Consider the following:
Let $f$ be a $L^1(\mathbb{R}^n)$ function and define the Borel measure $d\mu=fdx$, where $dx$ is the Lebesgue measure. Now, since $f$ is in $L^1(\mathbb{R}^n)$, we know that $d\mu$ is finite in all Borel sets (not just in compact sets). But is it true that $\mu$ is also a regular Borel measure and hence a Radon measure, or am i tripping balls :)?.
Any Borel measure on $\mathbb R^{n}$ which is finite on compact sets is regular. This is proved in Rudin's RCA.