In mathematics, Fatou's lemma establishes an inequality relating the integral (in the sense of Lebesgue) 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 of Fatou's lemma
Let f1, f2, f3, . . . be a sequence of non-negative measurable functions on a measure space (S,Σ,μ). Define the function f : S → [0, ∞] a.e. pointwise limit by
![{\displaystyle f(s)=\liminf _{n\to \infty }f_{n}(s),\qquad s\in S.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ce9801f4d3dcb9284bb7a4ef9dd8b7dc3d19bf8)
Then f is measurable and
![{\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu \,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e362c44a24a65c12d0564610460b1059c40c9cce)
Note: The functions are allowed to attain the value +∞ and the integrals may also be infinite.
Proof
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.
Direct proof
We will prove something a bit weaker here. Namely, we will allow fn to converge μ-almost everywhere on a subset E of S. We seek to show that
![{\displaystyle \int _{E}f\,d\mu \leq \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu \,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f012d2e4882722d05d5a408ecf56fda2b7221d04)
Let
.
Then μ(E-K)=0 and
![{\displaystyle \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 \mathbb {N} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d4907ff69cabc0e0d758ff0a959df55467d8e45)
Thus, replacing E by K we may assume that fn converge to f 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
![{\displaystyle \int _{E}\varphi \,d\mu \leq \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0079455bc4c47eca2db4de99d19fec9623d882b9)
We first consider the case when
.
Let a be the minimum non-negative value of φ (it exists since the integral of φ is infinite). Define
![{\displaystyle A=\{x\in E|\varphi (x)>a\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69b9d3434373015333593befc0931e9cf97fb0fc)
We must have that μ(A) is infinite since
![{\displaystyle \int _{E}\varphi \leq M\mu (A),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c15b450c02eae40ff80ba55f4ba3fee9e90fbc7)
where M is the (necessarily finite) maximum value of that φ attains.
Next, we define
![{\displaystyle A_{n}=\{x\in E|f_{k}(x)>a~\forall k\geq n\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bf1c4742450624afac34f585de959bd26fa28ab)
We have that
![{\displaystyle A\subseteq \bigcup _{n}A_{n}\Rightarrow \mu (\bigcup _{n}A_{n})=\infty .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5592c7f5a9c7f7e7567dcf85fa747872b1fde8e)
But An is a nested increasing sequence of functions and hence, by the continuity from below μ,
.
At the same time,
![{\displaystyle \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 ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59306097c1aa7ecde66df6ce0a462b4e7b07e214)
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
![{\displaystyle A_{n}=\{x\in E|f_{k}(x)>(1-\epsilon )\varphi (x)~\forall k\geq n\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d565ed3feb1af923752ca9624e37debd9420d54c)
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),
![{\displaystyle \lim _{n\rightarrow \infty }\mu (A-A_{n})=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51664d9bae31d3616de3896e0149f139e4bffbfa)
Thus, there exists n such that
![{\displaystyle \mu (A-A_{k})<\epsilon ,~\forall k\geq n.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adfafb6220ef3f26e0b540f15d4ab58c9c8f8e72)
Hence, for
![{\displaystyle \int _{E}f_{k}\,d\mu \geq \int _{A_{k}}f_{k}\,d\mu \geq (1-\epsilon )\int _{A_{k}}\varphi \,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d342ab0406162fc1789276c8e8cb4758adbac73)
At the same time,
![{\displaystyle \int _{E}\varphi \,d\mu =\int _{A}\varphi \,d\mu =\int _{A_{k}}\varphi \,d\mu +\int _{A-A_{k}}\varphi \,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/394ac55c221c49de4427a3c130158abb7ab30241)
Hence,
![{\displaystyle (1-\epsilon )\int _{A_{k}}\varphi \,d\mu \geq (1-\epsilon )\int _{E}\varphi \,d\mu -\int _{A-A_{k}}\varphi \,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b0f75a4dc357926d11a2f6011a4b73dfb7326c5)
Combining these inequalities gives that
![{\displaystyle \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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d4ca09a271678ced83c7379ffef7db9e2757b39)
Hence, sending ε to 0 and taking the liminf in n, we get that
![{\displaystyle \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu \geq \int _{E}\varphi \,d\mu ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/323e67920e302a3a2bcbd28b59421e23104e9523)
completing the proof.
Proof using the monotone convergence theorem
For every natural number k define pointwise the function
![{\displaystyle g_{k}=\inf _{n\geq k}f_{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/763f11d653adb23c7c69f432292ff633bd76af1a)
Then the sequence g1, g2, . . . of functions is increasing, meaning that gk ≤ gk+1 for all k, and converges pointwise to the limit inferior f.
For all k ≤ n we have gk ≤ fn, so that by the monotonicity of the integral
![{\displaystyle \int _{E}g_{k}\,d\mu \leq \int _{E}f_{n}\,d\mu ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd2ab5c704b565c675c72c2606264a1a92d51d7f)
hence
![{\displaystyle \int _{E}g_{k}\,d\mu \leq \inf _{n\geq k}\int _{E}f_{n}\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c67e6f98f1eba3c8852a1dc10b6181bada2fe0df)
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
![{\displaystyle \int _{E}f\,d\mu =\lim _{k\to \infty }\int _{E}g_{k}\,d\mu \leq \lim _{k\to \infty }\inf _{n\geq k}\int _{E}f_{n}\,d\mu =\liminf _{n\to \infty }\int _{E}f_{n}\,d\mu \,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa10878d8280f250d5e41ed472d30170d2ae437e)
Examples for strict inequality
Equip the space
with the Borel σ-algebra and the Lebesgue measure.
![{\displaystyle f_{n}(x)={\begin{cases}n&{\text{for }}x\in (0,1/n),\\0&{\text{otherwise.}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9ef78e41f0775a011a6b86c3673fd1fa0bc2b78)
![{\displaystyle f_{n}(x)={\begin{cases}{\frac {1}{n}}&{\text{for }}x\in [0,n],\\0&{\text{otherwise.}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf508e0cccac351160fdf49494d4f3711f15edfe)
These sequences
converge on
pointwise (respectively uniformly) to the zero function (with zero integral), but every
has integral one.
A counterexample
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{cases}-{\frac {1}{n}}&{\text{for }}x\in [n,2n],\\0&{\text{otherwise.}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/032ca178360ef3bcbb8ef9ee885942640791cad1)
This sequence converges uniformly on S to the zero function (with zero integral) and for every x ≥ 0 we even have fn(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 fn has integral −1, hence the inequality in Fatou's lemma fails.
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 an integrable function g on S such that fn ≤ g for all n, then
![{\displaystyle \limsup _{n\to \infty }\int _{S}f_{n}\,d\mu \leq \int _{S}\limsup _{n\to \infty }f_{n}\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/320399f58d1dbca4a04450838cf793e1e3910ac4)
Note: Here g integrable means that g is measurable and that
.
Proof
Apply Fatou's lemma to the non-negative sequence given by g – fn.
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 a non-negative integrable function g on S such that fn ≥ −g for all n, then
![{\displaystyle \int _{S}\liminf _{n\to \infty }f_{n}\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/77e10458fe89fce087b4791a9bfb10c1368ee396)
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
![{\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu \,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e362c44a24a65c12d0564610460b1059c40c9cce)
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
![{\displaystyle \lim _{k\to \infty }\int _{S}f_{n_{k}}\,d\mu =\liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cb0bf97d89a5f9520683116d9b13c5c47b036bf)
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)
![{\displaystyle \mu _{n}(E)\to \mu (E),~\forall E\in \Sigma .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23092924e00955a0993dcf4f11c7342df54c60e0)
Then, with fn non-negative integrable functions and f being their pointwise limit inferior, we have
![{\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu _{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9db14c2da494e8b8fea065007c2a34c6588b0a29)
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
![{\displaystyle \int _{E}f\,d\mu \leq \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu _{n}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76ef5707e07e000e63df0f18f797f75a5a96d6f8)
Let
.
Then μ(E-K)=0 and
![{\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} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be4dc99af3cdf404351037a90dd0141cfe86cdf1)
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
![{\displaystyle \int _{E}\phi \,d\mu =\lim _{n\to \infty }\int _{E}\phi \,d\mu _{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9f9b6277fb529b1750f5678ef5cb3b1237fb3d4)
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
![{\displaystyle \int _{E}\phi \,d\mu \leq \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4c8761fa8555fb368c566ff53c31d84f65e1829)
Let a be the minimum non-negative value of φ. Define
![{\displaystyle A=\{x\in E|\phi (x)>a\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd8af16ce5e15d47dc63f37e6a20ae150e20a57b)
We first consider the case when .
We must have that μ(A) is infinite since
![{\displaystyle \int _{E}\phi \,d\mu \leq M\mu (A),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48c1c7e345aee9b4fc30ec1ffcf69a8fb8860d0e)
where M is the (necessarily finite) maximum value of that φ attains.
Next, we define
![{\displaystyle A_{n}=\{x\in E|f_{k}(x)>a~\forall k\geq n\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bf1c4742450624afac34f585de959bd26fa28ab)
We have that
![{\displaystyle A\subseteq \bigcup _{n}A_{n}\Rightarrow \mu (\bigcup _{n}A_{n})=\infty .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5592c7f5a9c7f7e7567dcf85fa747872b1fde8e)
But An is a nested increasing sequence of functions and hence, by the continuity from below μ,
.
Thus,
.
At the same time,
![{\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 ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/748244e721e9e9bf4a6fc97d994686666a818887)
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
![{\displaystyle A_{n}=\{x\in E|f_{k}(x)>(1-\epsilon )\phi (x)~\forall k\geq n\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db48a41407009d31fe47e1483aa1f0fd59309599)
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),
![{\displaystyle \lim _{n\rightarrow \infty }\mu (A-A_{n})=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51664d9bae31d3616de3896e0149f139e4bffbfa)
Thus, there exists n such that
![{\displaystyle \mu (A-A_{k})<\epsilon ,~\forall k\geq n.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adfafb6220ef3f26e0b540f15d4ab58c9c8f8e72)
Therefore, since
![{\displaystyle \lim _{n\to \infty }\mu _{n}(A-A_{k})=\mu (A-A_{k}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3008516787bde00c7209e44d582321f4575ffc0c)
there exists N such that
![{\displaystyle \mu _{k}(A-A_{k})<\epsilon ,~\forall k\geq N.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f4a86d6ce313338d90a29c5ef269b8a238cd24e)
Hence, for
![{\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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75bbdf70e0b78022bb064700c69ee371c1ef9f89)
At the same time,
![{\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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f8766a0dd5e78f8f11985861052b079b509b233)
Hence,
![{\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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d379cd21c3d715cde28cf75a584227e123c8565)
Combining these inequalities gives that
![{\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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea96021fdca30b990eff9e6fa8903a12a1a27102)
Hence, sending ε to 0 and taking the liminf in n, we get that
![{\displaystyle \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{k}\geq \int _{E}\phi \,d\mu ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a6f13b83c00ce540dbfdaed2b0840a233a179f4)
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
; 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
and let
be a sub-σ-algebra. Then
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
![{\displaystyle Y_{k}=\inf _{n\geq k}X_{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82d611840b4aa4a052f03f91c7dd9492c85c25da)
Then the sequence Y1, Y2, . . . is increasing and converges pointwise to X.
For k ≤ n, we have Yk ≤ Xn, so that
almost surely
by the monotonicity of conditional expectation, hence
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
![{\displaystyle {\begin{aligned}\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}}]\\&\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13169cd09a4372a20eb250f3d617c3623bfacabc)
Let X1, X2, . . . be a sequence of random variables on a probability space
and let
be a sub-σ-algebra. If the negative parts
![{\displaystyle X_{n}^{-}:=\max\{-X_{n},0\},\qquad n\in {\mathbb {N} },}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09ee73d6bf62f88e5666a2a7450140a204aea8d8)
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
![{\displaystyle X:=\liminf _{n\to \infty }X_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c9821ab7e72c280f8228ab8252609f0b6271e04)
satisfies
![{\displaystyle \mathbb {E} [\max\{X,0\}\,|\,{\mathcal {G}}]=\infty ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11b354af5729d85d5b7f584bac0edc000ed0f6e6)
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} {\bigl [}X_{n}^{-}1_{\{X_{n}^{-}>c\}}\,|\,{\mathcal {G}}{\bigr ]}<\varepsilon \qquad {\text{for all }}n\in \mathbb {N} ,\,{\text{almost surely}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ca3bf9b4f2b18c99ab5fb278b2fa046d649f839)
Since
![{\displaystyle X+c\leq \liminf _{n\to \infty }(X_{n}+c)^{+},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83e30e216ad910f475896bbaa83a6134613920a3)
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
![{\displaystyle (X_{n}+c)^{+}=(X_{n}+c)+(X_{n}+c)^{-}\leq X_{n}+c+X_{n}^{-}1_{\{X_{n}^{-}>c\}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a26f5c992d59d067621b3a4cdb83edeaf821a28)
we have
almost surely,
hence
almost surely.
This implies the assertion.
References
- Royden, H.L. (1988). Real Analysis (3rd ed.).
External links