## 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

Một bài toán kinh điển trong luyện thi học sinh giỏi là bài này. Chứng minh rằng nếu $latex {\alpha}&fg=000000$ là số vô tỷ, dãy các số $latex {n\alpha -[n\alpha]}&fg=000000$, phần thập phân của $latex {n\alpha}&fg=000000$ với $latex {n\in \mathbb N}&fg=000000$ biên thiên, trù mật trong đoạn $latex {[0,1]}&fg=000000$. Lời giải dựa trên nguyên lý Dirichlet, còn được gọi là nguyên lý chuồng thỏ hoặc chuồng bồ câu tùy vào khu vực địa lý nơi bạn sinh sống.

Dùng chuỗi Fourier, Hermann Weyl chứng minh định đề mạnh hơn nhiều. Ông chứng minh rằng tập các phần thập phân $latex {n \alpha – [n \alpha]}&fg=000000$ phân bố đều trên đoạn $latex {[0,1]}&fg=000000$. Nếu lấy một đoạn con $latex {[a,b]}&fg=000000$ nằm giữa 0 và 1, xác suất để $latex {n \alpha – [n \alpha]}&fg=000000$ rơi vào trong đoạn này đúng bằng $latex {b-a}&fg=000000$.

Gọi $latex {I_{[a,b]}}&fg=000000$ là hàm đặc trưng của…

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## 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.