# Does $\chi_{p(\lim\limits_{n\to\infty} f_n)}=\lim\limits_{n\to\infty}\chi_{p(f_n)}$?

Is $\chi_{p(\displaystyle\lim_{n\to\infty} f_n)}=\displaystyle\lim_{n\to\infty}\chi_{p(f_n)}$, assuming $\lim f_n$ exists? Here $\chi_{p(f)}$ is $1$ for the set where the proposition $p$ on the function $f$ is true and $0$ for otherwise. Here the functions are $[0,1]\to\mathbb{R}$ and the limits are almost everywhere convergence.

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What sort of functions, and what sort of convergence? – Brian M. Scott Jan 6 '13 at 14:32
@Scott, I edited the question addressing your comment. – JDL Jan 6 '13 at 14:37
One last question: is that a.e. convergence with respect to Lebesgue measure? – Brian M. Scott Jan 6 '13 at 14:40
@Scott, yes it is. – JDL Jan 6 '13 at 14:41
@Edgar: Please avoid using \displaystyle in the title. – Asaf Karagila Jan 6 '13 at 22:48

## 2 Answers

No. Try $p=$ "being bounded" and $f_n(x)=\min\{n,1/x\}$.

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$\lim\chi_{f_n}=1$ and $\chi_{\lim f_n}=1$ a.e. So, I think your example is not a counter example. – JDL Jan 6 '13 at 15:30
See revised version. – Did Jan 6 '13 at 15:46

Or: $p(f)$ means "the function $f$ is identically zero", and $f_n(x) = 1/n$ for all $x \in [0,1]$ and all $n \in \mathbb N$.

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