Fatou's lemma

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, $$\operatorname{\mathcal B}_{\bar\R_{\geq 0}}$$ denotes the $$\sigma$$-algebra of Borel sets on $$[0,+\infty]$$.

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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 $${x\in X\setminus N}.$$ 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 $$\textstyle g_n(x)=\inf_{k\geq n}f_k(x)$$. Then: Since
 * 1) the sequence $$\{g_n(x)\}_n$$ is pointwise non-decreasing at any $x$ and
 * 2) $$g_n\leq f_n$$, $$\forall n \in \N$$.
 * $$f(x) =\liminf_{n\to\infty} f_n(x) = \sup_n \inf_{k\geq n} f_k(x) = \sup_n g_n(x)$$,

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}

\int_X f\,d\mu&= \int_X \sup_n g_n\,d\mu\\ &=\sup_n \int_X g_n\,d\mu\\ &=\liminf_{n\to \infty}\int_X g_n\,d\mu\\ &\leq \liminf_{n\to \infty}\int_X f_n\,d\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 $$g_n$$ are measurable.

Denote by $$\operatorname{SF}(f)$$ the set of simple $$(\mathcal{F}, \operatorname{\mathcal B}_{\R_{\geq 0}})$$-measurable functions $$s:X\to [0,\infty)$$ such that $$0\leq s\leq f$$ on $$X$$.

$$

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Now we turn to the main theorem

$$ $$ $$

$$ $$

$$

$$ $$ The proof is complete.

Examples for strict inequality
Equip the space $$S$$ with the Borel &sigma;-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{cases}n&\text{for }x\in (0,1/n),\\ 0&\text{otherwise.} \end{cases}$$
 * Example with uniform convergence: Let $$S$$ denote the set of all real numbers. Define

f_n(x)=\begin{cases}\frac1n&\text{for }x\in [0,n],\\ 0&\text{otherwise.} \end{cases}$$

These sequences $$(f_n)_{n\in\N}$$ 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

f_n(x)=\begin{cases}-\frac1n&\text{for }x\in [n,2n],\\ 0&\text{otherwise.} \end{cases}$$ 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 $S_{n}$, then fn(x) = 0. However, every function fn has integral &minus;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

\limsup_{n\to\infty}\int_S f_n\,d\mu\leq\int_S\limsup_{n\to\infty}f_n\,d\mu. $$

Note: Here g integrable means that g is measurable and that $$\textstyle\int_S g\,d\mu<\infty$$.

Sketch of proof
We apply linearity of Lebesgue integral and Fatou's lemma to the sequence $$g - f_n.$$ Since $$\textstyle\int_Sg\,d\mu < +\infty,$$ this sequence is defined $$\mu$$-almost everywhere and non-negative.

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 ≥ &minus;g for all n, then

\int_S \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\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
 * $$\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.$$

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
 * $$\lim_{k\to\infty} \int_S f_{n_k}\,d\mu=\liminf_{n\to\infty} \int_S f_n\,d\mu.$$

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 μn is a sequence of measures on the measurable space (S,Σ) such that (see Convergence of measures)
 * $$\forall E\in \mathcal{F} \colon\; \mu_n(E)\to \mu(E)$$.

Then, with fn non-negative integrable functions and f being their pointwise limit inferior, we have
 * $$ \int_S f\,d\mu \leq \liminf_{n\to \infty} \int_S f_n\, d\mu_n. $$


 * {| class="toccolours collapsible collapsed" width="90%" style="text-align:left"

!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
 * 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

\int_E f\,d\mu \le \liminf_{n\to\infty} \int_E f_n\,d\mu_n\,. $$ Let
 * $$ K=\{x\in E|f_n(x)\rightarrow f(x)\} $$.

Then μ(E-K)=0 and
 * $$ \int_{E}f\,d\mu=\int_{E-K}f\,d\mu,\int_{E}f_n\,d\mu=\int_{E-K}f_n\,d\mu ~\forall n\in \N. $$

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
 * $$ \int_{E}\phi\, d\mu=\lim_{n\to \infty} \int_{E} \phi\, d\mu_n. $$

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

\int_{E}\phi \,d\mu\leq \liminf_{n\rightarrow \infty} \int_{E}f_n\,d\mu_n $$ Let a be the minimum non-negative value of φ. Define

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

We first consider the case when $$\int_{E}\phi\, d\mu=\infty$$. We must have that μ(A) is infinite since
 * $$\int_{E}\phi\, d\mu \leq M\mu(A),$$

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

Next, we define

A_n=\{x\in E |f_k(x)>a~\forall k\geq n \}. $$ We have that

A\subseteq \bigcup_n A_n \Rightarrow \mu(\bigcup_n A_n)=\infty. $$ But An is a nested increasing sequence of functions and hence, by the continuity from below μ,

\lim_{n\rightarrow \infty} \mu(A_n)=\infty. $$. Thus,

\lim_{n\to\infty}\mu_n(A_n)=\mu(A_n)=\infty. $$. At the same time,

\int_E f_n\, d\mu_n \geq a \mu_n(A_n) \Rightarrow \liminf_{n\to \infty}\int_E f_n \, d\mu_n = \infty = \int_E \phi\, d\mu, $$ proving the claim in this case.

The remaining case is when $$\int_{E}\phi\, d\mu<\infty$$. 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-\epsilon)\phi(x)~\forall k\geq 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),

\lim_{n\rightarrow \infty} \mu(A-A_n)=0. $$ Thus, there exists n such that

\mu(A-A_k)<\epsilon ,~\forall k\geq n. $$ Therefore, since

\lim_{n\to \infty} \mu_n(A-A_k)=\mu(A-A_k), $$ there exists N such that

\mu_k(A-A_k)<\epsilon,~\forall k\geq N. $$ Hence, for $$k\geq N$$

\int_E f_k \, d\mu_k \geq \int_{A_k}f_k \, d\mu_k \geq (1-\epsilon)\int_{A_k}\phi\, d\mu_k. $$ At the same time,

\int_E \phi \, d\mu_k = \int_A \phi \, d\mu_k = \int_{A_k} \phi \, d\mu_k + \int_{A-A_k} \phi \, d\mu_k. $$ Hence,

(1-\epsilon)\int_{A_k} \phi \, d\mu_k \geq (1-\epsilon)\int_E \phi \, d\mu_k - \int_{A-A_k} \phi \, d\mu_k. $$ Combining these inequalities gives that

\int_{E} f_k \, d\mu_k \geq (1-\epsilon)\int_E \phi \, d\mu_k - \int_{A-A_k} \phi \, d\mu_k \geq \int_E \phi \, d\mu_k -  \epsilon\left(\int_{E} \phi \, d\mu_k+M\right). $$ Hence, sending ε to 0 and taking the liminf in n, we get that

\liminf_{n\rightarrow \infty} \int_{E} f_n \, d\mu_k \geq \int_E \phi \, 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 $$\scriptstyle(\Omega,\,\mathcal F,\,\mathbb P)$$; 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 $$\scriptstyle(\Omega,\mathcal F,\mathbb P)$$ and let $$\scriptstyle \mathcal G\,\subset\,\mathcal F$$ be a sub-σ-algebra. Then
 * $$\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]$$  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 Xn. 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 Yk ≤ Xn, 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 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} \mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr] &=\mathbb{E}[X|\mathcal G] =\mathbb{E}\Bigl[\lim_{k\to\infty}Y_k\,\Big|\,\mathcal G\Bigr] =\lim_{k\to\infty}\mathbb{E}[Y_k|\mathcal G]\\ &\le\lim_{k\to\infty} \inf_{n\ge k}\mathbb{E}[X_n|\mathcal G] =\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]. \end{align} $$

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


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

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}\bigl[X_n^-1_{\{X_n^->c\}}\,|\,\mathcal G\bigr]<\varepsilon,

\qquad\text{for all }n\in\mathbb{N},\,\text{almost surely}$$, then


 * $$\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]$$  almost surely.

Note: On the set where


 * $$X:=\liminf_{n\to\infty}X_n$$

satisfies


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

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}\bigl[X_n^-1_{\{X_n^->c\}}\,|\,\mathcal G\bigr]<\varepsilon

\qquad\text{for all }n\in\mathbb{N},\,\text{almost surely}.$$

Since


 * $$X+c\le\liminf_{n\to\infty}(X_n+c)^+,$$

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}\Bigl[\liminf_{n\to\infty}(X_n+c)^+\,\Big|\,\mathcal G\Bigr] \le\liminf_{n\to\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+\varepsilon$$  almost surely,

hence


 * $$\mathbb{E}[X\,|\,\mathcal G]\le

\liminf_{n\to\infty}\mathbb{E}[X_n\,|\,\mathcal G]+\varepsilon$$  almost surely.

This implies the assertion.