Lecture Notes

−Table of Contents

Rasmus Pallisgaard

Rasmus Pallisgaard

Hi everyone, I'm Rasmus and I'm an exchange student all the way from Denmark. Back home I study Machine Learning and have done research in the field of NLP, specifically studying multilingual models. I'm at Berkeley for a semester to study mathematics for a semester in order to get more familiar with the rigorous nature of maths (ML research is basically result driven with little theory to back it up - godspeed!). If you want to hear about whether AI will kill us all one day (It might, but then again so will global warming. edit: or Russia. Слава Україні!)

Notes

I'm gonna publish my notes on the lectures this weekend after I finish this weeks homework.

Homework

Resume of the Lebesque Integral We begin by covering the Riemann integral in short. Riemann integration, like all others, seek to measure the area under a graph. It does this using approximation by partitioning the area into rectangles. In this function the Riemann sum corresponds to $\sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ Letting $\Delta_{x_i}=x_i-x_{i-1}$ be the same value for all $i$ , and letting $x_{i-1}\leq t_i\leq x_i\forall i$ . If $a<x_i<b\forall i$ , then we can define the Riemann integral as $\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^nf(t_i)(x_i-x_{i-1}$ If the limit of this sum exists and is unique, then the function $f$ is Riemann integrable.

The Lebesque integral is based on the measurability of a functions undergraph. This undergraph is defined by $Uf=\{(x,y)\in\mathbb{R}\times[0,\infty):p\leq y<f(x)\}$ We say that the function $f$ is Lebesque measurable if the undergraph $Uf$ is measurable. If $f$ is Lebesque measurable, then we let $\int f=m(Uf)$ Note that $dx$ is missing. Because the Lebesque integral does not handle a limit of a sum of rectangles with width of $\Delta x_i$ , we omit the $dx$ .

Finally, a function is Lebesque integrable if the measure $m(Uf)$ is finite. Since the Lebesgue measure of $Uf$ can be infinity, we do by definition allow the Lebesque integral of $f$ to be infinite.

Homework 5

Homework 6 (There is a mistake in problem 1, b. I'll fix it and update the file.)

Resume of Lebesque measure theory

*This is gonna be a rough summarisation of all we've covered in Lebesque measure theory so far. It will probably not contain everything of importance and might have some gaps that I didn't think to cover. If you find gaps like these, please do write to me on Discord so I can review these. Thanks!*

Our venture into Lebesque measure theory begins with the definition of the outer measure - a measure of a subset $A$ found by $m^*(A)=\inf\left\{ \sum_k|B_k|:\{B_k\} \text{ is a covering of } A \text{ by open boxes} \right\}$ Useful from this definition and this section is the definition that sets with outer measure zero is called a zero set. Boxes are created from intervals $(a_i,b_i)$ since the measure of an interval is its end point minus its starting point. The proofs of useful properties, such as monotonicity, sub-additivity, etc. are most often proved using the $\epsilon$ trick.

A set $A\subset R$ is then measurable if it the division $A|A^c$ is so *clean* that for all subsets $X\subset R$ , $m^*(A)=m^*(X\cap A) + m^*(A\cap E^c)$ Although I personally like the Tao condition better: $m^*(A)=m^*(X\cap A) + m^*(A\setminus E)$ Here we see that by this definition, additivity is achieved. Throughout the course we have proved a lot of properties of measurable sets, including sub-additivities of outer measures, monotonicity, etc. As mentioned before, for this we often use the $\epsilon$ trick - a trick that is also used to prove measurability of a closed interval, a zero set, and a closed box in n dimensions.

We then went on to proof that the Lebesque measure is regular in the sense that a measurable set $E$ can be sandwiched between an $F_\sigma$ -set and a $G_\delta$ -set such that $F_\sigma\subset E\subset G_\delta$ . Here an $F_\sigma$ -set is a countable union of closed sets $F_\sigma=\cup^\infty_iF_i$ and a $G_\delta$ -set is a countable intersection of open sets $G_\delta=\cup^\infty_iG_i$ .

The proof of this uses the fact that can define a decreasing sequence of open sets from $\mathbb{R}$ such that the measure of these set sequences goes to the measure of $E$ . You can then define a closed increasing sequence from the complement of one of these sequences. The major step here is then to show that this complement set has the same measure as $E$ .

We then covered the theorems of products and slices. The theorem of measurable products says that if sets $A$ and $B$ are measurable, then $m(A\times B)=m(A)\cdot m(B)$ . The proof of this uses hulls and kernels of measurable sets (These are $F_\sigma$ -sets and $G_\delta$ -sets).

Afterwards we proved that if $E$ is measurable then it has measure zero iff almost every slice of $E$ has measure zero. A slice $E_x$ of a set $E\subset R^n\times R^k$ is defined as $E_x=\{y\in R^n:(x,y)\in E\}$ .

The prove of the above theorem is bit more involved, but essentially boils down to first finding that the set $E$ has the same measure as the set $E$ with all nonzero slices removed. Afterwards one seeks to prove that since all these slices has measure zero, then measure of $E$ is zero. We then, for a slice of any compact $K\subset E$ , surround it by a a long but thin compact box $W(x)$ . Since these boxes are thin, but not zero width, we can cover the set by a countable amount of these boxes. By disjointizing the widths we can find that the boxes have measure zero, so measure of $K$ is zero. Then inner measure is zero (see definition of inner measure with respect to closed subsets) is zero, and by measurability measure of $E$ is zero.

Proving the other direction, if $E$ has measure zero, then there exists a $G_\delta$ -set $G\supset E$ , The main step is to set up $X(\alpha)=\{x:m(G_x)>\alpha\}$ for each slice $G_x$ . One can then finalise the proof using the same disjointizing method on a compact set $K(x)$ contained in $G_x$ with $m(K(x))=m(G_x)$ and a neighbourhood $W(x)$ .