let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and $f_n, f:X\longrightarrow\mathbb{R}$ be measurable. show that $f_n\xrightarrow{\text{m}}f$ iff every subseq of $\{f_n\}$ has a subseq converging to $f$ a.e.($f_n\xrightarrow{\text{m}}f$ is the convergence in measure).

by the definition of the convergence in measure, $f_n\xrightarrow{\text{m}}f$ if for all $\alpha, \epsilon>0$, there exists $N>0$ s.t. $n>N\Longrightarrow\mu(\{|f_n-f>\alpha|\})<\epsilon$.

how do I show this without simply applying Egoroff's theorem?

btw, this is the 4th statement in Properties of this page.


If you have a sequence all of whose subsequences have a convergent subsequence to a common limit, that sequence is convergent.

Apply this fact, along with the fact that a.e. convergence implies convergence in measure, and you are there.

  • $\begingroup$ Thanks. I forgot a clause. $\endgroup$ – ncmathsadist Jun 21 '17 at 13:25

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