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Let $u$ be a subharmonic function in a domain $\Omega$ pf $\mathbb{C}$. The functions $u_{j} := \max(u, -j)$ still subharmonic. Let $\mu := \Delta u$ and $\mu_{j} := \Delta u_{j}$ be the associated Riesz measures (which are positive). Let $B$ be a borelian of $\Omega$.

Question : is it true that $$ \int_{B} 1_{\{\phi > -j\}}\mu_{j} \to \int_{B} 1_{\{\phi > -\infty\}}\mu\quad\text{ as }\quad j \to +\infty\;? $$


I try to prove it in the following way : let $B$ be a borelian. We then have
$$ \begin{split} 1_{\{\phi > -\infty\}}( \mu )(B) - 1_{\{\phi > -j\}}(\mu_{j})(B) & = 1_{\{\phi > -\infty\}}(\mu )(B) - 1_{\{\phi > -j\}}(\mu )(B) \\ &\quad + 1_{\{\phi > -j\}}(\mu)(B) - 1_{\{\phi > -j\}}(\mu_{j})(B) \end{split} $$ The quantity $1_{\{\phi > -\infty\}}(\mu)(B) - 1_{\{\phi > -j\} }(\mu)(B)$ converges to $0$ by monotone convergence. It then remains to estimate : $1_{\{\phi > -j\}}(\mu)(B) - 1_{\{\phi > -j\}}(\mu_{j})(B) = 1_{\{\phi > -j\}}(\mu - \mu_{j})(B)$. But I don't have enough information on the measures $\mu_{j}$ and $\mu$ to estimate this.

I wish you a good day.

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  • $\begingroup$ I wish you a good day too. $\endgroup$ Commented Aug 21, 2022 at 9:26

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This is not true. Take $u(z)=(2\pi)^{-1}\log|z|$, so that the Riesz measure $\mu=\delta_0$. Then the Riesz measure $\mu_j$ of $u_j\max\{ u,-j\}$ is the uniform measure on the circle $\{ z:|z|=e^{-j}\}$. Now let $B$ be the open right half-plane, so $\mu_j(B)=1/2$ while $\mu(B)=0$. If you the choose the closed right half-plane for $B$, then $\mu_j(B)=1/2$ while $\mu(B)=1$.

The correct statement is that $$\int\phi\, d\mu_j\to\int\phi\, d\mu,$$ for every continuous function with bounded support, that is the weak convergence of measures. This follows from three facts: 1. Laplace operator is continuous in the space $D'$ of Schwartz distributions, 2. For subharmonic functions, convergence in $D'$ is equivalent to convergence in $L^1_{\rm{loc}}$, and 3. For (positive) measures, convergence in $D'$ is equivalent to the weak convergence. The reference for these facts is L. Hormander, The analysis of linear partial differential operators, I.

Your statement with Borel sets can be true only if $\mu(\partial B)=0$.

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  • $\begingroup$ OK, thanks for this result. Could I hope that $\int_{B} 1_{\phi_{j} > -j} \psi\, d\mu_j\to\int_{B} 1_{\phi > -\infty }\psi\, d\mu,$ for all continuous test function $\psi$? Do you have references for the result you stated above? Here $B$ denotes a borelian. $\endgroup$ Commented Aug 22, 2022 at 7:58
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    $\begingroup$ @Analyse300: No, you cannot. Same counterexample. I added a reference. $\endgroup$ Commented Aug 22, 2022 at 10:48
  • $\begingroup$ OK. Thanks a lot. $\endgroup$ Commented Aug 22, 2022 at 17:37

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