Difference between revisions of "Measure Theory"
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As you can see, we can form a ring quite easily using {{M|\mathcal{J}^1}}, furthermore we can express things in this ring as disjoint unions! | As you can see, we can form a ring quite easily using {{M|\mathcal{J}^1}}, furthermore we can express things in this ring as disjoint unions! | ||
| − | We may now consider <math>R(\mathcal{J}^n)</math> - the [[Ring generated by | + | We may now consider <math>R(\mathcal{J}^n)</math> - the [[Ring generated by|ring generated by {{M|\mathcal{J}^n}}]], note that <math>\mathcal{J}^n\subset R(\mathcal{J}^n)</math> |
With the example of <math>\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n</math> you have probably already started to suspect that the "Lesbegue measure" or "n-dimensional version of volume" for <math>[[0,1]]</math> may well just be 1, this intuition is correct, but we're staying in the finite | With the example of <math>\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n</math> you have probably already started to suspect that the "Lesbegue measure" or "n-dimensional version of volume" for <math>[[0,1]]</math> may well just be 1, this intuition is correct, but we're staying in the finite | ||
Revision as of 20:10, 16 March 2015
First things
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 [math]\mathcal{J}^n[/math]
[math][[a,b))\in\mathcal{J}^n[/math] means [math][a_1,b_1)\times[a_2,b_2)\times\cdots\times[a_n,b_n)\in\mathcal{J}^n[/math]
We can clearly get a ring from this, but not a [ilmath]\sigma[/ilmath]-ring as for example:
[math]\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n[/math]
The Lebesgue measure on [math]\mathcal{J}^n[/math], which is [math]\lambda^n:\mathcal{J}^n\rightarrow[0,\infty][/math] where:
[math]\lambda^n\Big([[a,b))\Big)=\prod^n_{i=1}(b_i-a_i)[/math].
Forming a ring
So let us take the one dimensional case. Consider the following [math]\in\mathcal{J}^1[/math]
| 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 [math]R(\mathcal{J}^n)[/math] - the ring generated by [ilmath]\mathcal{J}^n[/ilmath], note that [math]\mathcal{J}^n\subset R(\mathcal{J}^n)[/math]
With the example of [math]\bigcup^\infty_{n=1}[[0,1-\tfrac{1}{n}))=[[0,1]]\notin\mathcal{J}^n[/math] you have probably already started to suspect that the "Lesbegue measure" or "n-dimensional version of volume" for [math][[0,1]][/math] may well just be 1, this intuition is correct, but we're staying in the finite deliberately right now.
Our first measure
Consider this: [math]\lambda_0^n:R(\mathcal{J}^n)\rightarrow[0,\infty][/math], 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!
Intuitively we know that we want [math]R(\mathcal{J}^n)[/math] to be the smallest ring we can have with [math]\mathcal{J}^n\subset R(\mathcal{J}^n)[/math] 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 [math]\mathcal{J}^n[/math]
However that is not good enough! We are being formal here! So
TODO: Finish this off
