# Non-regular measure can be represented by a regular measure

Let $X$ be a locally compact and Hausdorff space, and let $\mu$ be a positive measure on the Borel sets of $X$ (here $\mu$ is not necessarily regular).

Then the linear map $L : C_c(X) \to \Bbb C$ defined by $L : f \mapsto \int_X f \;d\mu$ can be represented, by Riesz theorem, as $$L(f) = \int_X f \;d\nu$$ where $\nu$ is a regular Borel measure (here $C_c(X)$ denotes the set of continuous functions on $X$ that are compactly supported).

My question is:

Is my previous reasoning correct? That is : have I proven that for every measure $\mu$ we can find a regular Borel measure $\nu$ with respect to which integration is the same as the integration w.r.t $\mu$?

This would mean (in some sense) that we can always only handle with regular measures. (Or am I misinterpreting something?)

In Folland, A Guide to Advanced Real Analysis, page 59-60 : "If $\mu$􏰼 is a Borel measure on $X$ such that 􏰼$\mu(K) < \infty$ for every compact $K$ of 􏱀$X$, then […] It is a fundamental fact, and a rich source of measures, that every positive linear functional on $C_c(X)$ is of this form, and that the measure 􏰼can be taken to be regular."

The map $L$ requires that $\mu$ is finite valued on compact sets to be well defined.
So, given a locally compact Hausdorff space $X$ and Borel measure $\mu$ that is finite on compact sets, there exists a regular Borel measure $\nu$ with the property that $$\int_X f \, d\mu = \int_X f \, d\nu$$ for every $f \in C_c(X)$.
This doesn't necessarily hold for all integrable Borel functions $f$, though, and in particular $\nu$ and $\mu$ are not necessarily the same measure.
If however every open set in $X$ is $\sigma$-compact then $\mu = \nu$. One conclusion of this is that a Borel measure that is finite on compact sets is necessarily regular on a locally compact Hausdorff space in which every open set is $\sigma$-compact.