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.
Standard statement In what follows, B R ¯ ≥ 0 {\displaystyle \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}} denotes the σ {\displaystyle \sigma } -algebra of Borel sets on [ 0 , + ∞ ] {\displaystyle [0,+\infty ]} .
Fatou's lemma remains true if its assumptions hold μ {\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 Fatou's lemma does not require the monotone convergence theorem, but the latter can be used to provide a quick and natural proof. A proof directly from the definitions of integrals is given further below.
Via the Monotone Convergence Theorem let g n ( x ) = inf k ≥ n f k ( x ) {\displaystyle \textstyle g_{n}(x)=\inf _{k\geq n}f_{k}(x)} . Then:
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 ) = sup n inf k ≥ n f k ( x ) = sup n g n ( x ) {\displaystyle f(x)=\liminf _{n\to \infty }f_{n}(x)=\sup _{n}\inf _{k\geq n}f_{k}(x)=\sup _{n}g_{n}(x)} , and infima and suprema of measurable functions are measurable we see that f {\displaystyle f} is measurable. By the Monotone Convergence Theorem and property (1), the sup and integral may be interchanged:
∫ X f d μ = ∫ X sup n g n d μ = sup n ∫ X g n d μ = lim inf n → ∞ ∫ X g n d μ ≤ lim inf n → ∞ ∫ X f n d μ , {\displaystyle {\begin{aligned}\int _{X}f\,d\mu &=\int _{X}\sup _{n}g_{n}\,d\mu \\&=\sup _{n}\int _{X}g_{n}\,d\mu \\&=\liminf _{n\to \infty }\int _{X}g_{n}\,d\mu \\&\leq \liminf _{n\to \infty }\int _{X}f_{n}\,d\mu ,\end{aligned}}}
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 and the fact that the functions f {\displaystyle f} and g n {\displaystyle g_{n}} are measurable. 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} .
Now we turn to the main theorem
The proof is complete.
Examples for strict inequality Equip the space S {\displaystyle S} with the Borel σ-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} define
f n ( x ) = { n for x ∈ ( 0 , 1 / n ) , 0 otherwise. {\displaystyle f_{n}(x)={\begin{cases}n&{\text{for }}x\in (0,1/n),\\0&{\text{otherwise.}}\end{cases}}}
Example with uniform convergence: Let S {\displaystyle S} denote the set of all real numbers. Define
f n ( x ) = { 1 n for x ∈ [ 0 , n ] , 0 otherwise. {\displaystyle f_{n}(x)={\begin{cases}{\frac {1}{n}}&{\text{for }}x\in [0,n],\\0&{\text{otherwise.}}\end{cases}}}
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 A suitable assumption concerning the negative parts of the sequence f1, f2, . . . of functions is necessary for Fatou's lemma, as the following example shows. Let S denote the half line [0,∞) with the Borel σ-algebra and the Lebesgue measure. For every natural number n define
f n ( x ) = { − 1 n for x ∈ [ n , 2 n ] , 0 otherwise. {\displaystyle f_{n}(x)={\begin{cases}-{\frac {1}{n}}&{\text{for }}x\in [n,2n],\\0&{\text{otherwise.}}\end{cases}}}
This sequence converges uniformly on S to the zero function and the limit, 0, is reached in a finite number of steps: for every x ≥ 0, if n > x, then 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
lim sup n → ∞ ∫ S f n d μ ≤ ∫ S lim sup n → ∞ f n d μ . {\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 μ < ∞ {\displaystyle \textstyle \int _{S}g\,d\mu <\infty } .
Sketch of proof We apply linearity of Lebesgue integral and Fatou's lemma to the sequence g − f n . {\displaystyle g-f_{n}.} Since ∫ S g d μ < + ∞ , {\displaystyle \textstyle \int _{S}g\,d\mu <+\infty ,} this sequence is defined μ {\displaystyle \mu } -almost everywhere and non-negative.
Extensions and variations of Fatou's lemma
Integrable lower bound Let f 1 , f 2 , … {\displaystyle f_{1},f_{2},\ldots } be a sequence of extended real-valued measurable functions defined on a measure space ( S , Σ , μ ) {\displaystyle (S,\Sigma ,\mu )} . If there exists an integrable function g {\displaystyle g} on S {\displaystyle S} such that f n ≥ − g {\displaystyle f_{n}\geq -g} for all n {\displaystyle n} , then
∫ S lim inf n → ∞ f n d μ ≤ lim inf n → ∞ ∫ S f n d μ . {\displaystyle \int _{S}\liminf _{n\to \infty }f_{n}\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .}
Proof Apply Fatou's lemma to the non-negative sequence given by f n + g {\displaystyle f_{n}+g} .
Pointwise convergence If in the previous setting the sequence f 1 , f 2 , … {\displaystyle f_{1},f_{2},\ldots } converges pointwise to a function f {\displaystyle f} μ {\displaystyle \mu } -almost everywhere on S {\displaystyle S} , then
∫ S f d μ ≤ lim inf n → ∞ ∫ S f n d μ . {\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu \,.}
Proof Note that f {\displaystyle f} has to agree with the limit inferior of the functions f n {\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 f 1 , f 2 , … {\displaystyle f_{1},f_{2},\ldots } converges in measure to a function f {\displaystyle f} .
Proof There exists a subsequence such that
lim k → ∞ ∫ S f n k d μ = lim inf n → ∞ ∫ S f n d μ . {\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 {\displaystyle f} , there exists a further subsequence, which converges pointwise to f {\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 μ {\displaystyle \mu } . Suppose that μ n {\displaystyle \mu _{n}} is a sequence of measures on the measurable space ( M , Σ ) {\displaystyle (M,\Sigma )} such that (see Convergence of measures)
∀ E ∈ F : μ n ( E ) → μ ( E ) {\displaystyle \forall E\in {\mathcal {F}}\colon \;\mu _{n}(E)\to \mu (E)} . Then, with f n {\displaystyle f_{n}} non-negative integrable functions and f {\displaystyle f} being their pointwise limit inferior, we have
∫ S f d μ ≤ lim inf n → ∞ ∫ S f n d μ n . {\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu _{n}.}
Asymptotically uniform integrable functions The following results use the notion asymptotically uniform integrable (a.u.i). A sequence { f n } n ∈ N {\displaystyle \{f_{n}\}_{n\in \mathbb {N} }} of measurable { R ∪ ± ∞ } {\displaystyle \{\mathbb {R} \cup \pm \infty \}} -valued functions is a.u.i with respect to a sequence of measures { μ n } n ∈ N {\displaystyle \{\mu _{n}\}_{n\in \mathbb {N} }} if
lim K → + ∞ lim sup n → ∞ ∫ M | f n ( s ) | I { s ∈ M : | f n ( s ) | ≥ K } μ n ( d s ) = 0 . {\displaystyle \lim _{K\rightarrow +\infty }\limsup _{n\rightarrow \infty }\int _{M}|f_{n}(s)|\mathbf {I} \{s\in M:|f_{n}(s)|\geq K\}\mu _{n}(ds)=0\,.}
Weakly converging measures A sequence of measures { μ n } n ∈ N {\displaystyle \{\mu _{n}\}_{n\in \mathbb {N} }} on a metric space M {\displaystyle M} converges weakly to a finite measure μ {\displaystyle \mu } on M if, for each bounded continuous function f {\displaystyle f} on M {\displaystyle M} ,
∫ M f ( s ) μ n ( d s ) → ∫ M f ( s ) μ ( d s ) as n → ∞ . {\displaystyle \int _{M}f(s)\mu _{n}(ds)\rightarrow \int _{M}f(s)\mu (ds)\quad {\text{as }}n\rightarrow \infty \,.}
Measures with convergence in total variation A sequence of finite measures { μ n } n ∈ N {\displaystyle \{\mu _{n}\}_{n\in \mathbb {N} }} on a measurable space ( M , Σ ) {\displaystyle (M,\Sigma )} converges in total variation to a measure μ {\displaystyle \mu } on ( M , Σ ) {\displaystyle (M,\Sigma )} if
sup { | ∫ M f ( s ) μ n ( d s ) − ∫ M f ( s ) μ ( d s ) | : f : M ↦ [ − 1 , 1 ] is measurable } → 0 as n → ∞ . {\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\}\rightarrow 0\quad {\text{as }}n\rightarrow \infty \,.}
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 ( Ω , 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 Let X1, X2, . . . be a sequence of non-negative random variables on a probability space ( Ω , F , P ) {\displaystyle \scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )} and let
G ⊂ F {\displaystyle \scriptstyle {\mathcal {G}}\,\subset \,{\mathcal {F}}} be a sub-σ-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 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 surely by 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 ]}&=\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}}}
Extension to uniformly integrable negative parts Let X1, X2, . . . be a sequence of random variables on a probability space ( Ω , F , P ) {\displaystyle \scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )} and let
G ⊂ F {\displaystyle \scriptstyle {\mathcal {G}}\,\subset \,{\mathcal {F}}} be a sub-σ-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 ε > 0 there exists a c > 0 such that
E [ X n − 1 { X n − > c } | G ] < ε , for all n ∈ N , almost surely {\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}}} , 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 → ∞