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I am currently self-studying Rudin's real complex analysis. The Riesz representation theorem in the book states:

Let $X$ be a locally compact Hausdorff topological space. Let $T:C_c(C)\rightarrow \mathbb{C}$ be a positive linear functional. Then there exists a $\sigma$-algebra $\mathscr{M}$ on $X$ that contains all the Borel sets and there exists a unique measure on the measurable space $(X,\mathscr{M})$ such that the following conditions hold (I will show the first condition only to be brief):

1) $T(f)=\int_X f\,d\mu$ for every $f\in C_c(X)$

2)... 3)....4)....

Question: Let $\mu$ be the measure constructed in the proof of the theorem. Is it true that for every $A\subseteq X$ that $\mu(\partial A)=0$?

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  • $\begingroup$ I've changed your \par into \partial; is it correct? $\endgroup$
    – egreg
    Oct 7, 2013 at 20:37
  • $\begingroup$ Assuming egreg's edit is right: No. Consider for example $T$ the Lebesgue integral. There are sets whose boundary has positive measure. $\endgroup$ Oct 7, 2013 at 20:38
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    $\begingroup$ Take $A =\Bbb Q$ for the Lebesgue measure on $X = \Bbb R$. $\endgroup$
    – Siméon
    Oct 7, 2013 at 20:38
  • $\begingroup$ @Ju'x OK. Thanks $\endgroup$
    – Amr
    Oct 7, 2013 at 20:41

1 Answer 1

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One example of a functional $T$ is given by choosing $x \in X$ and defining $T(f) := f(x)$. Then the measure $\mu$ is simply the $\delta$-function supported at $x$, whose value on a set $A$ is $1$ if $x \in A$ and $0$ otherwise.

So $\mu(\partial A)$ will be $0$ if $x \not\in \partial A$, but will be $1$ otherwise.


As a general remark, it's a good idea to keep $\delta$-functions in mind as an example of the kind of measure that can come up in the Riesz rep'n theorem, just to have examples in mind of a different nature than Lebesgue measure (although that is not necessary for answering your particular question).

Another class of examples is given by letting $Y$ be a closed subset of $X$, and $\nu$ be a Borel measure on $Y$. We can then define $T(f)$ by $T(f) = \int_Y f d\nu$. (Note that $f$ restricted to $Y$ is continuous with compact support, so the integral makes sense.) The resulting measure $\mu$ on $X$ will be supported on $Y$.

(Delta-functions are the special case when $Y$ is a point and $\nu$ gives the point a measure of $1$.)

So you could now take e.g. $A$ to be a ball in $X = \mathbb R^n$, and $Y = \partial A$ to be its boundary. Then for any non-zero Borel measure $\nu$ on $Y$, you would get a counterexample.

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