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If $f_n \rightarrow f$ in weak $L^p$, then is it true that $f_n \rightarrow f$ in measure?

Where the weak $L^p$ norm is given by: $[f]_p = \sup_{t>0} t \mu(\{x : |f(x)| > t \})^{\frac{1}{p}} < \infty$.

I've been having very little success finding a proof or counter example. It's hard for me to wrap my mind around the weak $L^p$ norm. I'd appreciate any insight!! Thanks

Davide Giraudo
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1 Answers1

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This follows from the inequality$$ \mu\left\{x: \lvert f_n(x)-f(x)\rvert \gt \varepsilon\right\}\leqslant \varepsilon^{-p} [f_n-f]_p^p. $$

Davide Giraudo
  • 181,608
  • So we have:

    $\mu\left{x: \lvert f_n(x)-f(x)\rvert \gt \varepsilon\right}\leqslant \varepsilon^{-p} [f_n-f]p^p=\epsilon^{-p} \sup{t>0} t \mu({x : |f(x)| > t })$

    Is there a reason this inequality is obvious?

     –  Mar 02 '20 at 18:00
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    In your comment, it should be $t^p$ and not $t$ appearing after the $\sup$. Perhaps that makes it more obvious? – Willie Wong Mar 02 '20 at 18:19