# 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?

• Consider $\chi_{[n,n+1]}$ on the real line. – zhw. Nov 2 '15 at 12:49
• 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 Nov 2 '15 at 13:45

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