## long article test 2

long article with excerp.

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[b]Regarding[/b] Remark 0, I don’t think $X_+ := \cup_n E_n$ works, If the first one $E_1$ is $X$ then the union is also $X$. There is an nontrivial example where $\mu(E_n)$ converges to $\sup \{ \mu(E): E \in {\mathcal X}\}$ while $\limsup E_n = X$ and $\liminf E_n = \emptyset$.
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