A sequence $\{f_n\}$ of measurable functions is said to be a Cauchy sequence in measure if, given $ϵ > 0$, there is an $N$ such that for all $m, n ≥ N$ we have $m\{x \in E : |f_n(x) − f_m(x)| \ge ϵ\} < ϵ$.

Show that if $\{f_n\}$ is Cauchy in measure, then there exists a measurable function $f$ to which the sequence $\{f_n\}$ converges in measure.

Idea: I need to show that $m\{x \in E : |f(x) − f_n(x)| \ge ϵ\} \rightarrow 0$ for some measurable function $f$. I was thinking that there exists $f(x) = \lim_{k\to \infty} f_{n_k}(x)$ where $f_{n_k}$ is a subsequence. But I am unclear as to where to proceed.


We essentially (as always) pass to a subsequence $\{g_n\}$ which is chosen such that if $E_j = \{x: |g_j(x) - g_{j+1}(x)| \geq 2^{-j}\}$ then $\mu(E_j) \leq 2^{-j}$. Let $F_k = \cup_{k}^\infty E_j$, so $\mu(F_k) \leq 2^{1-k}$ by subadditivity of $\mu$. For $x \notin F_k$ and $i \geq j \geq k$, you can show $$|g_j(x) - g_i(x)| \leq \sum_{l=j}^{i-1} |g_{l+1}(x) - g_l(x)| \leq 2^{1-j}$$ by definition of the subsequence.

So, $\{g_n\}$ is pointwise Cauchy on $F_k^C$. Let $F = \cap_1^\infty F_k$ (this is the limsup of the $E_j$'s), which has $\mu(F) = 0$. Then, on $F$, let $f=0$, and on $F^C$ let $f(x) = \lim_j g_j(x)$. Then, $g_j \to f $ a.e., and $|g_j(x) - f(x)| \leq 2^{1-j}$ for $x \in F_k^C$ and $j \geq k$. Then, note $\mu(F_k) \to 0$ and $g_j \to f$ in measure.

Finally, write $$\{|f_n(x) - f(x)|\geq \epsilon\} \subset \{|f_n(x) -g_j(x)| \geq 1/2 \epsilon\} \cup \{|g_j(x) -f(x)| \geq 1/2 \epsilon\}$$ and note that the right hand side can be made small with large $n$,$j$, and thus, you have convergence in measure.

This is taken from Theorem 2.30 in Folland's Real Analysis: Modern Techniques and their Applications 2e.

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    $\begingroup$ Sorry to bother you after 6 years, but could you explain why $g_j \to f$ in measure? $\endgroup$ – Sorfosh Feb 13 '20 at 19:01
  • $\begingroup$ @Sorfosh: Given $\epsilon > 0$, for $j$ large enough, we have that $2^{1 - j} < \epsilon.$ Then, $\{x : |g_j(x) - f(x)| > \epsilon\} \subset F_j$ and $\mu(F_j) \le 2^{1 - j}$. This shows that $$\lim_{j \to \infty}\mu\{x : |g_j(x) - f(x)| > \epsilon\} = 0.$$ $\endgroup$ – Aryaman Maithani Feb 21 at 7:56
  • $\begingroup$ I think the easy way to observe this is to take the limit $i \to \infty$ when you estimate $g_j (x) - g_i(x)$ $\endgroup$ – offonatangent May 24 at 18:21

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