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Let $X$ be a bounded domain of $\mathbb{R}^{n}$ ($n\geq2$), and $T$ a linear functional on the space $C_{0}(X)$ of continuous functions with compact support in $X$, that is positive (i.e. if $\phi\geq0$, then $T(\phi)\geq0$). We know that according to Riesz representation theorem there is a (unique) Borel measure $\mu_{T}$ such that $$T(\phi)=\int_{X}\phi d\mu_{T}$$ for all $\phi \in C_{0}(X)$. My question is: suppose $E$ is a Borel set with Lebesgue measure zero. Can we conclude that $\mu_{T}(E)=0$? If not, is there a sufficient condition in terms of Lebesgue measure of $E$ that will imply $\mu_{T}(E)=0$?

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Every positive regular Borel measure corresponds to some positive operator $T$. Hence you cannot expect special properties of $\mu$ like absolute continuity w.r.t Lebesgue measure. In particular $\mu$ may be singular w.r.t. Lebesgue measure

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