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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.

user0430
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1 Answers1

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We, essentially, (as always) pass to a subsequence $\{g_n\}_{n\in\Bbb N}$ which is chosen so that, if $E_j=\{x:|g_j(x)-g_{j+1}(x)|\ge 2^{-j}\},$ then $\mu(E_j)\le 2^{-j}.$

Let $F_k=\cup_{j=k}^\infty E_j,$ so $\mu(F_k)\le 2^{1-k}$ by subadditivity of $\mu.$

For $x\notin F_k$ and $i\ge j\ge k,$ you can show $$|g_j(x)-g_i(x)|\le\sum_{\ell=j}^{i-1}|g_{\ell+1}(x) - g_\ell(x)| \leq 2^{1-j}$$ by definition of the subsequence. So, $\{g_n\}_{n\in\Bbb N}$ is pointwise Cauchy on $F_k^C.$

Let $F=\cap_{k=1}^\infty F_k$ (this is the $\lim\sup$ of the $E_j$'s), which has $\mu(F)= 0.$

Then, on $F,$ let $f=0,$ and on $F^C,$ let $f(x):=\lim\limits_{j\to\infty}g_j(x)$.

Then, $g_j \to f $ a.e., and $|g_j(x)-f(x)|\le 2^{1-j}$ for $x \in F_k^C$ and $j \ge k.$ Then, note $\mu(F_k)\to 0$ and $g_j\to f$ in measure.

Finally, write $$\{|f_n(x)-f(x)|\ge\varepsilon\}\subset\left\{|f_n(x) -g_j(x)|\ge\frac\varepsilon2\right\}\cup\left\{|g_j(x) -f(x)|\ge\frac\varepsilon2\right\}$$ 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.

Matcha Latte
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Batman
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    Sorry to bother you after 6 years, but could you explain why $g_j \to f$ in measure? – Sorfosh Feb 13 '20 at 19:01
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    @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 $$\lim_{j \to \infty}\mu{x : |g_j(x) - f(x)| > \epsilon} = 0.$$ – Aryaman Maithani Feb 21 '21 at 07:56
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    I think the easy way to observe this is to take the limit $i \to \infty$ when you estimate $g_j (x) - g_i(x)$ – offonatangent May 24 '21 at 18:21
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    Note: one way to pass to such a subsequence is to take $N_j$ to be such that it satisfies the Cauchy condition with $\varepsilon = 2^{-j}$, and then take $g_j = f_{N_j}$. – mboratko Oct 27 '21 at 18:23