Fatou's lemma

The lemma of Fatou (after Pierre Fatou ) allowed in mathematics, the Lebesgue integral of the limit inferior of a sequence of functions by the limit inferior of the sequence of the corresponding Lebesgue integrals estimate upwards. It thus provides an indication of the interchangeability of limit processes.

Mathematical formulation

Let be a measure space. Non-negative, measurable functions For each result

Where on the left side of the limit inferior of the sequence is to be understood pointwise.

Analogue of this theorem holds also for the superior limit, if there is a non-negative integrable function with:

This can be summarized to the rule of thumb

Idea of ​​proof

To prove the lemma of Fatou for the limit inferior, one turns to the monotonically increasing sequence of functions

The set of the monotone convergence on. With the resulting equation and based on the monotonicity of the integral inequality

Are obtained from the rules for computing the limit:

For the lemma of Fatou with limit superior can be analogous procedure, since, by assumption, is integrable, so is integrable.

Examples of strict inequality

The reason space is respectively provided with the borel σ - algebra, and between the Lebesgue measure.

• Example of a probability space: Let the unit interval. Defining for all, and wherein the indicator function of the interval call.
• Example of uniform convergence: Let the set of real numbers. Define for all and. (Note that there is no integrable majorant in this example, and therefore the sup - part of the lemma of Fatou is not applicable. )

Each has integral one,

Therefore applies

The sequence converges to pointwise to the zero function

Therefore the integral is also zero

Therefore apply here the strict inequalities

Discussion of the conditions

On the condition of non-negativity of the individual functions can not be waived, as the following example shows: Let the half-open interval with the Borel σ - algebra and the Lebesgue measure. Defining for all. The sequence converges to ( even uniform ) against the null function ( with integral 0), but each has integral -1. is therefore