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Let $D$ be a domain of $\mathbb{R}^{m}$ and let $K(x)= \log|x|$ if $m=2$, and $K(x)=|x|^{2-m}$ if $m>2$. According to Riesz decomposition theorem (Hayman and Kennedy, "subharmonic functions", vol. 1, pg 104) if $u$ is subharmonic on $D$, then there is a unique Borel measure $\mu$ such that for all compact $E$ in $D$ we have $$u(x)=\int_{E}K(x-\zeta)d\mu(\zeta)+h(x)$$ where $h$ is harmonic on the interior of $E$.

Here is my question: is this equation valid for all $x\in D$ or all $x\in E$?

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    $\begingroup$ It holds only for $x\in E$. Since the measure is not restricted to $E$, the function does not have to be harmonic outside $E$. $\endgroup$ Sep 21 '20 at 23:29
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    $\begingroup$ It depends on the convention. You can also understand the claim as $u$ can be represented in the above form in $D$ with $h$ that is harmonic in the interior of $E$ (and subharmonic in the rest of the domain). The key point is that for every compact subset of $D$, the Riesz measure is finite, so the potential is well-defined but this may be not so in the entire domain $D$. $\endgroup$
    – fedja
    Sep 22 '20 at 1:57
  • $\begingroup$ But why do you say that h is subharmonic in the rest of the domain? $\endgroup$
    – M. Rahmat
    Sep 22 '20 at 15:31
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(You need a minus sign in front of $\int_E K(x-\zeta)\,d\mu(\zeta)$.)

The integral $\,-\!\!\int_E K(x-\zeta)\,d\mu(\zeta)$ is subharmonic throughout $D$ and harmonic in $D\setminus E$. If $u$ admits a harmonic majorant in $D$, hence a least such majorant (call it $k$), then $h$ (which depends on $E$) can be expressed as $\,-\!\!\int_{E^c} K(x-\zeta)\,d\mu(\zeta)+k(x)$ for $x\in D$.

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