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.
Problem 1. Let be a positive integer. Let be a familiy of sets that contains more than half of all subsets of an -element set . Prove that from we can select sets that form a separating family of , i.e., for any two distinct elements of there is a selected set containing exactly one of the two elements.
Problem 2. let and let be non-degenerate subintervals of the interval . Prove that
where the summation is over all pairs of indices such that and are not disjoint.
Problem 3. We have points in the plane, no three of them collinear. The points are colored with two colors. Prove that from the points we can form empty triangles (they have no colored points in their interiors) with pairwise disjoint interiors, such that all points occuring as vertices of the triangles have the same color.
Problem 4. For a positive integer , let be the number of sequences of positive integers such that and for . We make the convention . Let be the unique real number greater than such that . Prove that
- (i) .
- (ii) There exists no number such that .
Problem 5. Let be a non-real algebraic integer of degree two, and let be the set of irreducible elements of the ring . Prove that
Problem 6. Let be a representation of a finite -group over a field of characteristic . Prove that if the restriction of the linear map to a finite dimensional subspace of is injective, then the subspace spanned by the subspaces () is the direct sum of these subspaces.
Problem 7. Lef be a continuous function and let be arbitrary. Suppose that the Minkowski sum of the graph of and the graph of (i.e. the set has Lebesgue measure zero. Does it follow then that the function is of the form , with suitable constants ?
Problem 8. Let be a fixed integer. Calculate the distance
where runs over polynomials of degree less than with real coefficients and runs over functions of the form
defined on the closed interval , where and .
Problem 9. Let , and let be a convex body, i.e. a compact convex set with nonempty interior. Define the barycenter of the body with respect to the weight function by the usual formula
Prove that the translates of the body have pairwise distinct barycenters with respect to .
Problem 10. To each vertex of a given triangulation of the two dimensional sphere, we assign a convex subset of the plane. Assume that the three convex sets corresponding to the three vertices of any two dimensional face of the triangulation have at least one point in common. Show that there exist four vertices such that the corresponding convex sets have at least one point in common.
Problem 11. Let be a random variable that is uniformly distributed on the interval , and let
Show that, as , the limit distribution of is the Cauchy distribution with density function .
This is a cute problem I found this evening.
Suppose is a bijection such that if and , then .
Prove that if then .
Proof: The trick is to divide the pairs of positive integers into families with the same product.
Note that the -th column contains as many elements as the number of divisors of . Now we just just use a simple observation. Let be on the -th column (i.e. ). If then cannot be on one of the first columns. Indeed, the monotonicity property implies . The fact that is a bijection assures us that cover the first columns. Moreover, one element from the -th column is surely covered, namely . This means that
where we have denoted by the number of positive divisors of .
One of the questions I received in the comments to my old post on solving Laplace equation (in fact this is Poisson’s equation) using finite differences was how to apply this procedure on arbitrary domains. It is not possible to use this method directly on general domains. The main problem is the fact that, unlike in the case of a square of rectangular domain, when we have a general shape, the boudary can have any orientation, not only the orientation of the coordinate axes. One way to avoid approach this problem would be using the Finite Element Method. Briefly, you discretize the boundary, you consider a triangulation of the domain with respect to this discretization, then you consider functions which are polynomial and have support in a few number of triangles. Thus the problem is reduced to a finite dimensional one, which can be written as a matrix problem. The implementation is not straightforward, since you need to conceive algorithms for doing the discretization and triangulation of your domain.
One other approach is to consider a rectangle which contains the shape and add a penalization on the exterior of your domain . The problem to solve becomes something like:
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 .
Let be a decreasing sequence of positive numbers such that the series
is uniformly convergent. Then must satisfy .
Note that this result implies that the series is not uniformly convergent on . It is surprisngly similar to the following result:
Suppose that is a decreasing sequence of positive real numbers such that the series is convergence. Then . It is no surprise that the proofs of these two results are similar.
Here’s a nice problem inspired from a post on MathOverflow: link
We call a polygon shrinkable if any scaling of itself with a factor can be translated into itself. Characterize all shrinkable polygons.
It is easy to see that any star-convex polygon is shrinkable. Pick the point in the definition of star-convex, and any contraction of the polygon by a homothety of center lies inside the polygon.