# Measure Theory

$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$\require{AMScd}\newcommand{\d}[]{\mathrm{d}^{#1} }$

## Measures

To start with we define rings, for example consider the ring of all half-open-half-closed rectangles of dimension [ilmath]n[/ilmath], call this $\mathcal{J}^n$

$[[a,b))\in\mathcal{J}^n$ means $[a_1,b_1)\times[a_2,b_2)\times\cdots\times[a_n,b_n)\in\mathcal{J}^n$

We can clearly get a ring from this, but not a [ilmath]\sigma[/ilmath]-ring as for example:

$\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n$

The Lebesgue measure on $\mathcal{J}^n$, which is $\lambda^n:\mathcal{J}^n\rightarrow[0,\infty]$ where:

$\lambda^n\Big([[a,b))\Big)=\prod^n_{i=1}(b_i-a_i)$.

### Forming a ring

So let us take the one dimensional case. Consider the following $\in\mathcal{J}^1$

Example As disjoint union Measure
[ilmath][0,5)[/ilmath] (example:) [ilmath][0,1)\cup[1,2)\cup[2,3)\cup[3,4)\cup[4,5)[/ilmath] [ilmath]5[/ilmath]
[ilmath][0,5)-[2,5)[/ilmath] [ilmath][0,2)[/ilmath] [ilmath]2[/ilmath]
[ilmath][0,5)-[1,2)[/ilmath] [ilmath][0,1)\cup[2,5)[/ilmath] [ilmath]1+3=4[/ilmath]
[ilmath][0,5)-[0,1)[/ilmath] [ilmath][1,5)[/ilmath] [ilmath]4[/ilmath]
[ilmath][0,1)\cup[1,2)[/ilmath] [ilmath][0,2)[/ilmath] [ilmath]2[/ilmath]
[ilmath][0,1)\cup[3,4)[/ilmath] [ilmath][0,1)\cup[3,4)[/ilmath] [ilmath]1+1=2[/ilmath]
Using intersection (which can be done using [ilmath]-[/ilmath])
[ilmath][0,5)\cap[1,2)[/ilmath] [ilmath][1,2)[/ilmath] [ilmath]1[/ilmath]

As you can see, we can form a ring quite easily using [ilmath]\mathcal{J}^1[/ilmath], furthermore we can express things in this ring as disjoint unions!

We may now consider $R(\mathcal{J}^n)$ - the ring generated by [ilmath]\mathcal{J}^n[/ilmath], note that $\mathcal{J}^n\subset R(\mathcal{J}^n)$

With the example of $\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n$ you have probably already started to suspect that the "Lesbegue measure" or "n-dimensional version of volume" for $[[0,1]]$ may well just be 1, this intuition is correct, but we're staying in the finite deliberately right now.

## Our first measure

Consider this: $\lambda_0^n:R(\mathcal{J}^n)\rightarrow[0,\infty]$, we want to be able to "measure" things in our ring. The natural way to do this is to break them down in to separate things and add the measure of each bit!

We can clearly say: $\lambda_0^n(A)\mapsto\left\{\begin{array}{lr} \lambda^n(A) & \text{if }A\in\mathcal{J}^n\\ \sum^m_{i=1}\lambda^n(A_i) & \text{where }A=\bigudot^m_{i=1}A_i \end{array}\right.$, that is:

• If [ilmath]A[/ilmath] is just a rectangle, then its measure is the volume of that rectangle
• If [ilmath]A[/ilmath] is the disjoint union of some rectangles, then the volume of [ilmath]A[/ilmath] is the sum of the disjoint rectangles that make it up

Intuitively we know that we want $R(\mathcal{J}^n)$ to be the smallest ring we can have with $\mathcal{J}^n\subset R(\mathcal{J}^n)$ and using the logic described in the table above we can see that anything in this ring is the union of some (indeed finite) amount of sets in $\mathcal{J}^n$

However that is not good enough! We are being formal here!

We know that Every set in [ilmath]R(\mathcal{J}^n)[/ilmath] can be finitely covered by sets in [ilmath]\mathcal{J}^n[/ilmath], that is $\forall S\in R(\mathcal{J}^n)\exists \{S_i\}_{i=1}^m:S=\cup^m_{i=1}S_i$

But we do not know that every set in [ilmath]R(\mathcal{J}^n)[/ilmath] can be finitely covered by DISJOINT sets in [ilmath]\mathcal{J}^n[/ilmath]

Theorem to show that every set in [ilmath]R(\mathcal{J}^n)[/ilmath] may be covered by a disjoint finite selection in [ilmath]\mathcal{J}^n[/ilmath]

TODO: Todo finite ring proof for Lebesgue measure

Took me a while to work it out - will do later. Isn't that bad. Induction induction and more induction!

Now $\lambda_0^n(A)\mapsto\left\{\begin{array}{lr} \lambda^n(A) & \text{if }A\in\mathcal{J}^n\\ \sum^m_{i=1}\lambda^n(A_i) & \text{where }A=\bigudot^m_{i=1}A_i \end{array}\right.$ satisfies the definition of a Pre-measure