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My question is to prove this: Let $\mu$ be counting measure on $\mathbb{N}$. Then $f_n \to f$ in measure iff $f_n \to f$ uniformly.

This is a homework question, so if you give just hints for the right approach to the question (for the both directions), I will be very happy. Thanks.

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    It will be an empty set am I right? Then the set $E_{n,\epsilon} = \{x: |f_n - f| \geq \epsilon\} = \emptyset$. Then, I can sense that the uniform convergence is related to $E_{n,\epsilon}^c$ but... (Thanks!)2012-11-21

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  • $\Rightarrow$: fix $\delta>0$; we have $\mu\{k\in\Bbb N,|f_n(k)-f(k)|>\delta\}<1/2$ for $n$ large enough. As $\mu(A)<1$ if and only if $A$ is empty, for $n$ large enough and all $k$, $|f_n(k)-f(k)|<\delta$ so $f_n\to f$ uniformly.
  • $\Leftarrow$: fix $\delta>0$; and $n_0$ such that $\sup_{k\in\Bbb N}|f_n(k)-f(k)|<\delta$ whenever $n\geqslant n_0$. For $n\geqslant n_0$, what is $\mu\{k\in\Bbb N,|f_n(k)-f(k)|>\delta\}$?