# Pre-measure

## Definiton

$\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}[1][]{\mathrm{d}^{#1} }$ A pre-measure[1] is a measure on an algebra rather than a [ilmath]\sigma[/ilmath]-algebra, the properties are as follows:

• Here [ilmath]\mathcal{A} [/ilmath] is an algebra of sets (a system of subsets of [ilmath]X[/ilmath]) and [ilmath]\mu_0:\mathcal{A}\rightarrow[0,+\infty][/ilmath] such that:
1. $\mu_0(\emptyset)=0$ - the measure of the empty set is [ilmath]0[/ilmath]
2. $\mu_0\left(\bigudot_{i=1}^nA_i\right)=\sum_{i=1}^n\mu_0(A_i)$
• Where [ilmath](A_i)_{i=1}^n\subseteq\mathcal{A}[/ilmath] is a finite sequence
3. $\mu_0\left(\bigudot_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}\mu_0(A_n)$ whenever $\bigudot_{n=1}^\infty A_n\in\mathcal{A}$
• Where [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] is a sequence

Notice that property 2 implies property 3 (as we can make a finite sequence infinite just by defining the terms after the finite ones as the emptyset) so often just the following is given[Note 1]:

• $\mu_0(\emptyset)=0$ - the measure of the empty set is [ilmath]0[/ilmath] and
• $\mu_0\left(\bigudot_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}\mu_0(A_n)$ whenever $\bigudot_{n=1}^\infty A_n\in\mathcal{A}$

Note: It could be worth reading the discussion section of this page now if the motivation for this definition isn't clear

## Properties

Here [ilmath](X,\mathcal{A},\mu_0)[/ilmath] is a pre-measure space, and [ilmath]A,B\in\mathcal{A} [/ilmath]

• Finitely additive: if [ilmath]A\cap B=\emptyset[/ilmath] then [ilmath]\mu_0(A\udot B)=\mu_0(A)+\mu_0(B)[/ilmath]

Follows immediately from definition (property 2)

• Monotonic: [Note 2] if [ilmath]A\subseteq B[/ilmath] then [ilmath]\mu_0(A)\le\mu_0(B)[/ilmath]

TODO: Be bothered to write out

• If [ilmath]A\subseteq B[/ilmath] and [ilmath]\mu_0(A)<\infty[/ilmath] then [ilmath]\mu_0(B-A)=\mu_0(B)-\mu(A)[/ilmath]

TODO: Be bothered, note the significance of the finite-ness of [ilmath]A[/ilmath] - see Extended real value

• Strongly additive: [ilmath]\mu_0(A\cup B)=\mu_0(A)+\mu_0(B)-\mu_0(A\cap B)[/ilmath]

TODO: Be bothered

TODO: Again - be bothered

## Discussion

An algebra of sets is closed under finite union anyway, so the [ilmath]2^\text{nd} [/ilmath] part of the definition is fine. However there are infinite unions that are not closed on an algebra. There are some infinite unions which are however in the union.

### Example of infinite sequence not being in the algebra

Let [ilmath]\mathcal{I} [/ilmath] be the set of all half-open-half-closed intervals of [ilmath]\mathbb{R} [/ilmath]

• That is to say that [ilmath]\mathcal{I}:=\{[a,b)\vert\ a,b\in\mathbb{R}\wedge b\ge a\}[/ilmath] - with the convention that [ilmath][a,a)=\emptyset[/ilmath]

And let [ilmath]\mathcal{A}:=\mathcal{A}(\mathcal{I})[/ilmath] be the algebra generated by this set[Note 3].

• That is to say (informally, as this is just an example) [ilmath]\mathcal{A} [/ilmath] contains everything which is the union of finitely many half-open-half-closed intervals.
• Notice that:
• [ilmath][a,b)^c=(-\infty,a)\cup[b,+\infty)[/ilmath]
• [ilmath][a,d)-[b,c)[/ilmath] where [ilmath]a< b< c< d[/ilmath] is [ilmath][a,b)\cup[c,d)[/ilmath]
• It's closed under union and so forth.
• (It should be easy to convince yourself that this is an algebra)

Then it is easy to see that all finite unions are in this system. However consider:

• $\bigcup_{n=1}^\infty[0,1-\tfrac{1}{n})=[0,1]$ -which is not open, nor in [ilmath]\mathcal{A} [/ilmath]!
• But if this were finite: $\bigcup_{n=1}^m[0,1-\tfrac{1}{n})=[0,1-\tfrac{1}{m})\in\mathcal{A}$
• $\bigcap_{n=1}^\infty[-\tfrac{1}{n},\tfrac{1}{n}]=[0]$ which is again, not in [ilmath]\mathcal{A} [/ilmath]
• But of course if this were finite, then $\bigcap_{n=1}^m[-\tfrac{1}{n},\tfrac{1}{n}]=[-\tfrac{1}{m},\tfrac{1}{m})\in\mathcal{A}$

### Example of infinite sequence being in the algebra

This example constructs an infinite sequence from a finite one, a bit of a cheat but an example none the less!

• Given a finite sequence: [ilmath](B_i)_{i=1}^n\subseteq\mathcal{A}[/ilmath] we define:
• [ilmath](A_m)_{m=1}^\infty[/ilmath] with [ilmath]A_m=\left\{\begin{array}{lr}B_m & \text{if } m\le n\\ \emptyset & \text{otherwise}\end{array}\right.[/ilmath]
• Now [ilmath]\cup_{m=1}^\infty A_m=(\cup_{m=1}^n A_m)\bigcup(\cup_{m=n+1}^\infty A_m)=(\cup_{m=1}^n B_m)\bigcup(\cup_{m=n+1}^\infty\emptyset)=\cup_{m=1}^n B_m\in\mathcal{A}[/ilmath] by definition (Algebras are closed under finite union)
• Thus [ilmath]\cup_{m=1}^\infty A_m\in\mathcal{A}[/ilmath] - despite being infinite!

So it makes sense that when given an infinite sequence whose union happens to be in [ilmath]\mathcal{A} [/ilmath] - which is the definition of the domain of our pre-measure, we use our pre-measure to measure it!