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I tried this exercise all the past week, but a I can´t do it.

We know that if $\{f_n\}$ is Cauchy in measure, then, there is a measurable function $h$ such that $\{f_n\}$ converges in measure to $h$. Moreover, there is a subsequence that converges to $h$ almost uniformly and almost everywhere to $h$, but...

If $\{f_n\}$ is Cauchy in measure and there are subsequences $\{f_{n_k}\}$ and $\{f_{m_k}\}$ and measurable functions $f,g$ such that $\{f_{n_k}\}$ and $\{f_{m_k}\}$ converge to $f$ and $g$ almost everywhere, respectively, then $f=g$ almost everywhere.

I tried to show that $f=h=g$ a.e., but I can't solve it. Any help (or hint) is welcome.

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    $\begingroup$ If $\{f_n\}$ is Cauchy in measure, then, there is a measurable function $h$ such that $\{f_n\}$ converges in measure to $h$. Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$ such that there is a measurable function $f$ such that $\{f_{n_k}\}$ converges to $f$ a.e.. Since $\{f_{n_k}\}$ is a subsequence of $\{f_n\}$, $\{f_{n_k}\}$ converges to $h$ in measure. So, there is a subsequence $\{f_{n_{k_i}}\}$ of $\{f_{n_k}\}$ which converges to $h$ almost uniformly and a.e.. But, since $\{f_{n_{k_i}}\}$ is a subsequence of $\{f_{n_k}\}$, $\{f_{n_{k_i}}\}$ also converges to $f$ a.e.. So $f=h$ a.e.. $\endgroup$ – Ramiro Nov 11 '15 at 12:46
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    $\begingroup$ (In a similar way you can conclude $g=h$ almost everywhere. And so $g=f$ almost everywhere). $\endgroup$ – Ramiro Nov 11 '15 at 12:47
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Answered by Ramiro in a comment, quoted below:

If $\{f_n\}$ is Cauchy in measure, then there is a measurable function $h$ such that $\{f_n\}$ converges in measure to $h$. Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$ such that there is a measurable function $f$ such that $\{f_{n_k}\}$ converges to $f$ a.e. Since $\{f_{n_k}\}$ is a subsequence of $\{f_n\}$, $\{f_{n_k}\}$ converges to $h$ in measure. So, there is a subsequence $\{f_{n_{k_i}}\}$ of $\{f_{n_k}\}$ which converges to $h$ almost uniformly and a.e. But, since $\{f_{n_{k_i}}\}$ is a subsequence of $\{f_{n_k}\}$, $\{f_{n_{k_i}}\}$ also converges to $f$ a.e. So $f=h$ a.e.

In a similar way you can conclude $g=h$ almost everywhere. And so $g=f$ almost everywhere.

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