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Given a sequence $f_n \in L^p$ and $g \in L^p$, with $|f_n| \leq g$, I am trying to show that $f_n \to f$ in measure implies $f_n \to f$ in $L^p$.

Firstly, I know that if $f_n \to f$ in measure, then there is a subsequence $f_{n_i}$ such that $f_{n_i} \to f$ almost everywhere. Then I can use the dominated convergence theorem to show that $\lVert f_{n_i} - f_p\rVert \to 0$.

Now I am trying to show that $\lVert f_n - f\rVert_p \to 0$. My idea is to assume that $\lVert f_n - f\rVert_p \nrightarrow 0$ and then show that this contradicts the fact that $\lVert f_{n_i} - f\rVert_p \to 0$, but I am not sure of the details. Can anyone help me finish the argument?

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    $\begingroup$ In the title, you should add "with an hypothesis of domination". You are almost done: if $||f_n-f||_p$ doesn't converge to $0$, then we can find $\delta>0$ and a subsequence such that $||f_{n_k}-f||\geq \delta$. This subsequence still converges in measure to $f$, so by your previous argument we get a contradiction. $\endgroup$
    – Davide Giraudo
    Oct 18, 2012 at 19:01
  • $\begingroup$ @DavideGiraudo Ok, got it. Thanks! $\endgroup$
    – rt93
    Oct 18, 2012 at 19:16
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    $\begingroup$ You can answer your own question (hence it won't remain unanswered), and you will have your homework done properly. $\endgroup$
    – Davide Giraudo
    Oct 18, 2012 at 19:17
  • $\begingroup$ @DavideGiraudo : as one commenter makes note of below, we haven't used the boundedness of $f_n$ at all. Is this a problem? $\endgroup$
    – Alex
    Aug 8, 2017 at 18:08
  • $\begingroup$ Also, this question was featured on the UI-Urbana Champaign graduate qualifying exam from August 2015, if anyone cares to know. $\endgroup$
    – Alex
    Aug 8, 2017 at 18:09

1 Answer 1

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If $\lVert f_n - f\rVert_p \nrightarrow 0$, there exist a subsequence $ f_{n_i} $ such that $ \| f_{n_i} - f\| _p \ge \varepsilon $ for some $ \varepsilon >0 $. But $ f_{n_i} $ still converges in measure to $ f $. So, again There is a subsequence $ f_{n_{i_j}} $ of the $ f_{n_i} $ such that $ \|f_{n_{i_j}} - f\|_p \rightarrow 0$, and this is a contradiction.

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    $\begingroup$ How does the boundedness of $\|fn\|$ work in your argument? $\endgroup$
    – hil316
    Aug 10, 2016 at 3:30
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    $\begingroup$ To claim the fact that $\|f_{n_{i_j}} - f\|_p \rightarrow 0$ you would need to apply Dominated convergence theorem in $L^p$ setting, there you would need boundedness of $f_n$ $\endgroup$
    – Savannah
    Feb 11, 2017 at 6:56
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    $\begingroup$ you need this information to apply the domited convergence theorem. $\endgroup$
    – user29999
    Mar 14, 2017 at 22:09
  • $\begingroup$ how do you use the dominated convergence? $\endgroup$
    – the crazy
    Feb 10, 2019 at 18:28

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