# boundedness and convergence under integral implies weak convergence

I am trying to proof the following:

Let $$(\Omega,\mu)$$ be a measure space, $$1. Then weak convergence of $$(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.)

• What is $M{}{}{}$? May 28, 2020 at 16:44
• @Jose27 Oh sorry I forgot about that, I edited my question. May 28, 2020 at 16:51
• 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 $\varphi$. May 28, 2020 at 17:04
• Are you assuming that $\mu$ is $\sigma$-finite? May 28, 2020 at 17:09
• 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. May 28, 2020 at 17:18