### Master’s Thesis

The following link includes a draft of my master’s thesis. It is still in progress and may contain some typos and errors.

The following link includes a draft of my master’s thesis. It is still in progress and may contain some typos and errors.

In this article, we will investigate interplays between various kinds of convergence of sequence of random variables. Specifically, we will discuss convergence a.s., convergence in probability, convergence in th mean and convergence in distribution.

We can naturally formalize the notions of sequence of random variables being *Cauchy* in these contexts of convergence. We say: is *Cauchy in probability* if as ; is *Cauchy a.s.* if is Cauchy everywhere except a set of measure zero; is *Cauchy in * if as . Now we give some basic facts about the advantages of using Cauchy’s criterion.

Proposition 1a.s. if and only if for every as .

*Proof:* Let us consider the formulation of the set of points where ; that is

In fact, we can use the denseness of real numbers to write

So

Proposition 2is Cauchy a.s. if and only if as for every ; or equivalently as for every .

*Proof:* Using a similar argument, we obtain a chain of equivalences,

where the last equivalence is obtain from the following inequality

This completes the proof.

**Remark:** Similarly one can formulate another equivalent definition of Cauchy a.s., that is if for all as , then is Cauchy a.s. Let . Then there is an integer such that whenever . Sending gives which is just the previous proposition.

**1. Partial fractions**

Theorem 1 (Mittag-Leffler)Let be a sequence of complex numbers with and let be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points and the corresponding singular parts . Moreover, the most general meromorphic function of this kind can be written in the form

where ‘s are suitably chosen polynomials and is analytic in the whole plane.

*Proof:* The idea is to subtract a certain portion of analytic part of to ensure convergence without adding any poles. For each , the singular part is analytic in the ball . So we can obtain Taylor expansion of around the origin up to the degree . Using estimate for the remainder term, we have

where and denotes the maximum of in . In particular, this estimate implies that

For each , we may pick large enough to ensure , which can be done by making . In doing so, we obtain

Given , we shall show that the series converges uniformly in . We may split the series into two categories: the first one with and the second one with . Then represents a meromorphic function with poles at because it is a finite sum. In the latter category, since , we have in the disk.Thus above estimate implies

This completes the proof. Let us try to investigate several standard examples. Consider the function

Recall that and so . We then can work out a first few terms in the Laurant expansion at the origin which is

Thus the singular part of at 0 is . Moreover, as has period 1 namely for all , we can see that the singular part at integers are . Clearly converges on all compact subsets which do not contain integers. It follows that we may write

where is entire. Write we see that . From the uniform convergence of the series, we can see that converges to 0 as . Using the idendity

we see that uniformly as . In particular, this implies is bounded in the strip . By periodicity, is bounded in the entire complex plane. Then Liouville’s theorem says the reduces to a constant. In this case, must vanish as the limit is zero. Therefore, we have the identity

Integrate on both sides, we obtain that

In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of . In this case, subtracting all constant term is enough. That is which is comparable to . Thus we have

We may bracket terms of and together, and obtain

Now is necessary because both sides are odd.

Next we exploit this fact to investigate the sum

We may split into odd terms and even terms

Clearly we know that and . Thus in the limit we have

**2. Product representation**

An infinite product of complex numbers

is evaluated by taking limits of the partial products . WLOG, we say that converges if and only if at most a finite number of the factors are zero and if the partial products formed by the nonvanishing factors tend to a finite nonzero limit. By writing , we see that a necessary condition is that . Thus WLOG, we may write

so that is a necessary condition for convergence. If no factor is zero, it is natural to compare with

We then have the following result:

Lemma 2The infinite product converges simultaneously with the series whose terms represent the values of the principal branch of the logarithm.

*Proof:* Suppose converges. Denote the partial sum . Then clearly and by continuity of we have .

Lemma 3The product converges absolutely if and only if converges.

*Proof:* Recall that . Then if either the product of the series converges, we see that . Then we employ the following inequality

to derive the result. Next we prove Weierstrass products formula.

Theorem 4There exists an entire function with prescribed zeros provided in the infinitely many zeros case that . Every entire function with these zeros and no other zeros can be written in the form

where the product is taken over all , is an entire function.

*Proof:* First we shall show that the product converges. This is direct because

By taking , we see that for , the series as we may take sufficiently large enough to make .

Corollary 5Every function which is meromorphic in the whole plane is the quotient of two entire function.

*Proof:* Suppose is meromorphic in the whole plane. Let be an entire function with poles of for zeros. Then is an entire function. Then as desired. The product representation becomes more interesting if it is possible to pick a uniform . The previous proof shows that the product

converges to an entire function if the series converges for all . That is to say . Assume that is the smallest integer for which this series converges; the previous product expression is called the **canonical product** associated with the sequence and is the **genus** of the canonical product. If the exponent in the product representation of an entire function with

reduces to a polynomial, then is of finite genus. The genus of is equal to . For instance, an entire function of genus zero is of the form

with . The canonical representation of an entire function of genus 1 is either of the form

with ; or of the form

with .

Next we discuss an example: . It has zeros at . Since adn , we must take so that

In order to determine we form logarithmic derivative on both sides and find

This shows that and reduces to a constant. Since as , we must have identically. Hence we can write

**3. Hadamard product**

Recall that an entire function is said to of order is is the smallest number such that

for all as soon as is sufficiently large. Here denotes the maximum modulus of on . The genus and the order are closely related as seen by the following theorem.

Theorem 6The genus and the order of an entire function satisfies

*Proof:* Suppose that is an entire function of order . First we claim that for any given . This will address the issue of convergence of the canonical product. Given . Recall Jensen’s formula

and the fact

Apply these formulae with and we will get

This implies as . Moreover, note we also have the relation

This implies

Pick so that will show that .

In this post, we will discuss the product representation of entire functions. In order to do so, we first discuss the convergence of infinite products.

**1. Infinite products**

Given a sequence of complex numbers , we say the product converges if the limit exists.

Theorem 1If , then converges. Moreover, the product converges to 0 if and only if one of its factors is 0.

*Proof:* WLOG, we may assume that for all . Thus we may use power series to define logarithm so that . Then we obtain the following partial products

For each , we make the estimate

Using continuity of , we obtain the first result. To prove the second assertion, we note that if for all , then clearly the products converges to which is nonzero. Similarly we have convergence theorem for functions.

Theorem 2Suppose is a sequence of holomorphic functions in an open set . If for all we have

then converges to a holomorphic function. Moreover, if does not vanish in for any , then

**2. Weisterass product**

Theorem 3Given any sequence of complex numbers with as , there exists an entire function that vanishes at all and nowhere else. Any other entire function is of the form where is entire.

In general the function may not converge, so we need to insert extra factors to make it converge without adding any zeros. For each integer , we define and . Then we prove the following estimate:

Lemma 4If , then for some constant .

*Proof:* Since , we obtain the taylor expansion . Thus we have

Clearly . In particular, , thus

This provides a desired bound. *Proof:* Define . Fix , we will show that converges uniformly on . Decompose and we only need to consider the latter factor. Clearly for all and all and by previous lemma we see that the product converges uniformly in . Moreover, they never vanish in which implies the limit does not vanish also. Hence the function provides the desired function.

**3. Hadamard product**

In order to refine the result, we need some elementary results. First tool is Jensen’s formula.

Theorem 5Let be an open set contains the closure of a disk and suppose is holomorphic in , , and never vanishes on the circle . If denote the zeros of inside the disc, then

*Proof:* First note that if the theorem holds true for and then it holds true for . Consider which is holomorphic function in that vanishes nowhere. So we only need to show that the theorem is true for and . Since has no zeros we need to show . This is clearly true by mean-value property of the harmonic function . It remains to show

This boils down to show for any . Making change of variables shows this is equivalent to . Consider the function which is holomorphic and vanishes nowhere in the unit disk. Thus applying mean-value property to we have

Next we claim that if and does not vanish on the circle , then

This is immediate if we prove

Lemma 6If are the zeros of inside then

*Proof:* We start from the right-hand side.

**4. Functions of finite order**

Let be an entire function. We call has order of growth if

In particular, the order of is .

Theorem 7If is an entire function that has order of growth , then for sufficiently large . Moreover if are zeros of , then converges for any .

*Proof:* We shall use Jensen’s formula. Note that is an increasing function of . Now choose , then clearly we have

By Jensen’s formula we have

This proves the first part. Then we directly estimate

**5. Hadamard products**

Theorem 8Suppose is entire and has order of growth . Let be the integer such that . If are zeros of , then

where is a polynomial of degree , is the order of zero at .

*Proof:* The convergence of the canonical products is now easily solved. Using Lemma 2.2, we have . Thus the convergence of implies the convergence of the product. In order to show that is a polynomial of degree , we need more efforts.

Lemma 9We have the following estimates

Proof of Lemma.If , we directly have

If , then

Lemma 10For any , we have

except possibly for .

*Proof of Lemma. *Again we use usual decomposition

We first estimate the second product. Note that for all . By previous lemma, we have

Note that for some constant as . Thus we have

because . To estimate the first prouct we use previous lemma to obtain

Using a similar trick we have . Thus

It remains to estimate . But using the property that , we have

Now we consider

This completes the proof of the lemma.

Lemma 11There exists a sequence of complex numbers with such that

*Proof of Lemma.* First we note that converges so we may take . For each positive integer we may find such that the the disk does not intersect the forbidden areas in previous lemma. Suppose it is not the case, then there is a particular such that all balls of radius with will intersect the forbidden area. In particular, by rotating the those balls on the real line so that the diameter will lie in the positive real axis, we see that . This implies which is a contradiction.

Now we prove the Hadamard’s theorem. Let . Previously we’ve shown that converges to a holomorphic which has zeros as does and does not vanish everywhere else. Thus we have and in particular

This implies on . Finally we show that this is enough to guarantee that is a polynomial of degree less than .

Lemma 12Suppose is entire and satisfies

for a sequence of positive real numbers that tends to infinity. Then is a polynomial of degree .

*Proof of Lemma. *First we write . Let denote the circle centerer at the origin with radius . By Cauchy’s integral formula, we have whenver . This implies

and

which further implies

Thus

Recall that whenever . Thus for , we have

Letting shows that if . Hence is a polynomial of degree .

**Definition:** If there exists a 1-1 mapping of onto , we say that and can be put in *1-1 correspondence *, or that and have the same *cardinal number *, or briefly, that and are equivalent, and we write . This relation clearly has the following three properties:

- It is
*reflexive*: . - It is
*symmetric*: if , then . - It is
*transitive*: if and , then .

**Definition:** For any positive integer , let be the set whose elements are the integers, ; let be the set of all positive integers. For any set , we say:

- (a) is
*finite*if for some positive integer . - (b) is
*infinite*if is not finite. - (c) is
*countable*if . - (d) is
*uncountable*if is neither finite nor countable. - (d) is
*at most countable*if is finite or countable.

**Remark:** A finite set cannot be equivalent to one of its proper subsets. However, this is possible for infinite sets. In fact, we could restate the definition (b): is *infinite* if for some .

Theorem 1Every infinite subset of a countable set is countable.

*Proof:* Suppose and is infinite. Arrange the elements of which are all distinct in a sequence . Then we construct a sequence as follows:

Let be the smallest number such that . Note exists because is well-ordered. After choosing , we let be the smallest number greater than such that .

Then consider the function where . We then check is indeed a bijection. It is one-to-one because is a distinct sequence; then for any , we have . must be surjective for otherwise, there exists is not in the image of contradicting the fact that .

\hfill

Corollary 2Every subset of an at most countable set is at most countable.

*Proof:* Let be an at most countable set and . If is finite, then is must also be finite. So let us assume is countable. Then cannot be uncountable by theorem 1.

\hfill

Corollary 3An at most countable set that contains a countable subset is countable.

*Proof:* Let be at most countable and is countable. If is countable then we are done. So we need to prove that cannot be finite. Suppose is finite then there exists a bijection . Let . Then the composition must be injective. However, by pigeonhole’s principle we know this cannot happen, thus a contradiction.

\hfill

Theorem 4Countable unions of countable sets is countable.

Corollary 5At most countable unions of at most countable sets is at most countable.

Let be an arbitrary family of sets. We define the Cartesian product to be all the functions such that . We can think of as an evaluation of at the -th entry. For example if , is simply the first entry of the Cartesian product.

Theorem 6Finitely many Cartesian products of countable sets is countable.

Theorem 7Countable Cartesian products of countable sets is uncountable.

*Proof:* Let where is countable for each . Let be an enumeration. For each we pick two different element . Consider the following function:

Note obviously. However, note for all because differs from every at least at . Thus in not countable.

\hfill

Corollary 8Let be the set of all sequences whose elements are the digis 0 and 1. Then is uncountable.

*Proof:* Remember the set is indeed identified with the Cartisian product . Although we cannot directly apply theorem 7 because is finite, we still can use the same Cantor trick. Let be an enumeration of . We define the following function

Again obviously, but differs from for every . Thus is uncountable.

Theorem 9There exists no surjection from a set to .

*Proof:* We prove by contradiction. Let be a surjection. By definition, for any there exists such that . Consider the following set

So has an pre-image in . Let such that . It is either or . If it is the first case, then and by definition, but his contradicts the fact that . If it is the second case, we have , so and , contradiction again. Hence cannot be a surjection and the proof is complete.

\hfill

Finally the principle states that every convergent sequence nearly converges uniformly. Let’s figure out what the word “nearly” means in this context and this brings us to the Egorov’s theorem.

Theorem 1(Egorov)Suppose is a sequence of measurable functions defined on a measurable set with , and assume almost everywhere on . Given , we can find a closed set such that and uniformly on .

*Proof:* We may assume for every since we can always find a subset which differs with by a measure zero set such that the convergence holds everywhere on , then we just rename by . Define

Note for fixed we have the relation and as . Then by the continuity of measures we know which implies we can find an integer such that . By construction, we have whenever and . Then we choose so that and let .

First we notice the relation . Next for any , we can choose such that when . Now we indeed have completed our job as for any we can find an integer such that for all we \left|f_{j}(x)-f(x)\right|<\delta<\frac{1}{n} for all .

Finally, as a technical requirement, we can find a closed subset such that which leads us to the desired property .

The second principle asserts that every function is nearly continuous. Now we investigate the meaning of “*nearly*” in terms of mathematics.

Theorem 2(Lusin) Suppose f is measurable and finite valued on E with E of finite measure. Then for every there exists a closed set with

and such that is continuous.

*Proof:* Since by assumption, is measurable and finite valued on E, we can find a sequence of step functions, such that almost everywhere. Note that each is a step function which means it is continuous almost everywhere as it contains only at most countably many discontinuities which have Lebesgue measure 0. So without loss of generality, say for each we can find sets with so that is continuous outside . Now to proceed, we need something extra which is obtained from the third littlewood’s principle, i.e., Egrovff’s theorem. It enables us to find a set with such that uniformly on . Then for the integer N such that , we consider

For every , the step function is continuous on ; therefore, the uniform limit is also continous on . To complete the proof, notice that is also measurable so we can approximate from below by a closed set such that which gives us the desired .

**3. Measures and Outer measures**

Let be a measurable space.

**Definition:** A * measure* on is a function such that:

(1)

(2) For all families of disjoint measurable sets,

**Remarks:**

- is allowed
- Property (2) is called -additivity
- We can also derive

Proposition 6: Let , we have the following:

(1) If , then .

(2) If and , then

(3)

(4) If for all and for all , then .

(5) If for all and for all , then .

(6) If , then .

To be updated.