Several questions about Riesz–Markov–Kakutani representation theorem

Question

This is a list of questions about Riesz–Markov–Kakutani representation theorem .

1)If $f\in L^1(\mu)$, is it true that $\phi(f)=\int_Xfd\mu$, where $\mu$ is given by the theorem? I am quite sure it is correct, since $C_c(X)$ is dense in $L^1(\mu)$. And we can extract a subsequence that pointwise converges to $f$.

2)If $X$ is not locally compact, does the theorem still stands? If not, does there exist a non-regular measure such that $\phi(f)=\int_Xfd\mu$?

The property of locally compact is a must to use Urysohn lemma in the proof. The reason why Urysohn lemma (or partition unity) is widely used is because the way we construct the measure. So I am thinking if we can construct the same measure in another way that we can avoid Urysohn lemma.

3) If $X$ is not Hausdorff, what will happen as stated in 2)?

4)If $X$ is normal and not locally compact?

Answer 1

Well, as a start, if you drop local compactness, you lose uniqueness.

Suppose $X$ is Hausdorff and not locally compact. Then there exists a point $x \in X$ such that no open neighborhood of $U$ is contained in a compact set. Now if $f \in C_c(X)$ and $f(y) \ne 0$, then the set $U = f^{-1}(\mathbb{R} \setminus \{0\})$ is an open neighborhood of $y$ which is contained in the support of $f$, which by assumption is compact. As such, we must have $f(x)=0$ for every $f \in C_c(X)$. Thus if we take $\phi=0$ to be the zero functional, $\mu=0$ the zero measure, and $\nu = \delta_x$ a point mass at $x$, we have for every $f \in C_c(X)$ that $\phi(f) = \int f\,d\mu = \int f\,d\nu$.

For instance, if you take $X = \mathbb{Q}$, then every compact set has empty interior, so no point has an open neighborhood which is contained in a compact set. Hence $C_c(\mathbb{Q})$ consists only of 0. There are lots of Borel measures on $\mathbb{Q}$, but there is only one linear functional on $C_c(\mathbb{Q})$ (the zero functional), and it's induced by every measure.

Note that $\mathbb{Q}$ is metrizable, hence normal, and every measure on $\mathbb{Q}$ is regular. So that part is not the issue.

A similar example is to take $X$ to be an infinite-dimensional Banach space. Again we get $C_c(X) = 0$ by Riesz's lemma.

You can also lose uniqueness if you drop Hausdorff. Suppose $X$ is the line with two origins, call them $0$ and $0'$. It's easy to show that for every continuous $f : X \to \mathbb{R}$ we have $f(0) = f(0')$, and moreover the sets $\{0\}$ and $\{0'\}$ are Borel. So the point masses $\delta_0$ and $\delta_{0'}$ are two different measures which induce the same linear functional on $C_c(X)$.

I am not sure what happens to existence in either case.

Answer 2

For your first question, $C_c(X)$ is a subspace of $L^1(\mu)$, and $\phi$ is dominated by the $L^1(\mu)$ norm on $C_c(X)$. By the Hahn-Banach theorems, $\phi$ extends to some continuous linear functional $\Lambda$ on $L^1(\mu)$. For any $f\in L^1(\mu)$, we can find a sequence $\{f_n\}$ in $C_c(X)$ such that $\|f_n-f\|_1 \rightarrow 0$ as $n\rightarrow \infty$. By continuity of $\Lambda$, $$ \Lambda f = \lim_{n\rightarrow \infty} \int_Xf_nd\mu. $$ Also, $$ \lim_{n\rightarrow \infty} \bigg|\int_X(f_n-f)d\mu\bigg| \leq \lim_{n\rightarrow \infty} \int_X|f_n-f|d\mu = 0 $$ implies $$ \int_Xfd\mu = \lim_{n\rightarrow \infty} \int_{X}f_nd\mu, $$ hence $$ \Lambda f = \int_Xfd\mu. $$ This in fact shows that $\Lambda$ is the unique continuous extension of $\phi$ to $L^1(\mu)$. In other words, $C_c(X)$ and $L^1(\mu)$ have the same dual space.

Note that we never used pointwise convergence, nor do we need it. A direct application of Fatou's lemma, however, does show that $f_n(x)\rightarrow f(x)$ as $n\rightarrow \infty$ for a.e. $x \in X$.