I am trying to proof the following:

Let $(\Omega,\mu)$ be a measure space, $1<p<\infty$. Then weak convergence of $(f_n)$ to $f$ in $L^p(\Omega,\mu)$ is equivalent to $(f_n)$ being bounded and $\int_M f_n \to \int_M f$, for any measurable $M$ with finite $\mu$-volume.

I have a proven "$\Rightarrow$", by stating that weak convergence implies boundedness in general, and that integration over $M$ is a linear map to the underlying field i.e. a linear functional.
But I do not know how to approach the other direction. Of course I would be very happy to show this direction as well without making use of the isomorphism $(L^p)^* \to L^q$ for $p,q$ Hölder-conjugate, but I have not found a way yet.

(Unfortunately I don't know that much about measure theory, this is kind of a problem for me here.)

  • $\begingroup$ What is $M{}{}{}$? $\endgroup$
    – Jose27
    May 28, 2020 at 16:44
  • $\begingroup$ @Jose27 Oh sorry I forgot about that, I edited my question. $\endgroup$
    – Dominik
    May 28, 2020 at 16:51
  • 2
    $\begingroup$ Sketch: Write $\int_M f= \int_\Omega 1_Mf$, where $1_M$ is the indicator function of $M$ (i.e. it's one on $M$ and zero otherwise). Use this to show that $\int_\Omega \varphi f_n \to \int_\Omega \varphi f$ for every simple function $\varphi$. $\endgroup$
    – Jose27
    May 28, 2020 at 17:04
  • $\begingroup$ Are you assuming that $\mu$ is $\sigma$-finite? $\endgroup$
    – Michh
    May 28, 2020 at 17:09
  • 1
    $\begingroup$ We use density of simple functions in $L^q$ (that this works will require boundedness of the sequence). As for a proof without appealing to the Riesz representation of $(L^p)^*$ I'm not sure at the moment. $\endgroup$
    – Jose27
    May 28, 2020 at 17:18


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