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.

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

Theorem, Fatou's lemma. Given a measure space \((\Omega,\mathcal{F},\mu)\) and a set \(X \in \mathcal{F},\) let \(\{f_n\}\) be a sequence of \((\mathcal{F}, \operatorname{\mathcal B}_{\bar\R_{\geq 0}})\)-measurable non-negative functions \(f_n: X\to [0,+\infty]\). Define the function \(f: X\to [0,+\infty]\) by \(f(x) =\liminf_{n\to\infty} f_n(x),\) for every \(x\in X\). Then \(f\) is \((\mathcal{F}, \operatorname{\mathcal B}_{\bar\R_{\geq 0}})\)-measurable, and

\(\int_X f\,d\mu \le \liminf_{n\to\infty} \int_X f_n\,d\mu,\)

where the integrals and the limit inferior may be infinite.

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 \(N\).

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.

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

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

• 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.

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 n > x, then f_n(x) = 0. However, every function f_n has integral −1. Contrary to Fatou's lemma, this value is strictly less than the integral of the limit (0).

As discussed in § Extensions and variations of Fatou's lemma below, the problem is that there is no uniform integrable bound on the sequence from below, while 0 is the uniform bound from above.

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

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

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

Let \(f_1, f_2, \ldots\) be a sequence of extended real-valued measurable functions defined on a measure space \((S, \Sigma, \mu)\). If there exists an integrable function \(g\) on \(S\) such that \(f_n \ge -g\) for all \(n\), then

\(\int_S \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu.\)

If in the previous setting the sequence \(f_1, f_2, \ldots\) converges pointwise to a function \(f\) \(\mu\)-almost everywhere on \(S\), then

\(\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.\)

Measures with setwise convergence

In all of the above statements of Fatou's Lemma, the integration was carried out with respect to a single fixed measure \(\mu\). Suppose that \(\mu_n\) is a sequence of measures on the measurable space \((M, \Sigma)\) such that (see Convergence of measures)

\(\forall E\in \mathcal{F} \colon\; \mu_n(E)\to \mu(E)\).

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

\(\int_S f\,d\mu \leq \liminf_{n\to \infty} \int_S f_n\, d\mu_n.\)

Asymptotically uniform integrable functions

The following results use the notion asymptotically uniform integrable (a.u.i). A sequence \(\{f_n\}_{n\in\natnums}\) of measurable \(\{\R \cup \pm \infty\}\)-valued functions is a.u.i with respect to a sequence of measures \(\{\mu_n\}_{n\in\natnums}\) if

\(\lim_{K\rightarrow +\infty} \limsup_{n\rightarrow\infty} \int_M |f_n(s)| \mathbf{I}\{s\in M:|f_n(s)|\ge K\}\mu_n(ds)=0\,.\)

Weakly converging measures

Theorem, Denote \(f_n^-(s)=-\min\{ f_n(s),0 \}\). Let \(M\) be a metric space, \(\{\mu_n\}_{n\in\natnums}\) be a sequence of measures on \(M\) converging weakly to \(\mu\in\mathcal{M}(M)\), and \(\{f_n\}_{n\in\natnums}\) be a sequence of measurable \(\{\R \cup \pm \infty\}\)-valued functions on \(M\) such that \(\{f_n^-\}_{n\in\natnums}\) is a.u.i with respect to \(\{\mu_n\}_{n\in\natnums}\). Then

\(\int_M \liminf_{n\rightarrow\infty,s'\rightarrow s} f_n(s') \mu(ds) \le \liminf_{n \rightarrow \infty} \int_M f_n(s) \mu_n(ds)\,.\)

Theorem, Let \(M\) be a measurable space, \(\{\mu_n\}_{n\in\natnums}\) be a sequence of measures on \(M\) converging in total variation to a measure \(\mu\in\mathcal{M}(M)\), and \(\{f_n\}_{n\in\natnums}\) be a sequence of measurable \(\{\R \cup \pm \infty\}\)-valued functions on \(M\), and \(f\) be a measurable \(\{\R \cup \pm \infty\}\)-valued function. Assume that \(f\in L^1(M;\mu)\) and \(f_n\in L^1(M;\mu_n)\) Then

\(\liminf_{n\rightarrow\infty} \inf_{C\in\Sigma}\left( \int_C f_n(s) \mu_n(ds) - \int_C f(s) \mu(ds) \right) \ge 0 \,,\)

if and only if the following two statements hold:

(i) for each \(\varepsilon > 0\), \(\mu(\{s \in M : f_n(s) \leq f(s) - \varepsilon \}) \to 0\) as \(n \to \infty\), and, therefore, there exists a subsequence \(\{f_{n_k}\}_{k \in \natnums} \subset \{f_n\}_{n \in \natnums}\) such that \(f(s) \leq \liminf_{k \to \infty} f_{n_k}(s)\) for \(\mu\)-a.e. \(s \in M\);

(ii) \(\{f^-_n\}_{n \in \natnums}\) is a.u.i. with respect to \(\{\mu_n\}_{n \in \natnums}\).

Condensed: the full section is in Wikipedia.

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.

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.

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.