# A question about Riesz decomposition theorem

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 $$E$$.

I have two questions:

1) Suppose $$E$$ is a compact of $$D$$ with no interior. Then what happens to the above formula? Does it still hold?

2) Is it true that the function $$h$$ is given by the integral of $$u$$ over the boundary of $$\partial E$$ of $$E$$ with respect to the harmonic measure?

1) The formula is then meaningless: $$h$$ can be completely arbitrary (as there is no interior of $$E$$).
2) Yes, this is true. Consider a larger compact set $$E'$$ which is contained in $$D$$ and such that the interior of $$E'$$ contains $$D$$. Apply the decomposition theorem to $$E'$$ to get $$u(x) = \int_{E'} K(x - \xi) d\mu(\xi) + h'(x).$$ Then $$h(x) = h'(x) - \int_{E' \setminus E} K(x - \xi) d\mu(\xi).$$ Both $$h'$$ and $$K(x - \xi)$$ (for $$\xi \in E' \setminus E$$) are given as integrals with respect to the harmonic measure of $$E$$, and it remains to use Fubini's theorem.
• If $E$ is a compact with empty interior can we steel write formally $$u(x)=\int_{E}K(x-\zeta)d\mu{\zeta)+\int_{E}u(\zeta)d\omega_{x}(\zeta)$$ where $\omega_{x}$ is the harmonic measure at $x$? Of course it does not make sense to talk about harmonicity of the function defined by the harmonic mesure, but is such equation holds formally? – M. Rahmat Sep 11 '19 at 3:03
• Yes, the integral not only makes sense, but it can well be non-zero. Just imagine that $\mu$ is the Lebesgue measure, and $K$ is a fat Cantor set. – Mateusz Kwaśnicki Sep 11 '19 at 5:52