## A little encouragement

If you know who is David Aldous, you will laugh really hard after reading this extract from the report of a referee who rejected his paper. Even big name’s papers can be rejected, so don’t be discouraged young mathematicians

http://www.stat.berkeley.edu/~aldous/least_interesting_paper_ever.txt

## A useful identity in convergence of random series

Let be independent and let Then for any

The identity can be used to prove the following results:

1) P. Levy: Let be independent and let . If exists in probability, then it also exists almost surely.

2) Let be i.i.d. and If in probability, then in probability.

An application of Fourier analysis (in Vietnamese),

Author: Prof. Ngo Bao Chau

## Lecture 3: The outer measure generated by a measure

I’ll post lecture 1 and 2 soon.

In lecture 1 and 2, I defined a measure on a semiring and an outer measure on the power set of . Our goal is that we want to define a measurable set As we recall, a measurable set satifies the following condition: for any Notice that here is an outer measure.

In this lecture, a measure is extended to an outer measure As proved before, the collection of measurable sets with respect to this outer measure is a We will show that Thus, semirings are a basic collections of sets which a measure theory can be built on.

Let a measure space be fixed. The outer measure is defined as follows:

Theorem:is an outer measure.

**Proof: **

i) Let for Then Thus

ii) Let and We consider a sequence of sets in such that Then clearly, Thus, Further, we have This follows that Hence, is monotone.

iii) Let be a sequence of pairwise disjoint sets in If . Then clearly,

Thus we can assume Since each there exists a sequence of sets in such that and Hence, Furthermore,

we have This proves the -subadditivity of

The next theorem will show that is indeed an extension of from to

Theorem:Ifthen

*Proof:*

Let and for Then Thus,

Consider a sequence of sets in satisfying Since is a measure on semiring It is -subadditive on Therefore, we have Thus,

The outer measure generated by a measure have some nice characteristics which can be shown by the following theorem.

Theorem:Let The followings are equivalent:(1)

is -measurable.(2)

with any and(3)

with any and(4)

with any

**Proof:**

## Polynomial in a Field of Characteristic p

I’ll go ahead and it post it here as well:

Hey Ken, I’m trying to check if this proof is correct(it may not be the only way, but its one I thought of)

To Prove: Let be a field with . Let where . Show if is reducible, then has a root in (actually all).

This is actually an iff in the book, but this is the nontrivial direction.

My idea is the following. Let be a root of . Let be nonunit irreducible factors of . Now it is easy to see the set of all roots is the set . Thus, for some finite sets

Notice this means are both in . Let and . Then is a root for both and which are in . But since the degrees of these are ; respectively, the only way this is possible is if . But we could do this for any irreducible factor so each irreducible factor must have the same degree.

Hence, if with each irreducible, then for some . Since is prime either or must be . If then each root is in . If then is irreducible. Since is reducible the result follows.

After writing this out carefully it feels completely correct (of course I left out some easy details).

## Complex #34: sneaky use of Rouche’s theorem

**Problem #34 is: Prove that there does not exist a polynomial of the form such that for all such that .**

**Proof:** Suppose that there does exist such a polynomial. Note that for . By Rouche’s theorem, has the same number of zeros (with multiplicities) as inside the unit disk. But that is impossible, since has zeros and has at most zeros. QED.

## Complex #27: integrating over a long rectangle

**Problem 27: Show that has the same value on every straight line path (oriented in the direction) parallel to the real axis.**

Proof: Consider the rectangle , where is some given number and is a very large number. Since the integrand is entire, the integral over the boundary of is zero by the Cauchy Integral Theorem, and as , the horizontal pieces correspond to with being the line or the line , with one being negative and the other positive. Thus, the result is proved if we can show that the contributions from the vertical pieces go to zero as goes to .

Observe that the absolute value of the integrand at is , which on each vertical boundary piece is for . Integrating this over will give a value bounded by a constant times , which clearly goes to zero as .

QED.