$L^\infty(X,\mu)$ may fail to be a von Neumann algebra

After discussing here, we finally notice that $L^\infty(X,\mu)$ may fail to be a von Neumann algebra. Here is a counter example.

Let $X$ be an uncountable set, $\mu$ be a counting measure, and $\Sigma$ be the family of set $A$ such that $A$ or the complement of $A$ is countable. Let $S$ be a subset of $X$ such that both itself and its complement are uncountable, then the bounded subset $\{\chi_A|A \mbox{ is a countable subset of }S\}$ in $L^\infty(X,\mu)$ has no least upper bound.

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