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Mathematics and Trivia

The Riesz Representation Theorem – I

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Finally I figured out how to convert LaTeX to WordPress content. Thanks to LaTeX to WordPress. My first post is an experimental one. I have chosen Riesz representation theorem (in abstract integration). I have tried to keep the exposition simple. Any question/comment/criticism/suggestion is invited.

In this article we shall try to understand the Riesz representation theorem in measure theory. Anyone who is acquainted with Hilbert space basics knows that the Riesz representation theorem associates each linear functional to an inner product. Similarly, in the study of abstract integration the theorem associates to each positive linear functional a unique measure. The exact statement will appear later. This helps in establishing the Lebesgue measure on {\mathbb{R}^{n}}.

To begin with, we give some definitions and results that we shall be using. We shall be following the material and proof given in Rudin’s “Real and Complex Analysis” [Theorem 2.14], and this article is meant to fill the missing details.

{X} represents a topological space. {X} is Hausdorff, if any two distinct points in {X} can be separated by disjoint open sets in {X}. {X} is locally compact if every point {x\in X} has a neighborhood (an open set containing {x}) {V_{x}} whose closure is compact, i.e., {\overline{V}_{x}} is compact.

Let {f:X\rightarrow\mathbb{C}} be a complex function on {X}. Let  \displaystyle E=\left\{x:f(x)\neq 0\right\}. The support of {f} is then {\overline{E}} and is denoted by {supp(f)}. The collection of all continuous complex functions on {X} whose support is compact is denoted by {C_{c}(X)}. It is easy to verify that {C_{c}(X)} is a complex vector space.

The notation \displaystyle K\prec f means that {K} is a compact subset of {X}; {f\in C_{c}(X)}; {0\leq f\leq 1} and {f(x)=1} for all {x\in K}. Hence {K \subset supp(f).} The notation \displaystyle f\prec V means that {V} is open; {f\in C_{c}(X)}; {0 \leq f \leq 1} and {supp(f)\subset V}. The notation \displaystyle K\prec f\prec V means both the above statements hold true. Note that {K\prec f\prec V} implies {\chi_{K}\leq f\leq \chi_{V}}.

We shall use a major result – the Urysohn’s lemma. It is basically a statement about separating subsets by continuous function. Here we shall be using the following form. [Theorem 2.12]

Theorem 1. Suppose {X} is a locally compact Hausdorff space, {V} is open in {X}, {K\subset V}, and K is compact. Then there exists an {f\in C_{c}(X)}, such that \displaystyle K\prec f\prec V.

Interpreted in other way, we get the existence of a continuous function with compact support which vanishes outside {V} and assumes 1 on {K}. Moreover, {0\leq f\leq 1}. Using Urysohn’s lemma we get the existence of a partition of unity on {K}.

Theorem 2. Suppose {V_{1},...,V_{n}} are open subsets of a locally compact Hausdorff space {X}, {K} is compact, and \displaystyle K\subset V_{1},...,V_{n}. Then there exist functions {h_{i}\prec V_{i} \,(i=1,...,n)} such that \displaystyle h_{1}(x)+...+h_{n}(x)=1 \hspace{20pt} (x\in K)

The collection {\left\{h_{1},...,h_{n}\right\}} is called a partition of unity on {K} subordinate to the cover {\left\{V_{1},...,V_{n}\right\}}.

A linear functional is positive if {f(X)\subset [0,\infty]} then {\Lambda f\in [0,\infty].}

We now state the Riesz representation theorem.

Theorem 3. Let {X} be a locally compact Hausdorff space, and let {\Lambda} be a positive linear functional on {C_{c}(X)}. The there exists a {\sigma -}algebra {\mathcal{M}} in {X} which contains all Borel sets in {X}, and there exists a unique positive measure {\mu} on {\mathcal{M}} which represents {\Lambda} in the sense that

(a). {\Lambda f=\int_{X}f d\mu } for every {f\in C_{c}(X),}

and which has the additional properties:

(b) {\mu(K)<\infty} for every compact set {K\subset X}.

(c) For every {E\in\mathcal{M}}, we have \displaystyle \mu(E)=\inf\left\{\mu(V):E\subset V, V \text{open}\right\}

(d) For every open set {E} we have \displaystyle \mu(E)=\sup\left\{\mu(K):K\subset E, K \text{compact}\right\}. This also holds true if {E\in\mathcal{M}} with {\mu(E)<\infty}.

(e) If {E\in\mathcal{M}}, {A\subset E}, and {\mu(E)=0}, then {A\in\mathcal{M}}.

We shall prove this theorem in the next post.

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Written by Nirakar Neo

May 30, 2011 at 1:56 pm

Posted in Mathematics

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