Q:

Does pointwise convergence imply that a subsequence converges in measure?

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?

A:

(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

