In what follows, denotes the -algebra of Borel sets on .
Theorem — Fatou's lemma. Given a measure space and a set let be a sequence of -measurable non-negative functions . Define the function by setting for every .
Then is -measurable, and also , where the integrals may be infinite.
Fatou's lemma remains true if its assumptions hold -almost everywhere. In other words, it is enough that there is a null set such that the values are non-negative for every To see this, note that the integrals appearing in Fatou's lemma are unchanged if we change each function on .
Proof
Fatou's lemma does not require the monotone convergence theorem, but the latter can be used to provide a quick proof. A proof directly from the definitions of integrals is given further below.
In each case, the proof begins by analyzing the properties of . These satisfy:
the sequence is pointwise non-decreasing at any x and
, .
Since , we immediately see that f is measurable.
Via the Monotone Convergence Theorem
Moreover,
By the Monotone Convergence Theorem and property (1), the limit and integral may be interchanged:
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.
Denote by the set of simple-measurable functions such that on .
Monotonicity —
If everywhere on then
If and then
If f is nonnegative and , where is a non-decreasing chain of -measurable sets, then
Proof
1. Since we have
By definition of Lebesgue integral and the properties of supremum,
2. Let be the indicator function of the set It can be deduced from the definition of Lebesgue integral that
if we notice that, for every outside of Combined with the previous property, the inequality implies
3. First note that the claim holds if f is the indicator function of a set, by monotonicity of measures. By linearity, this also immediately implies the claim for simple functions.
Since any simple function supported on Sn is simple and supported on X, we must have
.
For the reverse, suppose g ∈ SF(f) with By the above,
Now we turn to the main theorem
Step 1 — is -measurable, for every , as is .
Proof
Recall the closed intervals generate the Borelσ-algebra. Thus it suffices to show, for every , that . Now observe that
Every set on the right-hand side is from , which is closed under countable intersections. Thus the left-hand side is also a member of .
Similarly, it is enough to verify that , for every . Since the sequence pointwise non-decreases,
.
Step 2 — Given a simple function and a real number , define
Since the pre-image of the Borel set under the measurable function is measurable, and -algebras are closed under finite intersection and unions, the first claim follows.
Step 2b. To prove the second claim, note that, for each and every ,
Step 2c. To prove the third claim, suppose for contradiction there exists
Then , for every . Taking the limit as ,
This contradicts our initial assumption that .
Step 3 — From step 2 and monotonicity,
Step 4 — For every ,
.
Proof
Indeed, using the definition of , the non-negativity of , and the monotonicity of Lebesgue integral, we have
.
In accordance with Step 4, as the inequality becomes
.
Taking the limit as yields
,
as required.
Step 5 — To complete the proof, we apply the definition of Lebesgue integral to the inequality established in Step 4 and take into account that :
These sequences converge on pointwise (respectively uniformly) to the zero function (with zero integral), but every 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
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 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 § 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 fn ≤ g for all n, then
Note: Here g integrable means that g is measurable and that .
Sketch of proof
We apply linearity of Lebesgue integral and Fatou's lemma to the sequence Since this sequence is defined -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
Proof
Apply Fatou's lemma to the non-negative sequence given by fn + g.
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
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)
Then, with fn non-negative integrable functions and f being their pointwise limit inferior, we have
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
Let
.
Then μ(E-K)=0 and
Thus, replacing E by E-K we may assume that fn converge to fpointwise on E. Next, note that for any simple function φ we have
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
Let a be the minimum non-negative value of φ. Define
We first consider the case when . We must have that μ(A) is infinite since
where M is the (necessarily finite) maximum value of that φ attains.
Next, we define
We have that
But An is a nested increasing sequence of functions and hence, by the continuity from below μ,
.
Thus,
.
At the same time,
proving the claim in this case.
The remaining case is when . We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define
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),
Thus, there exists n such that
Therefore, since
there exists N such that
Hence, for
At the same time,
Hence,
Combining these inequalities gives that
Hence, sending ε to 0 and taking the liminf in n, we get that
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
Extension to uniformly integrable negative parts
Let X1, X2, . . . be a sequence of random variables on a probability space and let be a sub-σ-algebra. If the negative parts
are uniformly integrable with respect to the conditional expectation, in the sense that, for ε > 0 there exists a c > 0 such that
,
then
almost surely.
Note: On the set where
satisfies
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
Since
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
almost surely.
Since
we have
almost surely,
hence
almost surely.
This implies the assertion.
References
Carothers, N. L. (2000). Real Analysis. New York: Cambridge University Press. pp. 321–22. ISBN0-521-49756-6.
Royden, H. L. (1988). Real Analysis (3rd ed.). London: Collier Macmillan. ISBN0-02-404151-3.
Weir, Alan J. (1973). "The Convergence Theorems". Lebesgue Integration and Measure. Cambridge: Cambridge University Press. pp. 93–118. ISBN0-521-08728-7.
March 31, 2023
fatou, lemma, confused, with, fatou, theorem, mathematics, establishes, inequality, relating, lebesgue, integral, limit, inferior, sequence, functions, limit, inferior, integrals, these, functions, lemma, named, after, pierre, fatou, used, prove, fatou, lebesg. Not to 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 Contents 1 Standard statement 1 1 Proof 1 1 1 Via the Monotone Convergence Theorem 1 1 2 From first principles 2 Examples for strict inequality 3 The role of non negativity 4 Reverse Fatou lemma 4 1 Sketch of proof 5 Extensions and variations of Fatou s lemma 5 1 Integrable lower bound 5 1 1 Proof 5 2 Pointwise convergence 5 2 1 Proof 5 3 Convergence in measure 5 3 1 Proof 5 4 Fatou s Lemma with Varying Measures 6 Fatou s lemma for conditional expectations 6 1 Standard version 6 1 1 Proof 6 2 Extension to uniformly integrable negative parts 6 2 1 Proof 7 ReferencesStandard statement EditIn what follows B R 0 displaystyle operatorname mathcal B mathbb R geq 0 denotes the s displaystyle sigma algebra of Borel sets on 0 displaystyle 0 infty Theorem Fatou s lemma Given a measure space W F m displaystyle Omega mathcal F mu and a set X F displaystyle X in mathcal F let f n displaystyle f n be a sequence of F B R 0 displaystyle mathcal F operatorname mathcal B mathbb R geq 0 measurable non negative functions f n X 0 displaystyle f n X to 0 infty Define the function f X 0 displaystyle f X to 0 infty by setting f x lim inf n f n x displaystyle f x liminf n to infty f n x for every x X displaystyle x in X Then f displaystyle f is F B R 0 displaystyle mathcal F operatorname mathcal B mathbb R geq 0 measurable and also X f d m lim inf n X f n d m displaystyle int X f d mu leq liminf n to infty int X f n d mu where the integrals may be infinite Fatou s lemma remains true if its assumptions hold m displaystyle mu almost everywhere In other words it is enough that there is a null set N displaystyle N such that the values f n x displaystyle f n x are non negative for every x X N displaystyle 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 displaystyle N Proof Edit Fatou s lemma does not require the monotone convergence theorem but the latter can be used to provide a quick proof A proof directly from the definitions of integrals is given further below In each case the proof begins by analyzing the properties of g n x inf k n f k x displaystyle textstyle g n x inf k geq n f k x These satisfy the sequence g n x n displaystyle g n x n is pointwise non decreasing at any x and g n f n displaystyle g n leq f n n N displaystyle forall n in mathbb N Since f x lim inf n f n x lim n inf k n f k x lim n g n x displaystyle f x liminf n to infty f n x lim n to infty inf k geq n f k x lim n to infty g n x we immediately see that f is measurable Via the Monotone Convergence Theorem Edit Moreover X f d m X lim n g n d m displaystyle int X f d mu int X lim n to infty g n d mu By the Monotone Convergence Theorem and property 1 the limit and integral may be interchanged X f d m lim n X g n d m lim inf n X g n d m lim inf n X f n d m displaystyle begin aligned int X f d mu amp lim n to infty int X g n d mu amp liminf n to infty int X g n d mu amp leq liminf n to infty int X f n d mu end aligned where the last step used property 2 From first principles Edit 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 Denote by SF f displaystyle operatorname SF f the set of simple F B R 0 displaystyle mathcal F operatorname mathcal B mathbb R geq 0 measurable functions s X 0 displaystyle s X to 0 infty such that 0 s f displaystyle 0 leq s leq f on X displaystyle X Monotonicity If f g displaystyle f leq g everywhere on X displaystyle X then X f d m X g d m displaystyle int X f d mu leq int X g d mu If X 1 X 2 F displaystyle X 1 X 2 in mathcal F and X 1 X 2 displaystyle X 1 subseteq X 2 then X 1 f d m X 2 f d m displaystyle int X 1 f d mu leq int X 2 f d mu If f is nonnegative and S i 1 S i displaystyle S cup i 1 infty S i where S 1 S i S displaystyle S 1 subseteq ldots subseteq S i subseteq ldots subseteq S is a non decreasing chain of m displaystyle mu measurable sets then S f d m lim n S n f d m displaystyle int S f d mu lim n to infty int S n f d mu Proof 1 Since f g displaystyle f leq g we have SF f SF g displaystyle operatorname SF f subseteq operatorname SF g By definition of Lebesgue integral and the properties of supremum X f d m sup s S F f X s d m sup s S F g X s d m X g d m displaystyle int X f d mu sup s in rm SF f int X s d mu leq sup s in rm SF g int X s d mu int X g d mu 2 Let 1 X 1 displaystyle mathbf 1 X 1 be the indicator function of the set X 1 displaystyle X 1 It can be deduced from the definition of Lebesgue integral that X 2 f 1 X 1 d m X 1 f d m displaystyle int X 2 f cdot mathbf 1 X 1 d mu int X 1 f d mu if we notice that for every s S F f 1 X 1 displaystyle s in rm SF f cdot mathbf 1 X 1 s 0 displaystyle s 0 outside of X 1 displaystyle X 1 Combined with the previous property the inequality f 1 X 1 f displaystyle f cdot mathbf 1 X 1 leq f implies X 1 f d m X 2 f 1 X 1 d m X 2 f d m displaystyle int X 1 f d mu int X 2 f cdot mathbf 1 X 1 d mu leq int X 2 f d mu 3 First note that the claim holds if f is the indicator function of a set by monotonicity of measures By linearity this also immediately implies the claim for simple functions Since any simple function supported on Sn is simple and supported on X we must have X f d m lim n S n f d m displaystyle int X f d mu geq lim n to infty int S n f d mu For the reverse suppose g SF f with X f d m ϵ X g d m displaystyle textstyle int X f d mu epsilon leq int X g d mu By the above X f d m ϵ X g d m lim n S n g d m lim n S n f d m displaystyle int X f d mu epsilon leq int X g d mu lim n to infty int S n g d mu leq lim n to infty int S n f d mu Now we turn to the main theorem Step 1 g n g n x displaystyle g n g n x is F B R 0 displaystyle mathcal F operatorname mathcal B mathbb R geq 0 measurable for every n 1 displaystyle n geq 1 as is f displaystyle f Proof Recall the closed intervals generate the Borel s algebra Thus it suffices to show for every t displaystyle t in infty infty that g n 1 t F displaystyle g n 1 t infty in mathcal F Now observe that g n 1 t x X g n x t k x X f k x t k f k 1 t displaystyle begin aligned g n 1 t infty amp left x in X mid g n x geq t right 3pt amp bigcap k left x in X mid f k x geq t right 3pt amp bigcap k f k 1 t infty end aligned Every set on the right hand side is from F displaystyle mathcal F which is closed under countable intersections Thus the left hand side is also a member of F displaystyle mathcal F Similarly it is enough to verify that f 1 0 t F displaystyle f 1 0 t in mathcal F for every t displaystyle t in infty infty Since the sequence g n x displaystyle g n x pointwise non decreases f 1 0 t n g n 1 0 t F displaystyle f 1 0 t bigcap n g n 1 0 t in mathcal F Step 2 Given a simple function s SF f displaystyle s in operatorname SF f and a real number t 0 1 displaystyle t in 0 1 define B k s t x X t s x g k x X displaystyle B k s t x in X mid t cdot s x leq g k x subseteq X Then B k s t F displaystyle B k s t in mathcal F B k s t B k 1 s t displaystyle B k s t subseteq B k 1 s t and X k B k s t displaystyle textstyle X bigcup k B k s t Proof Step 2a To prove the first claim write s as a weighted sum of indicator functions of disjoint sets s i 1 m c i 1 A i displaystyle s sum i 1 m c i cdot mathbf 1 A i Then B k s t i 1 m g k 1 t c i A i displaystyle B k s t bigcup i 1 m Bigl g k 1 Bigl t cdot c i infty Bigr cap A i Bigr Since the pre image g k 1 t c i displaystyle g k 1 Bigl t cdot c i infty Bigr of the Borel set t c i displaystyle t cdot c i infty under the measurable function g k displaystyle g k is measurable and s displaystyle sigma algebras are closed under finite intersection and unions the first claim follows Step 2b To prove the second claim note that for each k displaystyle k and every x X displaystyle x in X g k x g k 1 x displaystyle g k x leq g k 1 x Step 2c To prove the third claim suppose for contradiction there exists x 0 X k B k s t k X B k s t displaystyle x 0 in X setminus bigcup k B k s t bigcap k X setminus B k s t Then g k x 0 lt t s x 0 displaystyle g k x 0 lt t cdot s x 0 for every k displaystyle k Taking the limit as k displaystyle k to infty f x 0 t s x 0 lt s x 0 displaystyle f x 0 leq t cdot s x 0 lt s x 0 This contradicts our initial assumption that s f displaystyle s leq f Step 3 From step 2 and monotonicity lim n B n s t s d m X s d m displaystyle lim n int B n s t s d mu int X s d mu Step 4 For every s SF f displaystyle s in operatorname SF f X s d m lim k X g k d m displaystyle int X s d mu leq lim k int X g k d mu Proof Indeed using the definition of B k s t displaystyle B k s t the non negativity of g k displaystyle g k and the monotonicity of Lebesgue integral we have k 1 B k s t t s d m B k s t g k d m X g k d m displaystyle forall k geq 1 qquad int B k s t t cdot s d mu leq int B k s t g k d mu leq int X g k d mu In accordance with Step 4 as k displaystyle k to infty the inequality becomes t X s d m lim k X g k d m displaystyle t int X s d mu leq lim k int X g k d mu Taking the limit as t 1 displaystyle t uparrow 1 yields X s d m lim k X g k d m displaystyle int X s d mu leq lim k int X g k d mu as required Step 5 To complete the proof we apply the definition of Lebesgue integral to the inequality established in Step 4 and take into account that g n f n displaystyle g n leq f n X f d m sup s SF f X s d m lim k X g k d m lim inf k X g k d m lim inf k X f k d m displaystyle begin aligned int X f d mu amp sup s in operatorname SF f int X s d mu amp leq lim k int X g k d mu amp liminf k int X g k d mu amp leq liminf k int X f k d mu end aligned The proof is complete Examples for strict inequality EditEquip the space S displaystyle S with the Borel s algebra and the Lebesgue measure Example for a probability space Let S 0 1 displaystyle S 0 1 denote the unit interval For every natural number n displaystyle n definef n x n for x 0 1 n 0 otherwise displaystyle f n x begin cases n amp text for x in 0 1 n 0 amp text otherwise end cases dd Example with uniform convergence Let S displaystyle S denote the set of all real numbers Definef n x 1 n for x 0 n 0 otherwise displaystyle f n x begin cases frac 1 n amp text for x in 0 n 0 amp text otherwise end cases dd These sequences f n n N displaystyle f n n in mathbb N converge on S displaystyle S pointwise respectively uniformly to the zero function with zero integral but every f n displaystyle f n has integral one The role of non negativity EditA 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 s algebra and the Lebesgue measure For every natural number n define f n x 1 n for x n 2 n 0 otherwise displaystyle f n x begin cases frac 1 n amp text for x in n 2n 0 amp 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 gt x 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 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 EditLet f1 f2 be a sequence of extended real valued measurable functions defined on a measure space S S m If there exists a non negative integrable function g on S such that fn g for all n then lim sup n S f n d m S lim sup n f n d m displaystyle 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 S g d m lt displaystyle textstyle int S g d mu lt infty Sketch of proof Edit We apply linearity of Lebesgue integral and Fatou s lemma to the sequence g f n displaystyle g f n Since S g d m lt displaystyle textstyle int S g d mu lt infty this sequence is defined m displaystyle mu almost everywhere and non negative Extensions and variations of Fatou s lemma EditIntegrable lower bound Edit Let f1 f2 be a sequence of extended real valued measurable functions defined on a measure space S S m If there exists an integrable function g on S such that fn g for all n then S lim inf n f n d m lim inf n S f n d m displaystyle int S liminf n to infty f n d mu leq liminf n to infty int S f n d mu Proof Edit Apply Fatou s lemma to the non negative sequence given by fn g Pointwise convergence Edit If in the previous setting the sequence f1 f2 converges pointwise to a function f m almost everywhere on S then S f d m lim inf n S f n d m displaystyle int S f d mu leq liminf n to infty int S f n d mu Proof Edit 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 Edit The last assertion also holds if the sequence f1 f2 converges in measure to a function f Proof Edit There exists a subsequence such that lim k S f n k d m lim inf n S f n d m displaystyle 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 Edit In all of the above statements of Fatou s Lemma the integration was carried out with respect to a single fixed measure m Suppose that mn is a sequence of measures on the measurable space S S such that see Convergence of measures m n E m E E F displaystyle mu n E to mu E forall E in mathcal F Then with fn non negative integrable functions and f being their pointwise limit inferior we have S f d m lim inf n S f n d m n displaystyle int S f d mu leq liminf n to infty int S f n d mu n ProofWe will prove something a bit stronger here Namely we will allow fn to converge m almost everywhere on a subset E of S We seek to show that E f d m lim inf n E f n d m n displaystyle int E f d mu leq liminf n to infty int E f n d mu n Let K x E f n x f x displaystyle K x in E f n x rightarrow f x Then m E K 0 and E f d m E K f d m E f n d m E K f n d m n N displaystyle 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 mathbb 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 f we have E ϕ d m lim n E ϕ d m n displaystyle 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 f is any non negative simple function less than or equal to f then E ϕ d m lim inf n E f n d m n displaystyle 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 f Define A x E ϕ x gt a displaystyle A x in E phi x gt a We first consider the case when E ϕ d m displaystyle int E phi d mu infty We must have that m A is infinite since E ϕ d m M m A displaystyle int E phi d mu leq M mu A where M is the necessarily finite maximum value of that f attains Next we define A n x E f k x gt a k n displaystyle A n x in E f k x gt a forall k geq n We have that A n A n m n A n displaystyle 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 m lim n m A n displaystyle lim n rightarrow infty mu A n infty Thus lim n m n A n m A n displaystyle lim n to infty mu n A n mu A n infty At the same time E f n d m n a m n A n lim inf n E f n d m n E ϕ d m displaystyle 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 E ϕ d m lt displaystyle int E phi d mu lt infty We must have that m A is finite Denote as above by M the maximum value of f and fix e gt 0 Define A n x E f k x gt 1 ϵ ϕ x k n displaystyle A n x in E f k x gt 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 m A A n 0 displaystyle lim n rightarrow infty mu A A n 0 Thus there exists n such that m A A k lt ϵ k n displaystyle mu A A k lt epsilon forall k geq n Therefore since lim n m n A A k m A A k displaystyle lim n to infty mu n A A k mu A A k there exists N such that m k A A k lt ϵ k N displaystyle mu k A A k lt epsilon forall k geq N Hence for k N displaystyle k geq N E f k d m k A k f k d m k 1 ϵ A k ϕ d m k displaystyle 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 E ϕ d m k A ϕ d m k A k ϕ d m k A A k ϕ d m k displaystyle 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 ϵ A k ϕ d m k 1 ϵ E ϕ d m k A A k ϕ d m k displaystyle 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 E f k d m k 1 ϵ E ϕ d m k A A k ϕ d m k E ϕ d m k ϵ E ϕ d m k M displaystyle 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 e to 0 and taking the liminf in n we get that lim inf n E f n d m k E ϕ d m displaystyle 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 EditIn 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 W F P displaystyle scriptstyle Omega mathcal F mathbb P the integrals turn into expectations In addition there is also a version for conditional expectations Standard version Edit Let X1 X2 be a sequence of non negative random variables on a probability space W F P displaystyle scriptstyle Omega mathcal F mathbb P and let G F displaystyle scriptstyle mathcal G subset mathcal F be a sub s algebra Then E lim inf n X n G lim inf n E X n G displaystyle mathbb E Bigl liminf n to infty X n Big mathcal G Bigr leq 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 Edit 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 k X n displaystyle Y k inf n geq k X n Then the sequence Y1 Y2 is increasing and converges pointwise to X For k n we have Yk Xn so that E Y k G E X n G displaystyle mathbb E Y k mathcal G leq mathbb E X n mathcal G almost surelyby the monotonicity of conditional expectation hence E Y k G inf n k E X n G displaystyle mathbb E Y k mathcal G leq inf n geq 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 E lim inf n X n G E X G E lim k Y k G lim k E Y k G lim k inf n k E X n G lim inf n E X n G displaystyle begin aligned mathbb E Bigl liminf n to infty X n Big mathcal G Bigr amp 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 amp leq lim k to infty inf n geq k mathbb E X n mathcal G liminf n to infty mathbb E X n mathcal G end aligned Extension to uniformly integrable negative parts Edit Let X1 X2 be a sequence of random variables on a probability space W F P displaystyle scriptstyle Omega mathcal F mathbb P and let G F displaystyle scriptstyle mathcal G subset mathcal F be a sub s algebra If the negative parts X n max X n 0 n N displaystyle 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 e gt 0 there exists a c gt 0 such that E X n 1 X n gt c G lt e for all n N almost surely displaystyle mathbb E bigl X n 1 X n gt c mathcal G bigr lt varepsilon qquad text for all n in mathbb N text almost surely then E lim inf n X n G lim inf n E X n G displaystyle mathbb E Bigl liminf n to infty X n Big mathcal G Bigr leq liminf n to infty mathbb E X n mathcal G almost surely Note On the set where X lim inf n X n displaystyle X liminf n to infty X n satisfies E max X 0 G displaystyle 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 Edit Let e gt 0 Due to uniform integrability with respect to the conditional expectation there exists a c gt 0 such that E X n 1 X n gt c G lt e for all n N almost surely displaystyle mathbb E bigl X n 1 X n gt c mathcal G bigr lt varepsilon qquad text for all n in mathbb N text almost surely Since X c lim inf n X n c displaystyle X c leq 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 E X G c E lim inf n X n c G lim inf n E X n c G displaystyle mathbb E X mathcal G c leq mathbb E Bigl liminf n to infty X n c Big mathcal G Bigr leq liminf n to infty mathbb E X n c mathcal G almost surely Since X n c X n c X n c X n c X n 1 X n gt c displaystyle X n c X n c X n c leq X n c X n 1 X n gt c we have E X n c G E X n G c e displaystyle mathbb E X n c mathcal G leq mathbb E X n mathcal G c varepsilon almost surely hence E X G lim inf n E X n G e displaystyle mathbb E X mathcal G leq liminf n to infty mathbb E X n mathcal G varepsilon almost surely This implies the assertion References EditCarothers N L 2000 Real Analysis New York Cambridge University Press pp 321 22 ISBN 0 521 49756 6 Royden H L 1988 Real Analysis 3rd ed London Collier Macmillan ISBN 0 02 404151 3 Weir Alan J 1973 The Convergence Theorems Lebesgue Integration and Measure Cambridge Cambridge University Press pp 93 118 ISBN 0 521 08728 7 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