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 the French mathematician Pierre Fatou (1878 - 1929).
Fatou's lemma can be used to prove the Fatou–Lebesgue theorem and Lebesgue's dominated convergence theorem.
tandard statement of Fatou's lemma
If "f"1, "f"2, . . . is a sequence of non-negative measurable functions defined on a measure space ("S","Σ","μ"), then:
Note: On the left-hand side the limit inferior of the "f""n" is taken pointwise. The functions are allowed to attain the value infinity and the integrals may also be infinite.
Proof
Fatou's lemma is here proved using the monotone convergence theorem (it can be proved directly).
Let "f" denote the limit inferior of the "f""n". For every natural number "k" define pointwise the function:Then the sequence "g"1, "g"2, . . . is increasing and converges pointwise to "f".For "k" ≤ "n", we have "g""k" ≤ "f""n", so that:hence:Using the monotone convergence theorem, the last inequality, and the definition of the limit inferior, it follows that:
Examples for strict inequality
Equip the space with the Borel σ-algebra and the Lebesgue measure.
* Example for a probability space: Let denote the unit interval. For every natural number define:
* Example with uniform convergence: Let denote the set of all real numbers. Define :
These sequences converge on pointwise (respectively uniform) to the zero function (with zero integral), but every has integral one.
A counterexample
A suitable assumption concerning the negative parts of the sequence "f"1, "f"2, . . . 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 (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 "f"1, "f"2, . . . 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 "f""n" ≤ "g" for all "n", then:
Note: Here "g integrable" means that "g" is measurable and that