every

{x\inX\setminusN}.

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

stylegn(x)=infk\geqfk(x)

. Then:
  1. the sequence

\{gn(x)\}n

is pointwise non-decreasing at any and

gn\leqfn

,

\foralln\in\N

.Since

f(x)=\liminfn\toinftyfn(x)=\supninfk\geqfk(x)=\supngn(x)

, 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:

\begin{align} \intXfd\mu&=\intX\supngnd\mu\\ &=\supn\intXgnd\mu\\ &=\liminfn\to\intXgnd\mu\\ &\leq\liminfn\to\intXfnd\mu, \end{align}

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

gn

are measurable.

Denote by

\operatorname{SF}(f)

the set of simple
(l{F},\operatorname{lB}
\R\geq

)

-measurable functions

s:X\to[0,infty)

such that

0\leqs\leqf

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.

S=[0,1]

denote the unit interval. For every natural number

n

define

fn(x)=\begin{cases}n&forx\in(0,1/n),\\ 0&otherwise. \end{cases}

S

denote the set of all real numbers. Define
f
n(x)=\begin{cases}1n&forx\in
[0,n],\\ 0&otherwise. \end{cases}

These sequences

(fn)n\in\N

converge on

S

pointwise (respectively uniformly) to the zero function (with zero integral), but every

fn

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 :<math> f_n(x)=\begin{cases}-\frac1n&\text{for }x\in [n,2n],\\0&\text\endThis 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, 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

\limsupn\toinfty\intSfnd\mu\leq\intS\limsupn\toinftyfnd\mu.

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

style\intSgd\mu<infty

.

Sketch of proof

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence

g-fn.

Since

style\intSgd\mu<+infty,

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

\intS\liminfn\toinftyfnd\mu \le\liminfn\toinfty\intSfnd\mu.

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

\intSfd\mu\le\liminfn\toinfty\intSfnd\mu.

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

\limk\toinfty\intS

f
nk

d\mu=\liminfn\toinfty\intSfnd\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 μn is a sequence of measures on the measurable space (S,Σ) such that (see Convergence of measures)

\forallE\inl{F}\colon\mun(E)\to\mu(E)

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

\intSfd\mu\leq\liminfn\to\intSfnd\mun.

Fatou's lemma for conditional expectations

\scriptstyle(\Omega,lF,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

\scriptstyle(\Omega,lF,P)

and let

\scriptstylelG\subsetlF

be a sub-σ-algebra. Then

El[\liminfn\toinftyXn|lGr]\le\liminfn\toinftyE[Xn|lG]

   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 Xn. For every natural number k define pointwise the random variable

Yk=infn\geXn.

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

E[Yk|lG]\leE[Xn|lG]

   almost surelyby the monotonicity of conditional expectation, hence

E[Yk|lG]\leinfn\geE[Xn|lG]

   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

\begin{align} El[\liminfn\toinftyXn|lGr] &=E[X|lG] =El[\limk\toinftyYk|lGr] =\limk\toinftyE[Yk|lG]\\ &\le\limk\toinftyinfn\geE[Xn|lG] =\liminfn\toinftyE[Xn|lG]. \end{align}

Extension to uniformly integrable negative parts

Let X1, X2, . . . be a sequence of random variables on a probability space

\scriptstyle(\Omega,lF,P)

and let

\scriptstylelG\subsetlF

be a sub-σ-algebra. If the negative parts
-:=max\{-X
X
n,0\},   

n\in{N},

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

-1
El[X
->c\
\{X
n
}\,|\,\mathcal G\bigr]<\varepsilon, \qquad\textn\in\mathbb,\,\text, then

El[\liminfn\toinftyXn|lGr]\le\liminfn\toinftyE[Xn|lG]

   almost surely.

Note: On the set where

X:=\liminfn\toinftyXn

satisfies

E[max\{X,0\}|lG]=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

-1
El[X
->c\
\{X
n
}\,|\,\mathcal G\bigr]<\varepsilon\qquad\textn\in\mathbb,\,\text.

Since

X+c\le\liminfn\toinfty

+,
(X
n+c)

where x+ := max 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

E[X|lG]+c \leEl[\liminfn\toinfty

+|lGr] \le\liminf
(X
n\toinfty
+|lG]
E[(X
n+c)
   almost surely.

Since

+=(X
(X
n+c)+(X
-\le
n+c)

Xn+c+X

-1
->c\
\{X
n
},

we have

+|lG] \leE[X
E[(X
n|lG]+c+\varepsilon
   almost surely,

hence

E[X|lG]\le \liminfn\toinftyE[Xn|lG]+\varepsilon

   almost surely.

This implies the assertion.

References