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Given $\Omega\subset \mathbb R^N$ is open and let function $\varphi$: $\Omega\to [1,+\infty]$, $\varphi\in L^1_{loc}(\Omega)$ be given. Suppose $\mu$ is a finite Radon measure on $\Omega$ and we future have $$ \int_\Omega \varphi \,d\mu<\infty. $$ Then can I claim that $\varphi\mu$ is also a finite Radon measure? It looks true but I would like to have some details.

Thank you!


My try:

First of all, for each compact set $K\subset \Omega$, we have $$ \int_K \varphi\,d\mu\leq \int_\Omega \varphi\,d\mu<\infty. $$ Now I only need to show $\varphi\mu$ is Borel regular. Let $\nu:=\varphi \mu$, we have $\nu$ is Borel since $\varphi$ is Borel measurable and $\mu$ is Radon.

To finish my proof, I need to show for each $A\subset \Omega$, there exists a Borel set $B$ such that $A\subset B$ and $\nu(A)=\nu(B)$, and I got stuck here...


I found another similar post here... Please feel free to close mine.

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  • $\begingroup$ Why do you think $\int_\Omega \varphi \, d\mu < \infty$? You only have $\varphi \in L_{loc}^1(\Omega)$, not $\varphi \in L^1(\Omega)$. Furthermore, I don't think your definition of "Radon measure" is the standard one. Are you reading Evans & Gariepy? $\endgroup$ – PhoemueX Aug 7 '15 at 18:17
  • $\begingroup$ @PhoemueX Yes I am reading Evans & Gariepy. The part $\int_\Omega \varphi \,d\mu<\infty$ is the assumption. $\endgroup$ – spatially Aug 7 '15 at 18:19
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Disclaimer: This answer may be slightly off, since I am not that familiar anymore with the nonstandard definitions of Evans & Gariepy. Anyway...

Note that $\infty > \int_\Omega \varphi \, d\mu \geq \int_\Omega d\mu = \mu(\Omega)$.

Let $A \subset \Bbb{R}^N$. Since $\mu$ is Radon, there is $B \supset A \cap \Omega$ with $\mu(B) = \mu(A \cap \Omega)$. We can assume $B\subset \Omega$ (why?).

Since $B$ is Borel and hence $\mu$-measurable and of finite measure, we get

$$ \mu(B \setminus (A \cap \Omega)) = \mu(B) - \mu(A\cap \Omega) = 0 $$ and thus

$$ \nu((B \cup \Omega^c) \setminus A) = \int_{((B \cup \Omega^c) \setminus A) \cap \Omega} \varphi \, dx = \int_{B \setminus (A\cap \Omega)} \varphi \, dx = 0, $$ which easily yields $$ \nu(A) \leq \nu(B \cup \Omega^c) \leq \nu((B \cup \Omega^c) \setminus A) + \nu(A) = \nu(A), $$ so that $B \cup \Omega^c$ is a Borel set with $B \cup \Omega^c \supset A$ and $\nu(A) = \nu(B \cup \Omega^c)$.

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