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Let $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),dx)$ be the ordinary measure space and $w:\mathbb{R}^n\to [0,\infty)$ be a measurable function. Define $\mu$ by $$ \mu(A)=\int_{A}w(x)dx, \qquad A\in \mathcal{B}(\mathbb{R}^n). $$ Does measure $\mu$ become a Radon measure or at least Borel regular measure?

Definition of Radon measure: https://en.wikipedia.org/wiki/Radon_measure

Definition of Borel regular measure: https://en.wikipedia.org/wiki/Borel_regular_measure

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The measure $\mu$ will not be generally a Radon measure unless $w$ is locally integrable. For example, if you take $w(x)=|x|^{-n}$ (with $w(0)=0$), the measure will fail to be locally finite at the origin and is therefore not Radon.

It might be convenient to consider truncated weights $w_k(x)=\min(k,w(x))$ and the corresponding measures $\mu_k$. Then each $w_k$ is locally integrable, each $\mu_k$ is Radon, and $\lim_{k\to\infty}\mu_k(A)=\mu(A)$ for every Borel set $A$.

Borel regularity (as defined on the page you link to) is a property of outer measures. You defined the measure $\mu$ on the Borel $\sigma$-algebra, so the both the Borel property and the regularity property are satisfied trivially.

It makes more sense to ask whether the measure $\mu$ is inner and outer regular. It is a Borel measure by construction. If $w$ is the indicator function of a measurable set, then $\mu$ is is inner and outer regular. This applies also when $w$ is a simple function. Inner regularity for general measurable $w$ follows from standard arguments. I don't immediately see whether $\mu$ is necessarily outer regular if $w$ is not locally integrable.

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