Suppose we have a measure space $(\Omega, \Sigma, \mu)$ and measurable functions $f_n \colon \Omega \to \mathbb R$. Is it true, that if the sequence $f_n$ is convergent in measure, then it is necessarily a Cauchy sequence in measure?

EDIT: Cauchy sequence in measure means: $$(\forall \varepsilon > 0)(\forall \eta >0)(\exists n_0 \in \mathbb N)(\forall p, q \ge n_0)(\mu(\{x \in A: |f_p(x) - f_q(x)| \ge \varepsilon\}) < \eta)$$

  • $\begingroup$ cauchy in which topology, metric? $\endgroup$ – Quickbeam2k1 Jan 15 '15 at 19:51
  • 1
    $\begingroup$ I've added a clarification. $\endgroup$ – user207868 Jan 15 '15 at 20:01

They are actually equivalent conditions. As for the implication you're interested in, suppose $f_n\to f$ in measure. Let $\varepsilon,\eta > 0$. Then we can choose $N$ so that $n\geqslant N$ implies $$\mu(\{\omega\in\Omega : |f_n(\omega)-f(\omega)|\geqslant\varepsilon/2\})<\eta/2. $$ Now $$|f_n(\omega)-f_m(\omega)|\leqslant |f_n(\omega)-f(\omega)|+|f_m(\omega)-f(\omega)|, $$ so if $|f_n(\omega)-f_m(\omega)|\geqslant\varepsilon$, then $|f_n(\omega)-f(\omega)|+|f_m(\omega)-f(\omega)|\geqslant\varepsilon$. As $$\{\omega: |f_n(\omega)-f(\omega)|+|f_m(\omega)-f(\omega)|\geqslant \varepsilon\}\subset\{\omega : |f_n(\omega)-f(\omega)|\geqslant\varepsilon/2\}\cup\{\omega : |f_m(\omega)-f(\omega)|\geqslant\varepsilon/2\}, $$ for $n,m\geqslant N$ we have $$\mu(\{\omega : |f_n(\omega) - f_m(\omega)|\geqslant\varepsilon\})\leqslant\mu(\{\omega : |f_n(\omega)-f(\omega)|\geqslant\varepsilon/2\})+\mu(\{\omega : |f_m(\omega)-f(\omega)|\geqslant\varepsilon/2\})<\eta,$$ so that $f_n$ is Cauchy in measure.


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