# Understanding why the positive bounded linear functionals on $C_0(X)$ are given by integration against finite Radon measures.

Folland says in his chapter on the dual of $$C_0(X)$$ (continuous functions on $$X$$ vanishing at $$\infty$$):

We recall that for any LCH space $$X$$, $$C_0(X)$$ is the uniform closure of $$C_c(X)$$, and hence if $$\mu$$ is a Radon measure on $$X$$, the functional $$I(f)=\int f\ d\mu$$ extends continuously to $$C_0(X)$$ iff it is bounded with respect to the uniform norm. In view of the equality $$\mu(X)=\sup \big\{ \int f\ d\mu: f\in C_c(X), 0\le f\le 1 \big\}$$ together with the fact that $$|\int f\ d\mu| \le \int |f|\ d\mu$$, this happens precisely when $$\mu(X)<\infty$$, in which case $$\mu(X)$$ is the operator norm of $$I$$.

We have therefore identified the positive bounded linear functionals on $$C_0(X)$$: they are given by integration against finite Radon measures.

I feel like an idiot but I just can't seem to parse this passage. Can someone link me to another explanation of this? I'm specifically having trouble understanding why the first "iff" statement is true and why the finiteness of the measure is implied exactly. I just keep reading it and it's not sinking in.

Forgive me if this is not a good question.

For the second assertion, if $\mu(X)<\infty$ then the functional is bounded. Conversely if $\mu(X)=\infty$, there is a sequence $f_n$ with $0\leq f_n\leq 1$ and $\int f_nd\mu\nearrow\infty$; so the functional is unbounded.
• So one explicitly gets this sequence of $f_n$ by considering the sequence along which the supremum is unbounded and showing that such $f_n$ converge with respect to the uniform norm? I don't see how we are guaranteed this convergence a priori.
• We define $I \in C_c(X)^*$ by $I(f) = \int f \ \text{d}\mu$ and claim it can be extended continuously to $J \in C_0(X)^*$. But how do we know $J$ is still given by integration against $\mu$? Hahn-Banach guarantees $J(f) = I(f)$ for $f \in C_c(X)$, but what about if $f \in C_0(X) \setminus C_c(X)$? Commented May 6 at 23:49
• You don't take an arbitrary extension; you just define it as integration against $\mu$. Also note that when you have a bounded functional defined in a dense subspace, the extension to the whole space is unique. Commented May 7 at 0:03