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Let $U$ be a nonempty open subset of $\mathbb{R}^n$ and $\mu$ be a Radon measure on $U$. Define $$T_\mu(\phi) = \int_U \phi d\mu$$ for all $\phi \in D(U) = C_c^\infty(U)$. Prove that $T_\mu$ is a distribution on $U$ and that $supp(T_\mu) = supp(\mu)$.

I was able to shwo that $T_\mu$ is linear and continuous, and hence is a distribution. I'm having a bit of trouble showing that the supports are equal. My professor gave me a hint which was to look at any open subset $V \subseteq U$ and that $\mu(V) = 0$ iff $T_\mu = 0$ on $V$. I don't quite see how to use this hint to show the two supports are equal.

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By taking the complementary, showing that the supports are equal is equivalent to show that the biggest annihilating sets are equal.

So it is enough to show that an annihilating set for $\mu$ is an annihilating set for $T_{\mu}$ and conversly. That was the suggestion of your professor.

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  • $\begingroup$ What is an annihilating set? We've never used this term before $\endgroup$
    – Brenton
    May 19, 2016 at 17:40
  • $\begingroup$ For a distribution $T$ an annihilating set $U$ is a set for which $T(\varphi) =0$ for all $\varphi$ supported by $U$. The support of $T$ is by definition the complementary of the biggest annihilating set. (I.e the union of all annihilating sets) $\endgroup$
    – C. Dubussy
    May 19, 2016 at 17:53
  • $\begingroup$ So I need to show then that the biggest open set $U$ s.t. $T(\phi) = 0$ with $supp(\phi) \subset U$ is the same $U$ s.t. U is the union of all open sets $N$ with $\mu(N) = 0$? $\endgroup$
    – Brenton
    May 19, 2016 at 18:04
  • $\begingroup$ Yes but since it is a double inclusion, you can take only an open set and not obligatory the biggest. (If each open set is included in the other, than the union will also be.) $\endgroup$
    – C. Dubussy
    May 19, 2016 at 18:23

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