Fatou’s Lemma

In mathematics, Fatou’s lemma is an inequality that demonstrates a relationship between the Lebesgue integral of a sequence of functions’ limit inferior and the limit inferior of integrals of these functions. Pierre Fatou is honoured with the name of the lemma. The definition of Fatou’s lemma, strict inequality examples, reverse Fatou’s lemma and solved problems are all covered in-depth in this article.

Table of Contents:

• Fatou’s Lemma Definition
• Strict Inequality Examples
• Reverse Fatou’s Lemma
• Solved Example
• FAQs

Fatou’s Lemma Definition

Fatou’s Lemma states that “given a measure space (Ω, F, μ), and the set X, that belongs to F. Assume that {f_n} is a sequence of (F, B_R≥0) – measurable non-negative functions f_n: x→[0, +∞]. Now, define the function f: x→[0, +∞] by applying f(x) = lim inf_n→∞ f_n(x), for every x that belongs to X”.

Then f is (F, B_R≥0) – measurable, and

∫_x f dμ ≤ lim inf_n→∞ ∫_x f_n dμ, in which the integrals may be infinite.”

Fatou’s Lemma holds true if its assumptions hold μ, almost everywhere

Note: B_R≥0 represents the σ-algebra of Borel sets on [0, +∞].

Strict Inequality Examples

Connecting the space S with the Lebesgue measure and Borel σ-algebra, we have the following examples of strict inequality.

• Now, let us assume the probability space. Let S = [0, 1] represent the unit interval. Hence, for every natural number, n represents:

• Now, let us take an example of uniform convergence. Assume that S represents the set of all real numbers, and it is defined by

Reverse Fatou’s Lemma

Assume that f₁, f₂, and so on are a sequence of extended real-valued measurable functions defined on a measure space (S, Σ, μ). If a non-negative integrable function g on S exists such that f_n ≤ g for every n, then

lim sup_n→∞ ∫_S f_n dμ ≤ ∫_S lim sup_n→∞ f_n dμ

Solved Example

Example:

Assume that f_n = (1/n)1_[0,n] and f = 0. Prove that f_n converges uniformly to 0, f_n ∈ L¹(R,Mλ, λ) – that is, f_n is integrable with respect to the Lebesgue measure -yet ∫f dλ ≠ lim_n ∫ f_n dλ. Explain in detail why this does not contradict the Monotone Convergence Theorem or the Dominated Convergence Theorem. And also explain Fatou’s Lemma involved?

Solution:

Given,

For each x ∈ R, |fn(x)| ≤(1/n), so if ε > 0,

|fn(x)| ≤ ε for all n ≥ 1/ε

Hence, we can say that f_n→ 0 uniformly on R.

Now, for each n ∈ N, we have

∫f_n dλ = (1/n) λ([0, n]) = 1, such that f_n is integrable.

Hence,

0 = ∫ lim_n f_n dλ ≠ lim_n ∫ f_n dλ = 1

Because the sequence (f_n) is not Monotone, this does not contradict the Monotone Convergence Theorem. Furthermore, f_n(x) = 0 for n ≤ x for each x. When n ≥ x , the value leaps to 1/n > 0 and then drops.

We demonstrate that (f_n) is not dominated by an integrable function to prove that the Dominated Convergence Theorem is not violated. Take a look at this.

|f_n(x)| ≤ g(x), for all x.

Hence, for each n ∈ N, g(x) ≥ 1/n, for n − 1 < x ≤ n. So,

On integrating the above function, it gives

Thus, g can’t be integrable.

Fatou’s lemma must, of course, hold, as we can see.