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I am working to understand the relationships between the many modes of convergence. One of the true/false problems I am looking at is

If $f_n\rightarrow f$ pointwise, then a subsequence $f_{n_k}\rightarrow f$ in measure.

My initial thought is true if $\mu(X)<\infty$, since in that case $f_n\rightarrow f$ a.e. implies $f_n\rightarrow f$ in measure. Pointwise convergence is even better than almost everywhere convergence (right?), so the result should hold. If $\mu(X)=\infty$ then it is false?

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    $\begingroup$ Consider $\chi_{[n,n+1]}$ on the real line. $\endgroup$ – zhw. Nov 2 '15 at 12:49
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    $\begingroup$ As you said, if $\mu(X)<\infty$, then $f_n\rightarrow f$ a.e. implies $f_n\rightarrow f$ in measure. So we have $f_n\rightarrow f$ pointwise implies $f_n\rightarrow f$ a.e. which implies $f_n\rightarrow f$ in measure. And you don't even need to take any subsequence. HOWEVER, if $\mu(X)=\infty$, we may have $f_n\rightarrow f$ pointwise and NO subsequence $f_{n_k}\rightarrow f$ in measure. For a simple and nice example, consider $f_n=\chi_{[n,n+1]}$ on the real line, as suggested by @zhw. $\endgroup$ – Ramiro Nov 2 '15 at 13:45
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(This was answered by a comment)

As you said, if $\mu(X)<\infty$, then $f_n\rightarrow f$ a.e. implies $f_n\rightarrow f$ in measure. So we have $f_n\rightarrow f$ pointwise implies $f_n\rightarrow f$ a.e. which implies $f_n\rightarrow f$ in measure. And you don't even need to take any subsequence.

However, if $\mu(X)=\infty$, we may have $f_n\rightarrow f$ pointwise and NO subsequence $f_{n_k}\rightarrow f$ in measure. For a simple and nice example, consider $f_n=\chi_{[n,n+1]}$ on the real line, as suggested by zhw. -- Ramiro

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