2
$\begingroup$

Let $\mu$ be a finite finitely additive measure over the Borel sets of a locally compact topological space $X$. We know that for every bounded function $f$
$$\phi (f)=\int f d\mu. $$

It is known that $\phi(f)$ is a linear function. Suppose that one can also show that its restriction to $C_c(X)$ (continuous functions with the sup norm) is continuous. By they Riesz representation theorem there is a regular Borel measure $\tilde \mu$ such that

$$\int f d\mu= \int f d\tilde{\mu} $$ for every $f\in C_c(X)$.

My question is, do we also know that the equality holds for a larger space of functions? A place to start would be to show that it is true for simple functions.

$\endgroup$
4
  • 1
    $\begingroup$ How is $\phi(f)$ well defined for arbitrary bounded $f$? For example put $\mu=\mathcal{L}^1$ the one dimensional lebesgue measure on $\mathbb{R}$ and $f(x)=-1$ for $x<0$ and $f(x)=1$ for $x\geq0$. Then $\int f\, d\mu$ is not defined. $\endgroup$ Nov 21, 2019 at 15:19
  • $\begingroup$ Thanks. The finitely additive measure is finite. I added that to the statement. $\endgroup$
    – Condor5
    Nov 21, 2019 at 15:31
  • $\begingroup$ It all depends on the topological properties of your space and regularity of the measure. For instance, if the measure is outer regular (with open sets) and the space is normal, then the Urysohn Lemma (see en.wikipedia.org/wiki/Urysohn%27s_lemma) provides the equality for simple functions. $\endgroup$
    – Hugo
    Nov 21, 2019 at 20:10
  • $\begingroup$ Thanks @Hugo. However, I’m not sure I understand the whole argument. Could you explain a little more? $\endgroup$
    – Condor5
    Nov 22, 2019 at 1:39

0

You must log in to answer this question.

Browse other questions tagged .