## IMC 2016 – Day 2 – Problem 7

**Problem 7.** Today, Ivan the Confessor prefers continuous functions satisfying for all . Fin the minimum of over all preferred functions.

## SEEMOUS 2016 Problem 4 – Solution

## Variations on Fatou’s Lemma – Part 2

As we have seen in a previous post, Fatou’s lemma is a result of measure theory, which is strong for the simplicity of its hypotheses. There are cases in which we would like to study the convergence or lower semicontinuity for integrals of the type where converges pointwise to and converges to in some fashion, but not pointwise. For example, we could have that converges to in . In this case we could write the integral as where is the measure defined by . All measures considered in this post will be positive measures.

Certain hypotheses on the measures allow us to find a result similar to Fatou’s Lemma for varying measures. In the following, we define a type of convergence for the measures , named *setwise convergence*, which will allow us to prove the lower semicontinuity result. We say that converges setwise to if for every measurable set . The following proof is taken from Royden, H.L., *Real Analysis*, Chapter 11, Section 4. It is very similar to the proof of Fatou’s lemma given here.

**Theorem A.** Let be a sequence of measures defined on which converges setwise to a measure and a sequence of nonnegative measurable functions which converge pointwise (or almost everywhere in ) to the function . Then

## Variations on Fatou’s Lemma – Part 1

One of the most basic and powerful results of measure theory is Fatou’s Lemma. It provides information on the convergence of the integrals of functions which converge pointwise. The power of this lemma is the simplicity of its hypothesis. One of its interpretations is that the functional which to a positive function associates its integral is lower semicontinuous with respect to the pointwise convergence. Among other things, it allows us to prove the monotone convergence theorem, or Lebesgue’s dominated convergence theorem in just one line.

**Fatou’s Lemma** Suppose is a measure space and is a sequence of integrable functions converging pointwise almost everywhere to . Then is measurable and

*Proof:* One of the classical proofs of this result goes as follows. Using the definition

we note that it is enough to prove that

for every simple function .

Pick and . Suppose . Denote . Note that and . If we denote the support of , we may choose such that for every we have . Thus

Since is bounded above by a constant and , we have

Taking we have

Take and then supremum over to finish the proof. \hfill

**Monotone convergence** Suppose that is a sequence of function defined on such that

for every and . Denote . Then is measurable and

*Proof:* The comes from the monotonicity, while the is just Fatou’s Lemma.

**Dominated convergence** Suppose is a sequence of measurable functions defined on which converges almost everywhere to . Moreover, suppose there exists an integrable function such that for every . Then

*Proof:* Apply Fatou’s lemma to .

It is straightforward to see that the positivity condition can be replaced with a domination from below. If and is integrable, then apply Fatou’s lemma to to deduce that the conclusion still holds for . One may ask what happens with the part. If with integrable, then the reverse Fatou Lemma works. To see this, apply Fatou’s lemma to .

Of course, every one of the few hypotheses made for the Fatou’s lemma are essential. If we allow negative functions, then is a counterexample. converges to zero almost everywhere, while the integrals of are equal to . The inequality can be strict. Consider the similar functions .

Seeing that the hypotheses needed for Fatou lemma to hold are simple enough, we may try and see which generalizations of Fatou lemma hold? One interesting direction is to consider not only variable functions, but also variable measures. This will be treated in a future post.

## Generalized version of Lebesgue dominated convergence theorem

The following variant of the Lebesgue dominated convergence theorem may be useful in the case we can not dominate a sequence of functions by only one integrable function, but by a convergent sequence of integrable functions.

Let be a sequence of measurable functions defined on a measurable set with real values, which converges pointwise almost everywhere to . Suppose that there exists a sequence of positive, integrable functions such that the two following conditions hold:

- for almost every ;
- There exists an integrable function such that in .

Then

## SEEMOUS 2014

**Problem 1.** Let be a nonzero natural number and be a function such that . Let be distinct real numbers. If

prove that is not continuous.

**Problem 2.** Consider the sequence given by

Prove that the sequence is convergent and find its limit.

**Problem 3.** Let and such that , where and is the conjugate matrix of .

(a) Show that .

(b) Show that if then .

**Problem 4.** a) Prove that .

b) Find the limit