Some Math

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.

SDEs, Martingale Problems and Elliptic Operators

Cauchy’s criterion on almost sure convergence

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 {p}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: {X_n} is Cauchy in probability if {P(|X_m-X_n|>\epsilon)\rightarrow 0} as {m, n\rightarrow\infty}; {X_n} is Cauchy a.s. if {X_n} is Cauchy everywhere except a set of measure zero; {X_n} is Cauchy in {L^p} if {\mathop{\mathbb E} |X_n-X_m|^p\rightarrow 0} as {m,n\rightarrow\infty}. Now we give some basic facts about the advantages of using Cauchy’s criterion.

Proposition 1 {X_n\rightarrow X} a.s. if and only if {P(\sup_{k\geq n}|X_k-X|\geq\epsilon)\rightarrow 0} for every {\epsilon>0} as {n\rightarrow\infty}.

Proof: Let us consider the formulation of the set of points where {X_n\not\rightarrow X}; that is

\displaystyle \{\omega: X_n(\omega)\not\rightarrow X(\omega)\}=\cup_{\epsilon>0}\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}

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

\displaystyle \{\omega: X_n(\omega)\not\rightarrow X(\omega)\} =\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}

So

\displaystyle \begin{array}{rcl} X_n\xrightarrow{a.s.} X & \Leftrightarrow P(\{\omega: X_n(\omega)\not\rightarrow X(\omega)\}=0\\ & \Leftrightarrow P\left(\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}\right)=0\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq1/j\}\right)=0\;\;\forall j\in{\mathbb N}\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)=0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\cup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{k\geq n}\{\omega:|X_k(\omega)-X_n(\omega)|\geq\epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+ \end{array}

\Box

Proposition 2 {X_n} is Cauchy a.s. if and only if {P(\sup_{\substack{k\geq n\\l\geq n}}|X_k-X_l|\geq \epsilon)\rightarrow 0} as {n\rightarrow\infty} for every {\epsilon>0}; or equivalently {P(\sup_{k\geq 0}|X_{n+k}-X_n|\geq\epsilon)\rightarrow 0} as {n\rightarrow\infty} for every {\epsilon>0}.

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

\displaystyle \begin{array}{rcl} X_n \text{ is Cauchy a.s. }&\Leftrightarrow P(\{\omega: X_n \text{ is not Cauchy}\})=0\\ & \Leftrightarrow P\left(\cup_{j=1}^\infty\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq 1/j\}\right)=0\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq 1/j\}\right)=0\;\;\forall j\in{\mathbb N}\\ & \Leftrightarrow P\left(\cap_{n=1}^\infty\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)=0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\cup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{\substack{k\geq n\\l\geq n}}\{\omega:|X_k-X_l|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+\\ & \Leftrightarrow P\left(\sup_{\substack{k\geq 0}}\{\omega:|X_{n+k}-X_n|\geq \epsilon\}\right)\rightarrow 0\;\;\forall \epsilon\in{\mathbb R}^+ \end{array}

where the last equivalence is obtain from the following inequality

\displaystyle \sup_{k\geq 0}|X_{n+k}-X_n|\leq\sup_{\substack{k\geq 0\\l\geq 0}}|X_{n+k}-X_{n+l}|\leq 2\sup_{k\geq 0}|X_{n+k}-X_n|

This completes the proof. \Box

Remark: Similarly one can formulate another equivalent definition of Cauchy a.s., that is if {P(\sup_{m<k\leq n}|X_k-X_m|>\epsilon)\rightarrow 0} for all {\epsilon>0} as {n,m\rightarrow\infty}, then {X_n} is Cauchy a.s. Let {\eta>0}. Then there is an integer {N, M} such that {P(\sup_{m<k\leq n}|X_k-S_m|>\epsilon)<\eta} whenever {n>N, m>M}. Sending {N\rightarrow\infty} gives {P(\sup_{k>m}|X_k-X_m|>\epsilon)<\eta} which is just the previous proposition.

Notes: Probability by Durrett

I will update the notes constantly as I go.

Notes on probability

Study Notes

I will constantly update the following notes as I study.

Notes on Topology

Notes on Algebra

Partial fractions and infinite product

1. Partial fractions

Theorem 1 (Mittag-Leffler) Let {(a_n)_{n=1}^\infty} be a sequence of complex numbers with {\lim_{n\rightarrow\infty}a_n=\infty} and let {p_n(\cdot)} be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points {a_n} and the corresponding singular parts {p_n\left(\frac{1}{z-a_n}\right)}. Moreover, the most general meromorphic function of this kind can be written in the form

\displaystyle f(z)=\sum_{n=1}^\infty\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]+g(z)

where {q_n(\cdot)}‘s are suitably chosen polynomials and {g(\cdot)} is analytic in the whole plane.

Proof: The idea is to subtract a certain portion of analytic part of {p_n\left(\frac{1}{z-a_n}\right)} to ensure convergence without adding any poles. For each {n\in{\mathbb N}}, the singular part {p_n\left(\frac{1}{z-a_n}\right)} is analytic in the ball {B(0,\frac{|a_n|}{2})}. So we can obtain Taylor expansion {q_n(\cdot)} of {p_n\left(\frac{1}{z-a_n}\right)} around the origin up to the degree {d_n}. Using estimate for the remainder term, we have

\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum_{d_n+1}^\infty\frac{M_n}{(|a_n|/2)^k}|z|^k=M_n\left(\frac{2|z|}{|a_n|}\right)^{d_n+1}\frac{1}{1-\frac{|z|}{|a_n|/2}}=M_n\frac{\beta_n^{d_n+1}}{1-\beta_n}

where {\beta_n=\frac{2|z|}{|a_n|}} and {M_n} denotes the maximum of {|p_n(\cdot)|} in {|z|\leq\frac{|a_n|}{2}}. In particular, this estimate implies that

\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2M_n\beta_n^{d_n+1}\text{ whenever }|z|\leq\frac{|a_n|}{4}

For each {n}, we may pick {d_n} large enough to ensure {2M_n\beta_n^{d_n+1}\leq 2^{-n}}, which can be done by making {\frac{M_n}{2^{d_n}}\leq 2^{-n}}. In doing so, we obtain

\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2^{-n}\text{ whenever }|z|\leq\frac{|a_n|}{4}

Given {R> 0}, we shall show that the series converges uniformly in {|z|\leq R}. We may split the series {\sum_{n=1}^\infty\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|} into two categories: the first one with {|a_n|\leq 4R } and the second one with {|a_n|>4R}. Then {\sum_{|a_n|\leq 4R}\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]} represents a meromorphic function with poles at {a_n} because it is a finite sum. In the latter category, since {|a_n|>4R}, we have {|z|\leq\frac{|a_n|}{4}} in the disk.Thus above estimate implies

\displaystyle \sum_{|a_n|> 4R}\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum 2^{-n}

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

\displaystyle f(z)=\frac{\pi^2}{\sin^2\pi z}

Recall that {\sin\pi z=\pi z-\frac{(\pi z)^3}{3!}+\cdots} and so {\sin^2\pi z =(\pi z)^2-\frac{2(\pi z)^4}{3!}+\cdots}. We then can work out a first few terms in the Laurant expansion at the origin which is

\displaystyle f(z)=\frac 1 {z^2}-\frac{2\pi^2}{3!}+\text{higher order terms}

Thus the singular part of {f} at 0 is {\frac 1z^2}. Moreover, as {f} has period 1 namely {f(z+n)=f(z)} for all {n{\mathbb N}}, we can see that the singular part at integers are {\frac{1}{(z-n)^2}}. Clearly {\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}} converges on all compact subsets which do not contain integers. It follows that we may write

\displaystyle f(z)=\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}+g(z)

where {g(z)} is entire. Write {z=x+iy} we see that {\frac{1}{(z-n)^2}=\frac{1}{(x-n)^2-y^2+2iy(x-n)}}. From the uniform convergence of the series, we can see that {\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}} converges to 0 as {|y|\rightarrow 0}. Using the idendity

\displaystyle |\sin \pi z|^2=\cosh^2\pi y-\cos^2\pi x

we see that {f(z)\rightarrow 0} uniformly as {|y|\rightarrow 0}. In particular, this implies {g(z)} is bounded in the strip {0\leq x\leq 1}. By periodicity, {g(z)} is bounded in the entire complex plane. Then Liouville’s theorem says the {g(z)} reduces to a constant. In this case, {g(z)} must vanish as the limit is zero. Therefore, we have the identity

\displaystyle \frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}

Integrate on both sides, we obtain that

\displaystyle -\pi\cot\pi z=\sum\frac{-1}{z-n}

In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of {\frac{1}{z-n}}. In this case, subtracting all constant term is enough. That is {\sum_{n=-N}^N\frac{1}{z-n}+\frac{1}{n}=\sum_{n=-N}^N\frac{z}{n(z-n)}} which is comparable to {\frac 1 {n^2}}. Thus we have

\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac{1}{n}\right)+c

We may bracket terms of {n} and {-n} together, and obtain

\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\frac{2z}{z^2-n^2}+c

Now {c=0} is necessary because both sides are odd.

Next we exploit this fact to investigate the sum

\displaystyle \lim_{m\rightarrow\infty}\sum_{-m}^m\frac{(-1)^n}{z-n}=\frac 1z+\sum_{n=1}^\infty(-1)^n\frac{2z}{z^2-n^2}

We may split into odd terms and even terms

\displaystyle \sum_{-2k-1}^{2k+1}\frac{(-1)^n}{z-n}=\sum_{n=-k}^k\frac{1}{z-2n}-\sum_{n=-k-1}^{k}\frac{1}{z-1-2n}

Clearly we know that {\frac{\pi }{2}\cot\frac{\pi z}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-2n}} and {\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-1-2n}}. Thus in the limit we have

\displaystyle \frac{\pi }{2}\cot\frac{\pi z}{2}-\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\frac{\pi}{\sin\pi z}

2. Product representation

An infinite product of complex numbers

\displaystyle p_1p_2\cdots p_n=\prod_{n=1}^\infty p_n

is evaluated by taking limits of the partial products {P_n=p_1\cdots p_n}. WLOG, we say that {P_n} 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 {p_n=\frac{P_n}{P_{n-1}}}, we see that a necessary condition is that {p_n\rightarrow 1}. Thus WLOG, we may write

\displaystyle \prod_{n=1}^\infty(1+a_n)

so that {a_n\rightarrow 0} is a necessary condition for convergence. If no factor is zero, it is natural to compare with

\displaystyle \sum_{n=1}^\infty\log(1+a_n)

We then have the following result:

Lemma 2 The infinite product {\prod_{n=1}^\infty(1+a_n)} converges simultaneously with the series {\sum_{n=1}^\infty\log(1+a_n)} whose terms represent the values of the principal branch of the logarithm.

Proof: Suppose {\sum_{n=1}^\infty\log(1+a_n)} converges. Denote the partial sum {S_n}. Then clearly {P_n=e^{S_n}} and by continuity of {\exp} we have {P_n\rightarrow e^S}. \Box

Lemma 3 The product {\prod_{n=1}^\infty (1+a_n)} converges absolutely if and only if {\sum_{n=1}^\infty |a_n|} converges.

Proof: Recall that {\lim_{z\rightarrow 0}\frac{\lim(1+z)}{z}=1}. Then if either the product of the series converges, we see that {a_n\rightarrow 0}. Then we employ the following inequality

\displaystyle (1-\epsilon)|a_n|\leq |\log(1+a_n)|\leq (1+\epsilon)|a_n|

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

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

\displaystyle f(z)=z^me^{g(z)}\prod_{n=1}^\infty \left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}

where the product is taken over all {a_n\neq 0}, {g(z)} is an entire function.

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

\displaystyle \begin{array}{rcl} \log\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}& = \log(1-z/a_n)+\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}\\ & =-\sum_{k=m_n+1}^\infty\frac{(z/a_n)^{k}}{k} \end{array}

By taking {m_n=n}, we see that for {|z|\leq R}, the series {\sum_{k=n+1}^\infty\frac{(z/a_n)^{k}}{k}\leq \sum_{k=n+1}^\infty\left(\frac{R}{|a_n|}\right)^{k}<\infty} as we may take {|a_n|} sufficiently large enough to make {\frac{R}{|a_n|}<1}. \Box

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

Proof: Suppose {f(z)} is meromorphic in the whole plane. Let {g(z)} be an entire function with poles of {f(z)} for zeros. Then {h(z)=f(z)g(z)} is an entire function. Then {f(z)=\frac{h}{g}} as desired. \Box The product representation becomes more interesting if it is possible to pick a uniform {m_n=h}. The previous proof shows that the product

\displaystyle \prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}

converges to an entire function if the series {\sum_{n=1}^\infty\frac{1}{h+1}\left(\frac{R}{a_n}\right)^{h+1}} converges for all {R\in{\mathbb R}^+}. That is to say {\sum_{n=1}^\infty\frac{1}{|a_n|^{h+1}}<\infty}. Assume that {h} is the smallest integer for which this series converges; the previous product expression is called the canonical product associated with the sequence {(a_n)_{n=1}^\infty} and {h} is the genus of the canonical product. If the exponent {g(z)} in the product representation of an entire function {f(z)} with

\displaystyle z^me^{g(z)}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}

reduces to a polynomial, then {f} is of finite genus. The genus of {f} is equal to {\max(h,\deg(g))}. For instance, an entire function of genus zero is of the form

\displaystyle Cz^m\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)

with {\sum_{n=1}^\infty 1/|a_n|<\infty}. The canonical representation of an entire function of genus 1 is either of the form

\displaystyle Cz^me^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)e^{\frac{z}{a_n}}

with {\sum_{n=1}^\infty\frac{1}{|a_n|^2}<\infty, \sum_{n=1}^\infty\frac{1}{|a_n|}=\infty}; or of the form

\displaystyle Cz^m e^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)

with {\sum_{n=1}^\infty 1/|a_n|<\infty}.

Next we discuss an example: {\sin\pi z}. It has zeros at {z=\pm n}. Since {\sum\frac 1n=\infty }adn {\sum\frac 1{n^2}<\infty}, we must take {h=1} so that

\displaystyle \sin\pi z=ze^{g(z)}\prod_{n\neq 0}\left(1-\frac{z}{n}\right)e^{\frac{z}{n}}

In order to determine {g(z)} we form logarithmic derivative on both sides and find

\displaystyle \pi\cot\pi z=\frac{1}{z}+g'(z)+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac 1n\right)=g'(z)+\pi\cot\pi z

This shows that {g'(z)=0} and {g(z)} reduces to a constant. Since {\frac{\sin\pi z}{z}\rightarrow \pi} as {z\rightarrow 0}, we must have {e^{g(z)}=\pi} identically. Hence we can write

\displaystyle \sin\pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)

3. Hadamard product

Recall that an entire function is said to of order {\lambda} is {\lambda} is the smallest number such that

\displaystyle M(r)\leq e^{r^{\lambda+\epsilon}}

for all {\epsilon>0} as soon as {r} is sufficiently large. Here {M(r)} denotes the maximum modulus of {f(z)} on {|z|\leq r}. The genus and the order are closely related as seen by the following theorem.

Theorem 6 The genus {h} and the order {\lambda} of an entire function satisfies

\displaystyle h\leq \lambda\leq h+1

Proof: Suppose that {f} is an entire function of order {\lambda}. First we claim that {\sum_{n=1}^\infty|a_n|^{-(\lambda+\epsilon)}<\infty} for any given {\epsilon>0}. This will address the issue of convergence of the canonical product. Given {r>0}. Recall Jensen’s formula

\displaystyle \log|f(0)|=\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|+\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|d\theta

and the fact

\displaystyle \int_0^{r}\frac{n(x)}{x}dx=-\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|

Apply these formulae with {2r} and we will get

\displaystyle \begin{array}{rcl} n(r)\log 2&=\int_r^{2r}\frac{n(r)dx}{x}\\ & \leq \int_0^{2r}\frac{n(x)dx}{x}\\ & \leq \frac{1}{2\pi}\int_0^{2\pi}\log|f(2re^{i\theta})|d\theta-\log|f(0)| \end{array}

This implies {n(r)r^{-\lambda-\epsilon}\leq \frac{2}{\log 2}r^{\lambda-\lambda-\epsilon}-\frac{\log|f(0)|}{\log 2}r^{-\lambda-\epsilon}\rightarrow 0} as {r\rightarrow\infty}. Moreover, note we also have the relation

\displaystyle k\leq n(|z_k|)<|a_n|^{\lambda+\epsilon}

This implies

\displaystyle \sum\frac{1}{|a_n|^{h+1}}\leq\sum\frac{1}{n^{\frac{h+1}{\lambda+\epsilon}}}

Pick {\epsilon>0} so that {\lambda+\epsilon<h+1} will show that {\sum_{n=1}^\infty|a_n|^{-h-1}<\infty}. \Box

Hadamard products

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 {(a_n)_{n=1}^\infty}, we say the product {\prod_{n=1}^\infty(1+a_n)} converges if the limit {\lim_{N\rightarrow\infty}\prod_{n=1}^\infty(1+a_n)} exists.

Theorem 1 If {\sum_{n=1}^\infty|a_n|<\infty}, then {\prod_{n=1}^\infty(1+a_n)} converges. Moreover, the product converges to 0 if and only if one of its factors is 0.

Proof: WLOG, we may assume that {|a_n|\leq \frac 12} for all {n\in{\mathbb N}}. Thus we may use power series to define logarithm {\log(1+a_n)} so that {1+a_n=e^{\log(1+a_n)}}. Then we obtain the following partial products

\displaystyle \prod_{n=1}^N(1+a_n)=e^{\sum_{n=1}^N\log(1+a_n)}

For each {n\in{\mathbb N}}, we make the estimate

\displaystyle |\log(1+a_n)|\leq\sum_{k=1}^\infty\frac{|a_n|^k}{k}\leq \sum_{k=1}^\infty|a_n|(1+\frac 12+\frac 14+\cdots)\leq 2|a_n|

Using continuity of {e}, we obtain the first result. To prove the second assertion, we note that if {1+a_n\neq 0} for all {0}, then clearly the products converges to {e^{\text{something}}} which is nonzero. \Box Similarly we have convergence theorem for functions.

Theorem 2 Suppose {(F_n)_{n=1}^\infty} is a sequence of holomorphic functions in an open set {\Omega}. If for all {z\in\Omega} we have

\displaystyle |1-F_n(z)|\leq c_n\text{ and }\sum_{n=1}^\infty c_n<\infty

then {\prod_{n=1}^\infty F_n(z)} converges to a holomorphic function. Moreover, if {F_n} does not vanish in {\Omega} for any {n\in{\mathbb N}}, then

\displaystyle \frac{F'(z)}{F(z)}=\sum_{n=1}^\infty\frac{F_n'(z)}{F_n(z)}

2. Weisterass product

Theorem 3 Given any sequence {(a_n)_{n=1}^\infty} of complex numbers with {|a_n|\rightarrow\infty} as {n\rightarrow\infty}, there exists an entire function {f} that vanishes at all {z=a_n} and nowhere else. Any other entire function is of the form {f(z)e^{g(z)}} where {g} is entire.

In general the function {\prod_{n=1}^\infty(1-z/a_n)} may not converge, so we need to insert extra factors to make it converge without adding any zeros. For each integer {k\geq 0}, we define {E_0=(1-z)} and {E_k(z)=(1-z)e^{\sum_{n=1}^k\frac{z^k}{n}}}. Then we prove the following estimate:

Lemma 4 If {|z|\leq\frac 12}, then {|1-E_k(z)|\leq c|z|^{k+1}} for some constant {c>0}.

Proof: Since {|z|\leq \frac 12}, we obtain the taylor expansion {\log(1-z)=-\sum_{n=1}^\infty \frac{z^n}{n}}. Thus we have

\displaystyle E_k(z)=e^{\log(1-z)+\sum_{n=1}^k\frac{z^n}{n}}=e^{-\sum_{n=k+1}^\infty\frac{z^n}{n}}=e^w

Clearly {|w|\leq|z|^{n+1}|1+z^2+z^3+\cdots|\leq 2|z|^{n+1}}. In particular, {|w|\leq 1}, thus

\displaystyle |1-E_k(z)|=|1-e^w|\leq c_1|w|\leq 2c_1|z|^{n+1}

This provides a desired bound. \Box Proof: Define {h(z)=\prod_{k=0}^\infty E_k(z/a_n)}. Fix {R>0}, we will show that {h(z)} converges uniformly on {|z|\leq R}. Decompose {h(z)=\prod_{|a_n|\leq 2R} E_k(z/a_n)\prod_{|a_n|>2R}E_n(z/a_n)} and we only need to consider the latter factor. Clearly {|z/a_n|\leq \frac 12} for all {|z|\leq R} and all {|a_n|\geq 2R} and by previous lemma we see that the product converges uniformly in {|z|\leq R}. Moreover, they never vanish in {|z|\leq R} which implies the limit does not vanish also. Hence the function {z^m\prod_{n=1}^\infty E_n(z/a_n)} provides the desired function. \Box

3. Hadamard product

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

Theorem 5 Let {\Omega} be an open set contains the closure of a disk {D_R} and suppose {f} is holomorphic in {\Omega}, {f(0)\neq 0}, and never vanishes on the circle {C_R}. If {z_1,...,z_N} denote the zeros of {f} inside the disc, then

\displaystyle \log|f(0)|=\sum_{k=1}^N\log\left(\frac{|z_n|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|d\theta

Proof: First note that if the theorem holds true for {f_1} and {f_2} then it holds true for {f_1f_2}. Consider {g(z)=\frac{f(z)}{(z-a_1)\cdots(z-a_N)}} which is holomorphic function in {D_R} that vanishes nowhere. So we only need to show that the theorem is true for {g} and {z-w}. Since {g} has no zeros we need to show {\log|g(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|g(Re^{i\theta})|d\theta}. This is clearly true by mean-value property of the harmonic function {\log|g(z)|}. It remains to show

\displaystyle \begin{array}{rcl} \log|w|&=\log\left(\frac{|w|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|Re^{i\theta}-w|d\theta\\ & = \log\left(|w|\right)+\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-\frac{w}{R}|d\theta\\ \end{array}

This boils down to show {\frac{1}{2\pi}\int_0^{2\pi}\log|e^{i\theta}-a|d\theta=0} for any {|a|<1}. Making change of variables {\theta\rightarrow-\theta} shows this is equivalent to {\frac{1}{2\pi}\int_0^{2\pi}\log|e^{-i\theta}-a|d\theta=\frac{1}{2\pi}\int_0^{2\pi}\log|1-ae^{i\theta}|d\theta=0}. Consider the function {F(z)=1-az} which is holomorphic and vanishes nowhere in the unit disk. Thus applying mean-value property to {\log|F(z)|} we have

\displaystyle 0=\log|F(z)|=\frac{1}{2\pi}\int_0^{2\pi}\log|F(e^{i\theta})|d\theta=\frac{1}{2\pi}\int_0^{2\pi}\log|1-e^{i\theta}|d\theta

\Box Next we claim that if {f(0)\neq 0} and {f} does not vanish on the circle {C_R}, then

\displaystyle \int_0^R\frac{n(r)}{r}dr=\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|d\theta-\log|f(0)|

This is immediate if we prove

Lemma 6 If {z_1,...,z_N} are the zeros of {f} inside {D_R} then

\displaystyle \int_0^R\frac{n(r)}{r}dr=-\sum_{k=1}^N\log\left(\frac{|z_k|}{R}\right)

Proof: We start from the right-hand side.

\displaystyle -\sum_{k=1}^N\log\left(\frac{|z_k|}{R}\right)=-\sum_{k=1}^N\int_{R}^{|z_k|}\frac{dr}{r}=\sum_{k=1}^N\int_{|z_k|}^{R}\frac{dr}{r}=\int_{0}^{R}\sum_{k=1}^N1_{(|z_k|,R]}(r)\frac{dr}{r}=LHS

\Box

4. Functions of finite order

Let {f} be an entire function. We call {f} has order of growth {\leq\rho} if

\displaystyle |f(z)|\leq Ae^{B|z|^\rho}\text{ }\forall z\in\mathbb C

In particular, the order of {f} is {\rho_0=\inf \rho}.

Theorem 7 If {f} is an entire function that has order of growth {\leq\rho}, then {n(r)\leq Cr^\rho} for sufficiently large {r}. Moreover if {z_1, z_2, ...} are zeros of {f}, then {\sum_{n=1}^\infty|a_n|^{-s}} converges for any {s>\rho}.

Proof: We shall use Jensen’s formula. Note that {n(R)} is an increasing function of {R}. Now choose {R=2r}, then clearly we have

\displaystyle \int_{r}^{2r}\frac{n(x)dx}{x}\geq n(r)\int_{r}^{2r}\frac{dx}{x}=n(r)\log 2

By Jensen’s formula we have

\displaystyle \begin{array}{rcl} n(r)&\leq\frac{1}{\log 2}\int_0^{2r}\frac{n(x)dx}{x}\\ & \leq \frac{1}{2\pi\log 2}\int_0^{2\pi}\log|f(2re^{i\theta})|d\theta\\ & \leq \frac{1}{2\pi\log 2}\int_0^{2\pi}\log Ae^{Br^\rho}d\theta\leq Cr^\rho \end{array}

This proves the first part. Then we directly estimate

\displaystyle \begin{array}{rcl} \sum_{|a_k|\geq 1}^\infty|a_n|^{-s}&=\sum_{j=0}^\infty\sum_{2^j\leq|a_n|\leq 2^{j+1}}|a|^{-s}\\ &\leq \sum_{j=0}^\infty n(2^{j+1})2^{-js}\\ & \leq C\sum_{j=0}^\infty 2^{\rho(j+1)-js}\\ &\leq C'\sum_{j=0}^\infty 2^{(\rho-s)j}<\infty \end{array}

\Box

5. Hadamard products

Theorem 8 Suppose {f} is entire and has order of growth {\rho_0}. Let {k} be the integer such that {k\leq \rho_0<k+1}. If {a_1, a_2,...} are zeros of {f}, then

\displaystyle f(z)=z^m e^{g(z)}\prod_{n=1}^\infty E_k(z/a_n)

where {g(z)} is a polynomial of degree {\leq k}, {m} is the order of zero at {0}.

Proof: The convergence of the canonical products is now easily solved. Using Lemma 2.2, we have {|1-E_k(z/a_n)|\leq c_1|z/a_n|^{k+1}\leq c_2|a_n|^{-k-1}}. Thus the convergence of {\sum_{n=1}^\infty |a_n|^{-k-1}} implies the convergence of the product. In order to show that {g(z)} is a polynomial of degree {\leq k}, we need more efforts.

Lemma 9 We have the following estimates

\displaystyle |E_k(z)|\geq e^{-c|z|^{k+1}}\text{ if }|z|\leq\frac 12

\displaystyle |E_k(z)|\geq |1-z| e^{-c'|z|^{k}}\text{ if }|z|\geq\frac 12

Proof of Lemma. If {|z|\leq\frac 12}, we directly have

\displaystyle \begin{array}{rcl} |E_k(z)|&= |e^{\log(1-z)+\sum_{n=1}^k\frac{z^k}{n}}|\\ & = |e^{-\sum_{n=k+1}^\infty\frac{z^k}{n}}|\\ & \geq e^{-|\sum_{n=k+1}^\infty\frac{z^k}{n}|}\\& \geq e^{-c|z|^{k+1}} \end{array}

If {|z|\geq \frac 12}, then

\displaystyle \begin{array}{rcl} |E_k(z)|&=|1-z|||e^{\sum_{n=1}^k\frac{z^k}{n}}|\\ &\geq|1-z| e^{-\sum_{n=1}^k\frac{|z|^k}{n}}\\ &\geq |1-z|e^{-c|z|^k} \end{array}

Lemma 10 For any {\rho_0<s<k+1}, we have

\displaystyle \prod_{n=1}^\infty|E_k(z/a_n)|\geq e^{-c|z|^s}

except possibly for {z\in B_{|a_n|^{-k-1}}(a_n)}.

Proof of Lemma. Again we use usual decomposition

\displaystyle \begin{array}{rcl} \prod_{n=1}^\infty|E_k(z/a_n)|=\prod_{|a_n|\leq 2|z|}|E_k(z/a_n)|\prod_{|a_n|>2|z|}|E_k(z/a_n)| \end{array}

We first estimate the second product. Note that {|z/a_n|\leq \frac 12} for all {n\in{\mathbb N}}. By previous lemma, we have

\displaystyle \begin{array}{rcl} \prod_{|a_n|>2|z|}|E_k(z/a_n)|&\geq \prod_{|a_n|>2|z|}e^{-c|z/a_n|^{k+1}}\\ & = e^{-c|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-k-1}} \end{array}

Note that {|a_n|^{-k-1}=|a_n|^{-s}|a_n|^{s-k-1}\leq C|a_n|^{-s}|z|^{s-k-1}} for some constant {C>0} as {|a_n/z|^{s-k-1}\leq 2^{s-k-1}\leq C}. Thus we have

\displaystyle e^{-c|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-k-1}}\geq e^{-c'|z|^{k+1}\sum_{|a_n|>2|z|}|a_n|^{-s}|z|^{s-k-1}}\geq e^{-c''|z|^s}

because {\sum |a_n|^{-s}<\infty}. To estimate the first prouct we use previous lemma to obtain

\displaystyle \prod_{|a_n|\leq 2|z|}|E_k(z/a_n)|\geq \prod_{|a_n|\leq 2|z|}|1-z/a_n|e^{-c|z/a_n|^{k}}=|1-z/a_n|e^{-c|z|^k\sum_{|a_n|\leq 2|z|}|a_n|^{-k}}

Using a similar trick we have {|a_n|^{-k}=|a_n|^{-s}|a_n|^{s-k}\leq C|a_n|^{-s}|z|^{s-k}}. Thus

\displaystyle \prod_{|a_n|<2|z|}|E_k(z/a_n)|\geq \prod_{|a_n|\leq 2|z|}|1-z/a_n|e^{-c|z|^s}

It remains to estimate {|1-z/a_n|}. But using the property that {|z-a_n|\geq |a_n|^{-k-1}}, we have

\displaystyle |1-z/a_n|=|\frac{a_n-z}{a_n}|\geq \frac{|a_n|^{-k-1}}{|a_n|}=|a_n|^{-k-2}

Now we consider

\displaystyle \begin{array}{rcl} (k+2)\sum\log|a_n|&\leq (k+1)n(2|z|)\log 2|z|\leq C|z|^{s} \end{array}

This completes the proof of the lemma.

Lemma 11 There exists a sequence of complex numbers {(r_m)_{m=1}^\infty} with {|r_m|\rightarrow\infty} such that

\displaystyle |E_k(z/a_n)|\geq e^{-c|z|^s}\;\forall |z|=r_m

Proof of Lemma. First we note that {\sum|a_n|^{-k-1}} converges so we may take {\sum_{N}^\infty|a_n|<\frac {1}{ 10}}. For each positive integer {L} we may find {L\leq r\leq L+1} such that the the disk {|z|=r} does not intersect the forbidden areas in previous lemma. Suppose it is not the case, then there is a particular {L_0} such that all balls of radius {r} with {L_0\leq r<L_0+1} 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 {[L_0, L_0+1]\subset\cup_n [|a_n|-|a_n|^{-k-1},|a_n|+|a_n|^{-k-1}]}. This implies {2\sum_{N}^{\infty}|a_n|^{-k-1}\geq 1} which is a contradiction.

Now we prove the Hadamard’s theorem. Let {E(z)=z^m\prod_{n=1}^\infty E_k(z/a_n)}. Previously we’ve shown that {E(z)} converges to a holomorphic which has zeros as {f} does and does not vanish everywhere else. Thus we have {\frac{f(z)}{E(z)}=e^{g(z)}} and in particular

\displaystyle e^{\Re g(z)}=|e^{g(z)}|=\left|\frac{f(z)}{E(z)}\right|\leq \frac{A_1e^{B_1|z|^s}}{A_2e^{B_2|z|^{-s}}}=Ae^{B|z|^s}\text{ when }|z|=r_m

This implies {\Re g(z)\leq B|z|^s} on {|z|=r_m}. Finally we show that this is enough to guarantee that {g} is a polynomial of degree less than {s}.

Lemma 12 Suppose {g} is entire and {u=\Re g} satisfies

\displaystyle u(z)\leq Cr^s \text{ whenever }|z|=r

for a sequence of positive real numbers {r} that tends to infinity. Then {g} is a polynomial of degree {\leq s}.

Proof of Lemma. First we write {g(z)=\sum_{n=0}^\infty a_n z^n}. Let {C_r} denote the circle centerer at the origin with radius {r}. By Cauchy’s integral formula, we have {a_n=\frac{1}{2\pi i}\int_{C_r}\frac{g(\zeta)}{\zeta^{n+1}}d\zeta} whenver {n\geq 0}. This implies

\displaystyle a_n=\frac{1}{2\pi i}\int_0^{2\pi}\frac{g(re^{i\theta})ire^{i\theta}}{r^{n+1}e^{i(n+1)\theta}}d\theta=\frac{1}{2\pi r^n}\int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta\text{ whenever }n\geq 0

and

\displaystyle \int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta=0 \text{ whenever }n<0

which further implies

\displaystyle 0=\overline{\int_0^{2\pi }\frac{g(re^{i\theta})}{e^{in\theta}}d\theta}=\int_0^{2\pi}\overline{g(re^{i\theta})}e^{-in\theta}d\theta\text{ whenever }n\geq 0

Thus

\displaystyle a_n=\frac{1}{2\pi r^n}\int_0^{2\pi}(g(re^{i\theta})+\overline{g(re^{i\theta})})e^{-in\theta}d\theta=\frac{1}{\pi r^n}\int_0^{2\pi} u(re^{i\theta})e^{-in\theta}d\theta

Recall that {\int_0^{2\pi}e^{-in\theta}d\theta=0} whenever {n> 0}. Thus for {n\geq 1}, we have

\displaystyle \begin{array}{rcl} |a_n|&=\left|\frac{1}{\pi r^n}\int_0^{2\pi}(u(re^{i\theta})-Cr^s)e^{-in\theta}d\theta\right|\\ & \leq \frac{1}{\pi r^n}\int_0^{2\pi}|u(re^{i\theta}-Cr^s)|d\theta\\ & = \frac{1}{\pi r^n}\int_0^{2\pi}(Cr^s-u(re^{i\theta}))d\theta\\ & =2Cr^{s-n}-\frac{2u(0)}{r^n} \end{array}

Letting {r\rightarrow\infty} shows that {a_n=0} if {n>s}. Hence {g} is a polynomial of degree {\leq s}. \Box

Cardinality

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

  1. It is reflexive: {A\sim A}.
  2. It is symmetric: if {A\sim B}, then {B\sim A}.
  3. It is transitive: if {A\sim B} and {B\sim C}, then {A\sim C}.

Definition: For any positive integer {n}, let {J_{n}} be the set whose elements are the integers, {1,2,...,n}; let {J} be the set of all positive integers. For any set {A}, we say:

  1. (a) {A} is finite if {A\sim J_{n}} for some positive integer {n}.
  2. (b) {A} is infinite if {A} is not finite.
  3. (c) {A} is countable if {A\sim J}.
  4. (d) {A} is uncountable if {A} is neither finite nor countable.
  5. (d) {A} is at most countable if {A} 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): {A} is infinite if {A\sim B} for some {B\subset A}.

Theorem 1 Every infinite subset of a countable set {A} is countable.

Proof: Suppose {B\subseteq A} and {B} is infinite. Arrange the elements of {A} which are all distinct in a sequence {(a_{i})_{i=1}^{\infty}}. Then we construct a sequence {(n_{k})_{k=1}^{\infty}} as follows:

Let {n_{1}\in J} be the smallest number such that {a_{n_{1}}\in B}. Note {n_{1}} exists because {\mathbb{N}} is well-ordered. After choosing {n_{1},...,n_{k-1}}, we let {n_{k}} be the smallest number greater than {n_{k-1}} such that {a_{n_{k}}\in B}.

Then consider the function {f: J\rightarrow B} where {f(k)=a_{n_{k}}}. We then check {f} is indeed a bijection. It is one-to-one because {(a_{i})_{i=1}^{\infty}} is a distinct sequence; then for any {i\neq j}, we have {a_{n_{i}}\neq a_{n_{j}}}. {f} must be surjective for otherwise, there exists {b\in B} is not in the image of {(a_{i})_{i=1}^{\infty}} contradicting the fact that {B\subseteq A}.

\hfill\Box

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

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

\hfill\Box

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

Proof: Let {A} be at most countable and {B\subseteq A} is countable. If {A} is countable then we are done. So we need to prove that {A} cannot be finite. Suppose {A} is finite then there exists a bijection {f: A\rightarrow J_{n}}. Let {g: J\rightarrow B}. Then the composition {f\circ g:\mathbb{N}\rightarrow J_{n}} must be injective. However, by pigeonhole’s principle we know this cannot happen, thus a contradiction.

\hfill\Box

Theorem 4 Countable unions of countable sets is countable.

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

Let {\left\{X_{\alpha}|\alpha\in A\right\}} be an arbitrary family of sets. We define the Cartesian product {\prod_{\alpha\in A}X_{\alpha}} to be all the functions {\phi: A\rightarrow\bigcup_{\alpha\in A}X_{\alpha}} such that {\phi(\alpha)\in X_{\alpha}}. We can think of {\phi(\alpha)} as an evaluation of {\phi} at the {\alpha}-th entry. For example if {A=\mathbb{N}}, {\phi(1)} is simply the first entry of the Cartesian product.

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

Theorem 7 Countable Cartesian products of countable sets is uncountable.

Proof: Let {S:=\prod_{n=1}^{\infty} A_{n}} where {A_{n}} is countable for each {n\in J}. Let {(f_{i}: i\in J)} be an enumeration. For each {n\in J} we pick two different element {a_{n}, b_{n}\in A_{n}}. Consider the following function:

\displaystyle f(j) = \left\{ \begin{array}{lr} b_{j} & : \mathrm{if}\, f_{j}(j)=a_{j}\\ a_{j} & : \mathrm{otherwise} \end{array} \right.

Note {f\in S} obviously. However, note {f\neq f_{i}} for all {i\in J} because {f} differs from every {f_{i}} at least at {f_{j}(j)}. Thus {S} in not countable.

\hfill\Box

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

Proof: Remember the set {A} is indeed identified with the Cartisian product {\left\{0,1\right\}^{\mathbb{N}}}. Although we cannot directly apply theorem 7 because {\left\{0,1\right\}} is finite, we still can use the same Cantor trick. Let {(f_{i}: i\in J)} be an enumeration of {A}. We define the following function

\displaystyle f(j) = \left\{ \begin{array}{lr} 0 & : \mathrm{if}\, f_{j}(j)=1\\ 1 & : \mathrm{otherwise} \end{array} \right.

Again {f\in A} obviously, but {f} differs from {f_{i}} for every {i\in J}. Thus {A} is uncountable.

\Box

Theorem 9 There exists no surjection from a set {X} to {\mathcal{P}(X)}.

Proof: We prove by contradiction. Let {\varphi: X\rightarrow\mathcal{P}(X)} be a surjection. By definition, for any {A\in\mathcal{P}(X)} there exists {a\in X} such that {\varphi(x)=A}. Consider the following set

\displaystyle Y:=\left\{x\in X| x\notin\varphi(x)\right\}

So {Y} has an pre-image in {X}. Let {y\in X} such that {\varphi(y)=Y}. It is either {y\in Y} or {y\notin Y}. If it is the first case, then {y\in X} and {y\notin\varphi(y)=Y} by definition, but his contradicts the fact that {y\in Y}. If it is the second case, we have {y\notin Y}, so {y\in X} and {y\in\varphi(y)=Y}, contradiction again. Hence {\varphi} cannot be a surjection and the proof is complete.

\hfill\Box

Littlewood’s three principles (3)

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 {\left\{f_{n}\right\}_{i=1}^{\infty}} is a sequence of measurable functions defined on a measurable set {E} with {m(E)<\infty}, and assume {f_{i}\rightarrow f} almost everywhere on {E}. Given {\epsilon>0}, we can find a closed set {A_{\epsilon}\subseteq E} such that {m(E-A_{\epsilon})\leq\epsilon} and {f_{i}\rightarrow f} uniformly on {E}.

Proof: We may assume {f_{i}\rightarrow f} for every {x\in E} since we can always find a subset {E'\subseteq E} which differs with {E} by a measure zero set such that the convergence holds everywhere on {E'}, then we just rename {E'} by {E}. Define

\displaystyle E_{k}^{n}:=\left\{x\in E:\left|f_{j}(x)-f(x)\right|<\frac{1}{n}\;\forall j>k\right\}

Note for fixed {n} we have the relation {E_{k}^{n}\subseteq E_{k+1}^{n}} and {E_{k}^{n}\nearrow E} as {k\rightarrow\infty}. Then by the continuity of measures we know {\lim_{k\to \infty}m(E_{k}^{n})=m(E)} which implies we can find an integer {k_{n}\in\mathbb{N}} such that {m(K-E_{k_{n}}^{n})<\frac{1}{2^{n}}}. By construction, we have {\left|f_{j}(x)-f(x)\right|<\frac{1}{n}} whenever {j>k_{n}} and {x\in E_{k}^{n}}. Then we choose {N\in\mathbb{N}} so that {\sum_{n=N}^{\infty}<\frac{\epsilon}{2}} and let {\hat{A}_{\epsilon}:=\bigcap_{n\geq N}E_{k}^{k_{n}}}.
First we notice the relation {m(E-\hat{A}_{\epsilon})=m(\bigcup_{n\geq N}E-E_{k}^{k_{n}})\leq\sum_{n=N}^{\infty}(E-E_{k}^{k_{n}}<\frac{\epsilon}{2})}. Next for any {\delta>0}, we can choose {N'\in\mathbb{N}} such that {\frac{1}{n}<\delta} when {n>N'}. Now we indeed have completed our job as for any {\delta>0} we can find an integer {k_{n}} such that for all {j>k_{n}} we \left|f_{j}(x)-f(x)\right|<\delta<\frac{1}{n} for all {x\in\hat{A}_{\epsilon}}.

Finally, as a technical requirement, we can find a closed subset {A_{\epsilon}\subseteq\hat{A}_{\epsilon}} such that {m(\hat{A}_{\epsilon}-A_{\epsilon})<\frac{\epsilon}{2}} which leads us to the desired property {m(E-A_{\epsilon})<\epsilon}.                                                                                                                                                                           \Box

Littlewood’s three principles (2)

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 {\epsilon>0} there exists a closed set {F_{\epsilon}} with

\displaystyle F_{\epsilon}\subseteq E\;\mathrm{and}\;m(E-F_{\epsilon})\leq\epsilon

and such that {f|_{F_{\epsilon}}} is continuous.

Proof: Since by assumption, {f} is measurable and finite valued on E, we can find a sequence of step functions, {f_{n}} such that {f_{n}\rightarrow f} almost everywhere. Note that each {f_{n}} 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 {n} we can find sets {E_{n}} with {m(E_{n})\leq\frac{1}{2^{n}}} so that {f_{n}} is continuous outside {E_{n}}. 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 {A_{\frac{\epsilon}{3}}} with {m(E-A_{\frac{\epsilon}{3}})\leq\frac{\epsilon}{3}} such that {f_{n}\rightarrow f} uniformly on {A_{\frac{\epsilon}{3}}}. Then for the integer N such that {\sum_{n\geq N}\frac{1}{2^{n}}\leq\frac{\epsilon}{3}}, we consider

\displaystyle F':=A_{\frac{\epsilon}{3}}-\bigcup_{n\geq N} E_{n}

For every {n\geq N}, the step function {f_{n}} is continuous on {F'}; therefore, the uniform limit {f} is also continous on {F'}. To complete the proof, notice that {F'} is also measurable so we can approximate from below by a closed set {F_{\epsilon}} such that {m(F'-F_{\epsilon})\leq\frac{\epsilon}{3}} which gives us the desired {m(E-F_{\epsilon})\leq\epsilon}.

A Mild Glimpse Into Measure Theory (2)

3. Measures and Outer measures

 

Let {(E,\mathscr{M})} be a measurable space.

Definition: A measure on {(E,\mathscr{M})} is a function {\mu:\mathscr{M}\rightarrow [0,\infty]} such that:
(1) {\mu(\emptyset)=0}
(2) For all families {\left\{A_{n}\right\}_{n\in\mathbb{N}}} of disjoint measurable sets,

\displaystyle \mu(\bigcup_{n\in\mathbb{N}} A_{n})=\sum_{n\in\mathbb{N}}\mu(A_{n})

Remarks:

  • {\mu(A)=+\infty} is allowed
  • Property (2) is called {\sigma}-additivity
  • We can also derive {\mu(\bigcup_{n=1}^{k} A_{n})=\sum_{n=1}^{k}\mu(A_{n})}

 

Proposition 6 : Let {A,B\in\mathscr{M}}, we have the following:
(1) If {A\subseteq B}, then {\mu(A)\leq\mu(B)}.
(2) If {\mu(A)<+\infty} and {A\subseteq B}, then {\mu(A\setminus B)=\mu(A)-\mu(B)}
(3) {\mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B)}
(4) If {A_{n}\in\mathscr{M}} for all {n\in\mathbb{N}} and {A_{n}\subseteq A_{n+1}} for all {n\in\mathbb{N}}, then {\mu(\bigcup_{n\in\mathbb{N}}A_{n})=\lim_{n\rightarrow\infty}\mu(A_{n})}.
(5) If {B_{n}\in\mathscr{M}} for all {n\in\mathbb{N}} and {B_{n}\supseteq B_{n+1}} for all {n\in\mathbb{N}}, then {\mu(\bigcap_{n\in\mathbb{N}}B_{n})=\lim_{n\rightarrow\infty}\mu(B_{n})}.
(6) If {A_{n}\in\mathscr{M}}, then {\mu(\bigcup_{n\in\mathbb{N}}A_{n})\leq\sum_{n\in\mathbb{N}}A_{n}}.

To be updated.