Fatou's lemma

Examples for strict inequality

Equip the space ${\displaystyle S}$ with the Borel σ-algebra and the Lebesgue measure.

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

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

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

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

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

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

This sequence converges uniformly on S to the zero function 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.

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 f_n ≤ g for all n, then

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

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

Sketch of proof

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence ${\displaystyle g-f_n.}$ Since ${\displaystyle {\int }_{S}gd\mu <+\mathrm{\infty },}$ this sequence is defined ${\displaystyle \mu }$-almost everywhere and non-negative.

Extensions and variations of Fatou's lemma

Integrable lower bound

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

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

Proof

Apply Fatou's lemma to the non-negative sequence given by ${\displaystyle f_n+g}$.

Pointwise convergence

If in the previous setting the sequence ${\displaystyle f_1,f_2,\dots }$ converges pointwise to a function ${\displaystyle f}$ ${\displaystyle \mu }$-almost everywhere on ${\displaystyle S}$, then

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

Proof

Note that ${\displaystyle f}$ has to agree with the limit inferior of the functions ${\displaystyle f_n}$ 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 ${\displaystyle f_1,f_2,\dots }$ converges in measure to a function ${\displaystyle f}$.

Proof

There exists a subsequence such that

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

Since this subsequence also converges in measure to ${\displaystyle f}$, there exists a further subsequence, which converges pointwise to ${\displaystyle f}$ almost everywhere, hence the previous variation of Fatou's lemma is applicable to this subsubsequence.

Fatou's Lemma with Converging Measures

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 ${\displaystyle \mu }$. Suppose that ${\displaystyle {\mu }_n}$ is a sequence of measures on the measurable space ${\displaystyle (M,\mathrm{\Sigma })}$ such that (see Convergence of measures)

${\displaystyle \mathrm{\forall }E\in \mathcal{F}:{\mu }_n(E)\to \mu (E)}$.

Then, with ${\displaystyle f_n}$ non-negative integrable functions and ${\displaystyle f}$ being their pointwise limit inferior, we have

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

Asymptotically uniform integrable functions

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

${\displaystyle \underset{K\to +\mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim sup}{\int }_M|f_n(s)|\mathbf{I}\{s\in M:|f_n(s)|\ge K\}{\mu }_n(ds)=0.}$

Weakly converging measures

A sequence of measures ${\displaystyle \{{\mu }_n{\}}_{n\in \mathbb{N}}}$ on a metric space ${\displaystyle M}$ converges weakly to a finite measure ${\displaystyle \mu }$ on M if, for each bounded continuous function ${\displaystyle f}$ on ${\displaystyle M}$,

${\displaystyle {\int }_{M}f(s){\mu }_n(ds)\to {\int }_{M}f(s)\mu (ds)\text{as }n\to \mathrm{\infty }.}$

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

${\displaystyle {\int }_M\underset{n\to \mathrm{\infty },s^{\prime }\to s}{lim inf}{f}_n(s^{\prime })\mu (ds)\le \underset{n\to \mathrm{\infty }}{lim inf}{\int }_M{f}_n(s){\mu }_n(ds).}$

Measures with convergence in total variation

A sequence of finite measures ${\displaystyle \{{\mu }_n{\}}_{n\in \mathbb{N}}}$ on a measurable space ${\displaystyle (M,\mathrm{\Sigma })}$ converges in total variation to a measure ${\displaystyle \mu }$ on ${\displaystyle (M,\mathrm{\Sigma })}$ if

${\displaystyle sup\left\{\left|{\int }_{M}f(s){\mu }_n(ds)-{\int }_{M}f(s)\mu (ds)\right|:f:M\mapsto [-1,1]\text{ is measurable}\right\}\to 0\text{as }n\to \mathrm{\infty }.}$

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

${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}\underset{C\in \mathrm{\Sigma }}{inf}\left({\int }_C{f}_n(s){\mu }_n(ds)-{\int }_{C}f(s)\mu (ds)\right)\ge 0,}$

if and only if the following two statements hold:

(i) for each ${\displaystyle \epsilon >0}$, ${\displaystyle \mu (\{s\in M:f_n(s)\le f(s)-\epsilon \})\to 0}$ as ${\displaystyle n\to \mathrm{\infty }}$, and, therefore, there exists a subsequence ${\displaystyle \{f_{n_k}{\}}_{k\in \mathbb{N}}\subset \{f_n{\}}_{n\in \mathbb{N}}}$ such that ${\displaystyle f(s)\le \underset{k\to \mathrm{\infty }}{lim inf}{f}_{n_k}(s)}$ for ${\displaystyle \mu }$-a.e. ${\displaystyle s\in M}$;

(ii) ${\displaystyle \{f_n^{-}{\}}_{n\in \mathbb{N}}}$ is a.u.i. with respect to ${\displaystyle \{{\mu }_n{\}}_{n\in \mathbb{N}}}$.

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 ${\displaystyle (\mathrm{\Omega },\mathcal{F},\mathbb{P})}$; 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 ${\displaystyle (\mathrm{\Omega },\mathcal{F},\mathbb{P})}$ and let ${\displaystyle \mathcal{G}\subset \mathcal{F}}$ be a sub-σ-algebra. Then

${\displaystyle \mathbb{E}[\underset{n\to \mathrm{\infty }}{lim inf}{X}_n|\mathcal{G}]\le \underset{n\to \mathrm{\infty }}{lim inf}\mathbb{E}[X_n|\mathcal{G}]}$ 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

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

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

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

by the monotonicity of conditional expectation, hence

${\displaystyle \mathbb{E}[Y_k|\mathcal{G}]\le \underset{n\ge k}{inf}\mathbb{E}[X_n|\mathcal{G}]}$ 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

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

Extension to uniformly integrable negative parts

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

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

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

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

then

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

Note: On the set where

${\displaystyle X:=\underset{n\to \mathrm{\infty }}{lim inf}{X}_n}$

satisfies

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

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

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

Since

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

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

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

Since

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

we have

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

hence

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

This implies the assertion.