TITLE: A question about Rudin's proof of Lusin's theorem
QUESTION [3 upvotes]: In page 56 of Rudin's Real and Complex Analysis, it's stated:

[I]f $f$ is a complex measurable function and $B_n=\{x:|f(x)|>n\}$, then $\bigcap B_n= \varnothing$.

My question is why?

REPLY [3 votes]: By definition, a complex function is a map $X\to\Bbb C$. If $f$ is complex, then the set $\bigcap B_n$ is the set of all points $x$ such that $|f(x)| > n$ for every $n\in\Bbb N$. There is no complex number $z$ with the property that $|z|>n$ for every natural number $n$. Hence there is no such $x\in\bigcap B_n$, so $\bigcap B_n = \varnothing$.
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