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Let $f_1, f_2, \ldots$, and $g$ be measurable complex-valued functions on the open unit disk. We say that the sequence $f_1, f_2, \ldots$ converges in measure to $g$ if, for all $\epsilon, \mu >0$, the Lebesgue measure of $\{z : |f_i(z)-g(z)| > \epsilon\}$ is eventually smaller than $\mu$.

Assume that $f_1, f_2, \ldots$ is a sequence of meromorphic functions that converges in measure to $g$. Does it follow that $g$ is meromorphic modulo a set of measure zero?

This question may be too elementary for mathoverflow. I haven't thought seriously about complex analysis in quite some time.

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    $\begingroup$ Can you get a counterexample with something like $f_n = \sum_{i=1}^n \frac{c_i}{z_i}$, where $c_i$ is some sequence of constants decaying sufficiently fast, and $z_i$ is a sequence of distinct points that has an accumulation point? The limiting function couldn't be meromorphic because its singularities would not be isolated. $\endgroup$
    – Nate Eldredge
    Commented Dec 4, 2023 at 6:51
  • $\begingroup$ @NateEldredge You mean $f_n(z) = \sum_{k=1}^n c_k/(z-z_k)$? $\endgroup$
    – Christophe Leuridan
    Commented Dec 5, 2023 at 8:46
  • $\begingroup$ @ChristopheLeuridan: Yes I did, thanks for the correction. $\endgroup$
    – Nate Eldredge
    Commented Dec 5, 2023 at 14:47

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