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| In [[mathematics]], '''Fatou's lemma''' establishes an [[inequality (mathematics)|inequality]] relating the [[integral]] (in the sense of [[Lebesgue integration|Lebesgue]]) of the [[limit superior and limit inferior|limit inferior]] of a [[sequence]] of [[function (mathematics)|function]]s to the limit inferior of integrals of these functions. The [[Lemma (mathematics)|lemma]] is named after [[Pierre Fatou]].
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| Fatou's lemma can be used to prove the [[Fatou–Lebesgue theorem]] and Lebesgue's [[dominated convergence theorem]].
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| ==Standard statement of Fatou's lemma==
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| Let ''f''<sub>1</sub>, ''f''<sub>2</sub>, ''f''<sub>3</sub>, . . . be a sequence of [[non-negative number|non-negative]] [[measurable function|measurable]] functions on a [[measure space]] (''S'',''Σ'',''μ''). Define the function ''f'' : ''S'' → <nowiki>[</nowiki>0, ∞<nowiki>]</nowiki> a.e. pointwise limit by
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| :<math>
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| f(s) =\liminf_{n\to\infty} f_n(s),\qquad s\in S.
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| </math> | |
| Then ''f '' is measurable and
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| :<math>
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| \int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.
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| </math>
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| | |
| '''Note:''' The functions are allowed to attain the value [[Extended real number line|+∞]] and the integrals may also be infinite.
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| ===Proof===
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| 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]].
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| <div style="clear:both;width:90%;" class="NavFrame">
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| <div class="NavHead" style="background-color:#FFFAF0; text-align:left; font-size:larger;">Direct proof</div>
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| <div class="NavContent" style="text-align:left;display:none;">
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| We will prove something a bit weaker here. Namely, we will allow ''f''<sub>n</sub> to converge μ-[[almost everywhere]] on a subset E of S. We seek to show that
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| :<math>
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| \int_E f\,d\mu \le \liminf_{n\to\infty} \int_E f_n\,d\mu\,.
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| </math>
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| Let
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| :<math> K=\{x\in E|f_n(x)\rightarrow f(x)\} </math>.
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| Then ''μ(E-K)=0'' and
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| :<math> \int_{E}f\,d\mu=\int_{K}f\,d\mu,~~~\int_{E}f_n\,d\mu=\int_{K}f_n\,d\mu ~\forall n\in \N. </math>
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| Thus, replacing ''E'' by ''K'' we may assume that ''f''<sub>n</sub> converge to ''f'' [[pointwise convergence|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
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| :<math>
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| \int_{E}\varphi \,d\mu\leq \liminf_{n\rightarrow \infty} \int_{E}f_n\,d\mu
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| </math>
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| We first consider the case when <math>\int_{E}\varphi=\infty</math>.
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| Let ''a'' be the minimum non-negative value of ''φ'' (it exists since the integral of ''φ'' is infinite). Define
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| :<math>
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| A=\{x\in E |\varphi(x)>a\}
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| </math> | |
| We must have that ''μ(A)'' is infinite since
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| :<math>\int_{E}\varphi\leq M\mu(A),</math>
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| where ''M'' is the (necessarily finite) maximum value of that ''φ'' attains.
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| Next, we define
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| :<math>
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| A_n=\{x\in E |f_k(x)>a~\forall k\geq n \}.
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| </math>
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| We have that
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| :<math>
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| A\subseteq \bigcup_n A_n \Rightarrow \mu(\bigcup_n A_n)=\infty.
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| </math>
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| But ''A<sub>n</sub>'' is a nested increasing sequence of functions and hence, by the continuity from below ''μ'',
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| :<math>
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| \lim_{n\rightarrow \infty} \mu(A_n)=\infty.
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| </math>.
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| At the same time,
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| :<math>
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| \int_E f_n\, d\mu \geq a \mu(A_n) \Rightarrow \liminf_{n\to \infty} \int_E f_n \, d\mu = \infty = \int_E \varphi\, d\mu,
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| </math>
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| proving the claim in this case.
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| The remaining case is when <math>\int_{E}\varphi<\infty</math>. We must have that ''μ(A)'' is finite. Denote, as above, by ''M'' the maximum value of ''φ'' and fix ''ε>0.'' Define
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| :<math>
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| A_n=\{x\in E|f_k(x)>(1-\epsilon)\varphi(x)~\forall k\geq n\}.
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| </math>
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| Then ''A<sub>n</sub>'' is a nested increasing sequence of sets whose union contains ''A.'' Thus, ''A-A<sub>n</sub>'' 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),
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| :<math>
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| \lim_{n\rightarrow \infty} \mu(A-A_n)=0.
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| </math>
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| Thus, there exists n such that
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| :<math>
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| \mu(A-A_k)<\epsilon ,~\forall k\geq n.
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| </math>
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| Hence, for <math>k\geq n</math>
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| :<math>
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| \int_E f_k \, d\mu \geq \int_{A_k}f_k \, d\mu \geq (1-\epsilon)\int_{A_k}\varphi\, d\mu.
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| </math>
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| At the same time,
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| :<math>
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| \int_E \varphi \, d\mu = \int_A \varphi \, d\mu = \int_{A_k} \varphi \, d\mu + \int_{A-A_k} \varphi \, d\mu.
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| </math>
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| Hence,
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| :<math>
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| (1-\epsilon)\int_{A_k} \varphi \, d\mu \geq (1-\epsilon)\int_E \varphi \, d\mu - \int_{A-A_k} \varphi \, d\mu.
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| </math>
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| Combining these inequalities gives that
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| :<math>
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| \int_{E} f_k \, d\mu \geq (1-\epsilon)\int_E \varphi \, d\mu - \int_{A-A_k} \varphi \, d\mu \geq \int_E \varphi \, d\mu - \epsilon\left(\int_{E} \varphi \, d\mu+M\right).
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| </math>
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| Hence, sending ''ε'' to 0 and taking the liminf in n, we get that
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| :<math>
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| \liminf_{n\rightarrow \infty} \int_{E} f_n \, d\mu \geq \int_E \varphi \, d\mu,
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| </math>
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| completing the proof.
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| </div>
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| </div>
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| <div style="clear:both;width:90%;" class="NavFrame">
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| <div class="NavHead" style="background-color:#FFFAF0; text-align:left; font-size:larger;">Proof using the monotone convergence theorem</div>
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| <div class="NavContent" style="text-align:left;display:none;">
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| For every natural number ''k'' define pointwise the function
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| :<math>g_k=\inf_{n\ge k}f_n.</math>
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| Then the sequence ''g''<sub>1</sub>, ''g''<sub>2</sub>, . . . of functions is increasing, meaning that ''g<sub>k</sub>'' ≤ ''g''<sub>''k''+1</sub> for all ''k'', and converges pointwise to the limit inferior ''f''.
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| For all ''k'' ≤ ''n'' we have ''g<sub>k</sub>'' ≤ ''f<sub>n</sub>'', so that by the monotonicity of the integral
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| :<math>\int_E g_k\,d\mu\le\int_E f_n\,d\mu,</math>
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| hence
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| :<math>
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| \int_E g_k\,d\mu
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| \le\inf_{n\ge k}\int_E f_n\,d\mu.
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| </math>
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| 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
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| :<math>
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| \int_E f\,d\mu
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| =\lim_{k\to\infty}\int_E g_k\,d\mu
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| \le\lim_{k\to\infty} \inf_{n\ge k}\int_E f_n\,d\mu
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| =\liminf_{n\to\infty} \int_E f_n\,d\mu\,.
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| </math>
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| </div>
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| </div>
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| ==Examples for strict inequality==
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| Equip the space <math>S</math> with the [[Borel algebra|Borel σ-algebra]] and the [[Lebesgue measure]].
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| * Example for a [[probability space]]: Let <math>S=[0,1]</math> denote the [[unit interval]]. For every [[natural number]] <math>n</math> define
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| ::<math>
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| f_n(x)=\begin{cases}n&\text{for }x\in (0,1/n),\\
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| 0&\text{otherwise.}
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| \end{cases}</math>
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| * Example with [[uniform convergence]]: Let <math>S</math> denote the set of all [[real number]]s. Define
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| ::<math>
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| f_n(x)=\begin{cases}\frac1n&\text{for }x\in [0,n],\\
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| 0&\text{otherwise.}
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| \end{cases}</math>
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| These sequences <math>(f_n)_{n\in\N}</math> converge on <math>S</math> pointwise (respectively uniformly) to the [[zero function]] (with zero integral), but every <math>f_n</math> has integral one.
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| ==A counter example==
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| A suitable assumption concerning the negative parts of the sequence ''f''<sub>1</sub>, ''f''<sub>2</sub>, . . . 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
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| :<math>
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| f_n(x)=\begin{cases}-\frac1n&\text{for }x\in [n,2n],\\
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| 0&\text{otherwise.}
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| \end{cases}</math>
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| This sequence converges uniformly on ''S'' to the zero function (with zero integral) and for every ''x'' ≥ 0 we even have ''f<sub>n''</sub>(''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<sub>n</sub>'' has integral −1, hence the inequality in Fatou's lemma fails.
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| ==Reverse Fatou lemma==
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| Let ''f''<sub>1</sub>, ''f''<sub>2</sub>, . . . be a sequence of [[extended real number line|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''<sub>''n''</sub> ≤ ''g'' for all ''n'', then
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| :<math>
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| \limsup_{n\to\infty}\int_S f_n\,d\mu\leq\int_S\limsup_{n\to\infty}f_n\,d\mu.
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| </math>
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| '''Note:''' Here ''g integrable'' means that ''g'' is measurable and that <math>\textstyle\int_S g\,d\mu<\infty</math>.
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| ===Proof===
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| Apply Fatou's lemma to the non-negative sequence given by ''g'' – ''f''<sub>''n''</sub>.
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| ==Extensions and variations of Fatou's lemma==
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| ===Integrable lower bound===
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| Let ''f''<sub>1</sub>, ''f''<sub>2</sub>, . . . 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''<sub>''n''</sub> ≥ −''g'' for all ''n'', then
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| :<math>
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| \int_S \liminf_{n\to\infty} f_n\,d\mu
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| \le \liminf_{n\to\infty} \int_S f_n\,d\mu.\
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| </math> | |
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| ====Proof====
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| Apply Fatou's lemma to the non-negative sequence given by ''f''<sub>''n''</sub> + ''g''.
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| ===Pointwise convergence===
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| If in the previous setting the sequence ''f''<sub>1</sub>, ''f''<sub>2</sub>, . . . [[Pointwise convergence|converges pointwise]] to a function ''f'' ''μ''-[[almost everywhere]] on ''S'', then
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| :<math>\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.</math>
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| ====Proof====
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| Note that ''f'' has to agree with the limit inferior of the functions ''f''<sub>''n''</sub> almost everywhere, and that the values of the integrand on a set of measure zero have no influence on the value of the integral.
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| ===Convergence in measure===
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| The last assertion also holds, if the sequence ''f''<sub>1</sub>, ''f''<sub>2</sub>, . . . [[Convergence in measure|converges in measure]] to a function ''f''.
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| ====Proof====
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| There exists a subsequence such that
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| :<math>\lim_{k\to\infty} \int_S f_{n_k}\,d\mu=\liminf_{n\to\infty} \int_S f_n\,d\mu.\ </math>
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| 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.
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| ===Fatou's Lemma with Varying Measures===
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| 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<sub>n</sub> is a sequence of measures on the measurable space (''S'',''Σ'') such that (see [[Convergence of measures]])
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| :<math>\mu_n(E)\to \mu(E),~\forall E\in \Sigma. </math>
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| Then, with ''f<sub>n</sub>'' non-negative integrable functions and ''f'' being their pointwise limit inferior, we have
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| :<math> \int_S f\,d\mu \leq \liminf_{n\to \infty} \int_S f_n\, d\mu_n. </math>
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| :{| class="toccolours collapsible collapsed" width="90%" style="text-align:left"
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| !Proof
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| |-
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| |We will prove something a bit stronger here. Namely, we will allow ''f''<sub>n</sub> to converge μ-[[almost everywhere]] on a subset E of S. We seek to show that
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| :<math>
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| \int_E f\,d\mu \le \liminf_{n\to\infty} \int_E f_n\,d\mu_n\,.
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| </math>
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| Let
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| :<math> K=\{x\in E|f_n(x)\rightarrow f(x)\} </math>.
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| Then ''μ(E-K)=0'' and
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| :<math> \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. </math> | |
| Thus, replacing ''E'' by ''E-K'' we may assume that ''f''<sub>n</sub> converge to ''f'' [[pointwise convergence|pointwise]] on E. Next, note that for any simple function ''φ'' we have
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| :<math> \int_{E}\phi\, d\mu=\lim_{n\to \infty} \int_{E} \phi\, d\mu_n. </math>
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| 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
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| :<math>
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| \int_{E}\phi \,d\mu\leq \liminf_{n\rightarrow \infty} \int_{E}f_n\,d\mu_n
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| </math>
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| Let ''a'' be the minimum non-negative value of ''φ.'' Define
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| :<math>
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| A=\{x\in E |\phi(x)>a\}
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| </math>
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| | |
| We first consider the case when <math>\int_{E}\phi\, d\mu=\infty</math>.
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| We must have that ''μ(A)'' is infinite since
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| :<math>\int_{E}\phi\, d\mu \leq M\mu(A),</math>
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| where ''M'' is the (necessarily finite) maximum value of that ''φ'' attains.
| |
| | |
| Next, we define
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| :<math>
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| A_n=\{x\in E |f_k(x)>a~\forall k\geq n \}.
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| </math>
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| We have that
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| :<math> | |
| A\subseteq \bigcup_n A_n \Rightarrow \mu(\bigcup_n A_n)=\infty.
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| </math>
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| But ''A<sub>n</sub>'' is a nested increasing sequence of functions and hence, by the continuity from below ''μ'',
| |
| :<math>
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| \lim_{n\rightarrow \infty} \mu(A_n)=\infty.
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| </math>.
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| Thus,
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| :<math>
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| \lim_{n\to\infty}\mu_n(A_n)=\mu(A_n)=\infty.
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| </math>.
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| At the same time,
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| :<math>
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| \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,
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| </math>
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| proving the claim in this case.
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| | |
| The remaining case is when <math>\int_{E}\phi\, d\mu<\infty</math>. We must have that ''μ(A)'' is finite. Denote, as above, by ''M'' the maximum value of ''φ'' and fix ''ε>0.'' Define
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| :<math>
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| A_n=\{x\in E|f_k(x)>(1-\epsilon)\phi(x)~\forall k\geq n\}.
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| </math>
| |
| Then ''A<sub>n</sub>'' is a nested increasing sequence of sets whose union contains ''A.'' Thus, ''A-A<sub>n</sub>'' 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),
| |
| :<math>
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| \lim_{n\rightarrow \infty} \mu(A-A_n)=0.
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| </math>
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| Thus, there exists n such that
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| :<math>
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| \mu(A-A_k)<\epsilon ,~\forall k\geq n.
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| </math>
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| Therefore, since
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| :<math>
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| \lim_{n\to \infty} \mu_n(A-A_k)=\mu(A-A_k),
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| </math>
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| there exists N such that
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| :<math>
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| \mu_k(A-A_k)<\epsilon,~\forall k\geq N.
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| </math>
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| Hence, for <math>k\geq N</math>
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| :<math>
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| \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.
| |
| </math>
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| At the same time,
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| :<math>
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| \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.
| |
| </math>
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| Hence,
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| :<math>
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| (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.
| |
| </math>
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| Combining these inequalities gives that
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| :<math>
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| \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).
| |
| </math>
| |
| Hence, sending ''ε'' to 0 and taking the liminf in n, we get that
| |
| :<math>
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| \liminf_{n\rightarrow \infty} \int_{E} f_n \, d\mu_k \geq \int_E \phi \, d\mu,
| |
| </math>
| |
| completing the proof.
| |
| |}
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| | |
| ==Fatou's lemma for conditional expectations==
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| In [[probability theory]], by a change of notation, the above versions of Fatou's lemma are applicable to sequences of [[random variables]] ''X''<sub>1</sub>, ''X''<sub>2</sub>, . . . defined on a [[probability space]] <math>\scriptstyle(\Omega,\,\mathcal F,\,\mathbb P)</math>; the integrals turn into [[expected value|expectation]]s. In addition, there is also a version for [[conditional expectation]]s.
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| | |
| ===Standard version===
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| Let ''X''<sub>1</sub>, ''X''<sub>2</sub>, . . . be a sequence of non-negative random variables on a probability space <math>\scriptstyle(\Omega,\mathcal F,\mathbb P)</math> and let
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| <math>\scriptstyle \mathcal G\,\subset\,\mathcal F</math> be a sub-[[σ-algebra]]. Then
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| :<math>\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]</math> [[almost surely]].
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| | |
| '''Note:''' Conditional expectation for non-negative random variables is always well defined, finite expectation is not needed.
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| ====Proof====
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| 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|monotone convergence theorem for conditional expectations]] has to be applied.
| |
| | |
| Let ''X'' denote the limit inferior of the ''X''<sub>''n''</sub>. For every natural number ''k'' define pointwise the random variable
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| :<math>Y_k=\inf_{n\ge k}X_n.</math>
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| Then the sequence ''Y''<sub>1</sub>, ''Y''<sub>2</sub>, . . . is increasing and converges pointwise to ''X''.
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| For ''k'' ≤ ''n'', we have ''Y''<sub>''k''</sub> ≤ ''X''<sub>''n''</sub>, so that
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| :<math>\mathbb{E}[Y_k|\mathcal G]\le\mathbb{E}[X_n|\mathcal G]</math> almost surely
| |
| by the [[Conditional_expectation#Basic_properties|monotonicity of conditional expectation]], hence
| |
| :<math>\mathbb{E}[Y_k|\mathcal G]\le\inf_{n\ge k}\mathbb{E}[X_n|\mathcal G]</math> almost surely,
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| because the countable union of the exceptional sets of probability zero is again a [[null set]].
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| Using the definition of ''X'', its representation as pointwise limit of the ''Y''<Sub>''k''</sub>, the monotone convergence theorem for conditional expectations, the last inequality, and the definition of the limit inferior, it follows that almost surely
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| :<math>
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| \begin{align}
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| \mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]
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| &=\mathbb{E}[X|\mathcal G]
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| =\mathbb{E}\Bigl[\lim_{k\to\infty}Y_k\,\Big|\,\mathcal G\Bigr]
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| =\lim_{k\to\infty}\mathbb{E}[Y_k|\mathcal G]\\
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| &\le\lim_{k\to\infty} \inf_{n\ge k}\mathbb{E}[X_n|\mathcal G]
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| =\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G].
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| \end{align}
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| </math>
| |
| | |
| ===Extension to uniformly integrable negative parts===
| |
| Let ''X''<sub>1</sub>, ''X''<sub>2</sub>, . . . be a sequence of random variables on a probability space <math>\scriptstyle(\Omega,\mathcal F,\mathbb P)</math> and let
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| <math>\scriptstyle \mathcal G\,\subset\,\mathcal F</math> be a sub-[[σ-algebra]]. If the negative parts
| |
| | |
| :<math>X_n^-:=\max\{-X_n,0\},\qquad n\in{\mathbb N},</math>
| |
| | |
| are uniformly integrable with respect to the conditional expectation, in the sense that, for ''ε'' > 0 there exists a ''c'' > 0 such that
| |
| | |
| :<math>\mathbb{E}\bigl[X_n^-1_{\{X_n^->c\}}\,|\,\mathcal G\bigr]<\varepsilon,
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| \qquad\text{for all }n\in\mathbb{N},\,\text{almost surely}</math>,
| |
| then | |
| | |
| :<math>\mathbb{E}\Bigl[\liminf_{n\to\infty}X_n\,\Big|\,\mathcal G\Bigr]\le\liminf_{n\to\infty}\,\mathbb{E}[X_n|\mathcal G]</math> almost surely.
| |
| | |
| '''Note:''' On the set where
| |
| | |
| :<math>X:=\liminf_{n\to\infty}X_n</math>
| |
| | |
| satisfies
| |
| | |
| :<math>\mathbb{E}[\max\{X,0\}\,|\,\mathcal G]=\infty,</math>
| |
| | |
| 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
| |
| | |
| :<math>\mathbb{E}\bigl[X_n^-1_{\{X_n^->c\}}\,|\,\mathcal G\bigr]<\varepsilon
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| \qquad\text{for all }n\in\mathbb{N},\,\text{almost surely}.</math>
| |
| | |
| Since
| |
| | |
| :<math>X+c\le\liminf_{n\to\infty}(X_n+c)^+,</math>
| |
| | |
| where ''x''<sup>+</sup> := 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
| |
| | |
| :<math>\mathbb{E}[X\,|\,\mathcal G]+c
| |
| \le\mathbb{E}\Bigl[\liminf_{n\to\infty}(X_n+c)^+\,\Big|\,\mathcal G\Bigr]
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| \le\liminf_{n\to\infty}\mathbb{E}[(X_n+c)^+\,|\,\mathcal G]</math> almost surely.
| |
| | |
| Since
| |
| | |
| :<math>(X_n+c)^+=(X_n+c)+(X_n+c)^-\le X_n+c+X_n^-1_{\{X_n^->c\}},</math>
| |
| | |
| we have
| |
| | |
| :<math>\mathbb{E}[(X_n+c)^+\,|\,\mathcal G]
| |
| \le\mathbb{E}[X_n\,|\,\mathcal G]+c+\varepsilon</math> almost surely,
| |
| | |
| hence
| |
| | |
| :<math>\mathbb{E}[X\,|\,\mathcal G]\le
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| \liminf_{n\to\infty}\mathbb{E}[X_n\,|\,\mathcal G]+\varepsilon</math> almost surely.
| |
| | |
| This implies the assertion.
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| | |
| ==References==
| |
| *{{cite book
| |
| | last = Royden
| |
| | first = H.L.
| |
| | title = Real Analysis
| |
| | edition = 3rd
| |
| | year = 1988
| |
| }}
| |
| | |
| ==External links==
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| *{{planetmath reference|id=3678|title=Fatou's lemma}}
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| [[Category:Inequalities]]
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| [[Category:Lemmas]]
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| [[Category:Measure theory]]
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| [[Category:Real analysis]]
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| [[Category:Articles containing proofs]]
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