Radon measure and a non-L1 function This is a part of the exercise 7.17 in Folland's Real Analysis:
Suppose $\mu$ is a positive Radon measure on a locally compact Hausdorff space $X $ with $\mu (X)=\infty. $ Show that there exists $f\in C_0 (X) $ such that $\int_X f d\mu=\infty. $
I guess if we can choose a sequence of disjoint open sets having compact subsets such that the sum of measure of cpt subsets is $\infty $, then done. But the space is not normal in general.. I'm totally confused with it.

Since $\mu$ is Radon, it is inner regular on open sets, in particular on $X$.
Thus, there is a sequence of compact sets $K_n$ with $\mu(K_n) > 4^n$.
By the Urysohn-Lemma, there is a function $f_n \in C_c(X)$ with $0 \leq f_n \leq 1$ and with $f_n \equiv 1 $ on $K_n$.
Now the $M$-test shows that
$$
f := \sum_n 2^{-n} f_n
$$
converges uniformly and hence to a function in $C_0$ (since $C_0$ is closed with respect to uniform convergence).
But the monoton convergence theorem shows
$$
\int f \, d\mu = \sum_n 2^{-n} \int f_n \, d\mu \geq \sum_n 2^{-n} \mu(K_n) = \sum_n 2^n = \infty.
$$
