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Suppose $\phi:\mathbb{R}\rightarrow \mathbb{R}$ is strictly increasing, continuous function and $f_n$ is sequence of measurable function. Prove that if $\phi(f_n)$ is Cauchy in $L^1$, then there exists a subsequence $f_{n_i}$ converges almost everywhere.

How can I prove it? I can only get a subsequence of $\phi(f_{n_i})$ converges almost everywhere.

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  • $\begingroup$ Good start. Now use the assumptions on $\phi$. To begin, consider a sequence $g_n\colon \Omega \to \mathbb R$ such that $\phi(g_n)$ converges everywhere. (Here I call $\Omega$ the measure space on which the functions $f_n$ are defined). Prove that $g_n$ converges everywhere. Later you will show that the same holds for sequences converging almost everywhere. $\endgroup$ – Giuseppe Negro Feb 1 '18 at 13:03
  • $\begingroup$ Can i prove in this way? Suppose $\phi(g_n)$ converges to $g$ a.e. and $g_n$ has no convergent subsequence. Fix $x$, there are $N$ and $\epsilon_0$ such that $|g_n(x)-g_m(x)| >\epsilon_0$ for all $n$,$m>N$. Then we pick a subsequence $g_{n_k}$ from $g_n$ such that $g_{n_k}$ is increasing and $g_{n_k}-g_{n_l}>\epsilon_0$ for all $k>l>N$. $\endgroup$ – mnmn1993 Feb 1 '18 at 14:54
  • $\begingroup$ Let $u$ be given, then by continuity and monotonicity of $\phi$, if $0\leq u-v < \delta$, then $\phi(u)-\phi(v)< \epsilon$. Here we choose $\delta $ to be maximal such that if $\delta<u-v$,then $\phi(u)-\phi(v)\geq \epsilon$, this is possible by monotonicity of $\phi$. Choosing $\epsilon$ small enough such that $g_{n_k}(x)-g_{n_N}(x)>\epsilon_0>\delta$, we have $\phi(g_{n_k})-\phi(g_{n_N})\geq \epsilon$ for all $k> N$. Contradicting that $phi(g_{n_k}(x))$ is convergent. $\endgroup$ – mnmn1993 Feb 1 '18 at 14:59
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    $\begingroup$ Maybe that works, I don't know, unfortunately I have no time to check right now. I had something simpler in mind. Notice that Phi is invertible with a continuous inverse. $\endgroup$ – Giuseppe Negro Feb 1 '18 at 15:04
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    $\begingroup$ That's a property of strictly monotone functions, not a general fact (inverses of continuous functions might be discontinuous). That link won't help, look elsewhere $\endgroup$ – Giuseppe Negro Feb 1 '18 at 16:03

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