I have read that given a Radon measure $\mu$ in $\mathbb{R}^d$, the operator $T_{\mu}(\phi)=\int_{\mathbb{R}^d} \phi d \mu$, $\phi \in C_c^\infty(\mathbb{R}^d),$ defines a distribution. However, when I try to prove this result I find the following difficulties related to integration respect to a Radon measure:
- If $\phi\in C_c^\infty(\mathbb{R}^d)$, do we have $\int_{\mathbb{R}^d} \phi d \mu<\infty$?
- If $\phi\in C_c^\infty(\mathbb{R}^d)$, can we affirm that $\int_{\mathbb{R}^d} \phi d \mu\leq \max{|\phi|}\int_{\mathbb{R}^d} d \mu\ $.
In general, I would like to know which properties cease to be true when we change Lebesgue measure to an arbitrary Radon measure.
Hint: by definition, a Radon measure is finite on compact sets, and the subscript $c$ in $C_c(\mathbb R^d)$ means supported on a compact set.