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[math]\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}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math] 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. [math]\mu_0(\emptyset)=0[/math] - the measure of the empty set is [ilmath]0[/ilmath]
    2. [math]\mu_0\left(\bigudot_{i=1}^nA_i\right)=\sum_{i=1}^n\mu_0(A_i)[/math]
      • Where [ilmath](A_i)_{i=1}^n\subseteq\mathcal{A}[/ilmath] is a finite sequence
    3. [math]\mu_0\left(\bigudot_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}\mu_0(A_n)[/math] whenever [math]\bigudot_{n=1}^\infty A_n\in\mathcal{A}[/math]
      • 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]:

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

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


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

  • Subadditive: [ilmath]\mu_0(A\cup B)\le\mu_0(A)+\mu_0(B)[/ilmath]

TODO: Again - be bothered


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:

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

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!

See also


  1. Infact not stating 2 explicitly is the most common case. I only state it because it is intuitive to define it as a property we want. We are assured of closed-under-finite-union already, so we can measure over that. We then extend this to countably infinite.
  2. Sometimes stated as monotone (it is monotone in Measures, Integrals and Martingales in fact!)
  3. Not standard notation - made up for this example


  1. Measures, Integrals and Martingales - Rene L. Schilling