Difference between revisions of "Extending pre-measures to outer-measures"

Latest revision as of 16:59, 17 August 2016

Caution:This page is currently being written and is not ready for being used as a reference, it's a notes quality page

Statement

Given a pre-measure, $\bar{\mu}$, on a ring of sets, $\mathcal{R}$, we can define a new function, $\mu^*$ which is^[1]:

• an extension of $\bar{\mu}$ and
• an outer-measure (on the hereditary $\sigma$-ring generated by $\mathcal{R}$, written $\mathcal{H}_{\sigma_R}(\mathcal{R})$)

Given by:

• $\mu^*:\mathcal{H}_{\sigma_R}(\mathcal{R})\rightarrow\bar{\mathbb{R} }_{\ge0}$
  • $$\mu^*:A\mapsto\text{inf}\left\{\left.\sum^\infty_{n=1}\bar{\mu}(A_n)\right\vert(A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}$$ - here $\text{inf}$ denotes the infimum of a set.

The statement of the theorem is that this $\mu^*$ is indeed an outer-measure

Proof

For brevity we define the following shorthands:

1. $$\alpha_A:=\left\{(A_n)_{n=1}^\infty\ \Big\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}$$
2. $$\beta_A:=\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \Big\vert\ (A_n)_{n=1}^\infty\in\alpha_A \right\}$$

Now we may define $\mu^*$ as:

• $\mu^*:A\mapsto\text{inf}(\beta_A)$

Proof that $\mu^*$ is an extension of $\bar{\mu}$

• Let $A\in\mathcal{R}$ be given
  • In order to prove $\bar{\mu}(A)=\mu^*(A)$ we need only prove $[\bar{\mu}(A)\ge\mu^*(A)\wedge\bar{\mu}(A)\le\mu^*(A)]$^{[Note 1]}
    1. Part 1: $\bar{\mu}(A)\ge\mu^*(A)$
       • Consider the sequence $({ A_n })_{ n = 1 }^{ \infty }$ given by $A_1:=A$ and $A_i:=\emptyset$ for $i>1$, so the sequence $A,\emptyset,\emptyset,\ldots$.
         • Clearly $A\subseteq\bigcup^\infty_{n=1}A_n$ (as $\bigcup^\infty_{n=1}A_n=A$)
         • As such this $({ A_n })_{ n = 1 }^{ \infty } \in\alpha_A$
         • This means $\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A$ (as $({ A_n })_{ n = 1 }^{ \infty } \in\alpha_A$ and $\beta_A$ is the sum of all the pre-measures Template:WRT $\bar{\mu}$ of the sequences of sets in $\alpha_A$)
         • Recall that the infimum of a set is, among other things, a lower bound of the set. So:
           • for $\text{inf}(S)$ (for a set, $S$) we see:
             • $\forall s\in S[\text{inf}(S)\le s]$ - this uses only the lower bound part of the infimum definition.
         • By applying this to $\text{inf}(\beta_A)\big(=\mu^*(A)\big)$ we see:
           • $\mu^*(A):=\text{inf}(\beta_A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)$
             • as $\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A$ and $\text{inf}(S)$ remember and
             • By definition of a (pre-)measure, $\mu(\emptyset)=0$, so: $\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)+\bar{\mu}(\emptyset)+\bar{\mu}(\emptyset)+\cdots=\bar{\mu}(A)$
       • We have shown $\mu^*(A)\le\bar{\mu}(A)$ as required
    2. Part 2: $\bar{\mu}(A)\le\mu^*(A)$
       • SEE NOTEPAD. Define $\gamma_A:=\left\{\bar{\mu}(A)\right\}$, then using the (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set we see $\forall x\in\beta_A\exists y\in\gamma_A[y\le x]$ - we may now pass to the infimum.

Proof that $\mu^*$ is $\sigma$-subadditive

• Let $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R})$ be given. We want to show that $\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)$
  • Let $\epsilon>0$ (with $\epsilon\in\mathbb{R}$) be given.
    • We will now define a new family of sequences. For each $A_n$ we will construct the sequence $({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R}$ of sets such that:
      1. $\forall n\in\mathbb{N}[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}]$ and
      2. $\forall n\in\mathbb{N}[\sum^\infty_{m=1}\bar{\mu}(A_{nm})\le\mu^*(A_n)+\epsilon\frac{1}{2^n}]$
    • Let $n\in\mathbb{N}$ be given (we will now define $({ A_{mn} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R}$)
      • Recall that $\mu^*(A_n):=\text{inf}(\beta_{A_n})$
      • Any value greater than the $\text{inf}(\beta_{A_n})$, say $w$, is not a lower bound so there must exist an element in $\beta_{A_n}$ less that $w$ (so $w$ cannot be a lower bound)
        • Choose $w:=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}$
          • As $\epsilon>0$ and $\frac{1}{2^n}>0$ we see $\frac{\epsilon}{2^n}>0$, thus $\mu^*(A_n)<\mu^*(A_n)+\frac{\epsilon}{2^n}$
      • By the definition of infimum:
        • $\exists s\in\beta_{A_n}[w>\text{inf}(\beta_{A_n})\implies s< w]$
      • If $s\in\beta_{A_n}$ then:
        • $\exists(B_n)_{n=1}^\infty\in\alpha_{A_n}$ such that $s=\sum^\infty_{n=1}\bar{\mu}(B_n)$.
      • As $s<w=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}=\mu^*(A_n)+\frac{\epsilon}{2^n}$ and $s=\sum^\infty_{n=1}\bar{\mu}(B_n)$ we see:
        • $\sum^\infty_{n=1}\bar{\mu}(B_n)<\mu^*(A_n)+\frac{\epsilon}{2^n}$
      • Caution:This doesn't show that $$A_n\subseteq\bigcup_{m=1}^\infty A_{nm}$$ - don't forget!
      • Define a new sequence, $({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R}$ to be the sequence $({ B_n })_{ n = 1 }^{ \infty } \in\alpha_{A_n}$ we just showed to exist
    • Since $n\in\mathbb{N}$ was arbitrary for each $A_n\in(A_k)_{k=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})$ we now have a new sequence: $({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R}$ such that:
      • $$\forall n\in\mathbb{N}\left[\sum^\infty_{m=1}\bar{\mu}(A_{nm})<\mu^*(A_n)+\frac{\epsilon}{2^n}\right]$$ and $$\forall n\in\mathbb{N}\left[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}\right]$$
    • Recall now that a union of subsets is a subset of the union, thus:
      • $$\bigcup_{n=1}^\infty A_n\subseteq \bigcup_{n=1}^\infty\left(\bigcup_{m=1}^\infty A_{nm}\right)$$
    • So $$\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\left(\sum_{m=1}^\infty \bar{\mu}(A_{nm})\right)<\sum_{n=1}^\infty\left(\mu^*(A_n)+\frac{\epsilon}{2^n}\right)$$ $$=\sum^\infty_{n=1}\mu^*(A_n)+\sum^\infty_{n=1}\frac{\epsilon}{2^n}$$
      • Note that $\sum^\infty_{n=1}\frac{\epsilon}{2^n}=\epsilon\sum^\infty_{n=1}\frac{1}{2^n}$ and that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ is a classic example of a geometric series, we see easily that:
        • $\epsilon\sum^\infty_{n=1}\frac{1}{2^n}=1\epsilon=\epsilon$ thus:
    • $$\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon$$
  • Since $\epsilon>0$ (with $\epsilon\in\mathbb{R}$ was arbitrary we see:
    • $\forall\epsilon>0\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon\right]$
  • Recall that $\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff\left(a\le b\right)$ (from the epsilon form of inequalities)
  • Thus: $$\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)$$
• Since $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R})$ was arbitrary we have shown that:
  • $$\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)\right]$$

This completes the proof that $\mu^*$ is $\sigma$-subadditive

Caveats

1. Halmos starts with a set $A\in\mathcal{H}_{\sigma R}(\mathcal{R})$ and a sequence $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R})$ such that:
   • $A\subseteq\bigcup_{n=1}^\infty A_n$

   where as I just start with a sequence, as $\mathcal{H}_{\sigma R}(\mathcal{R})$ is a $\sigma$-algebra, their union is also in $\mathcal{H}_{\sigma R}(\mathcal{R})$

2. Warning:I never consider the case where a measure measures a set to be infinite. Where this happens things like $\infty<\infty$ make no sense

The rest

Still to do:

1. $\mu^*$ being monotonic with respect to set inclusion and the usual ordering on the reals.
2. $\mu^*(\emptyset)=0$ - this can come from the extension part as $\bar{\mu}$ has this property already

Notes

1. Jump up ↑ This is called the trichotomy rule or something, I should link to the relevant part of a partial order here

References

1. ↑ ^{Jump up to: 1.0 1.1 1.2} Measure Theory - Paul R. Halmos