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if $\Omega$ consists of a finite number atoms then $L^1(\Omega)$ is a finite dimension. why you can show this?

(An atom in $(X, \Omega, \mu)$ is a measurable set $E$ such that $\mu(E)>0$ and for every measurable subset $F\subset E$ either $\mu(F)=0$ or $\mu(F)=\mu(E)$.)

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  • $\begingroup$ This might help: math.stackexchange.com/questions/866312/… $\endgroup$ – Prahlad Vaidyanathan Jun 17 '15 at 8:35
  • $\begingroup$ I dont really understand the question. Do you mean that $L^1$ is finite dimensional iff we can write $\Omega$ as a finite union of atoms? $\endgroup$ – PhoemueX Jun 17 '15 at 9:43
  • $\begingroup$ this is in the question : The space $L^1(\Omega)$ is never reflexive except in the trivial case where$\Omega$ consists of a finite number of atoms—and then $L^1(\Omega)$ is finite-dimensional $\endgroup$ – shahab Jun 17 '15 at 9:55
  • $\begingroup$ The statement "if $\Omega$ has finite atoms" is not the same as saying that $\Omega$ is finite and every $\omega \in \Omega$ is an atom. $\endgroup$ – DisintegratingByParts Jun 17 '15 at 15:43
  • $\begingroup$ question is relation between $L^1(\Omega)$ and $\Omega$ $\endgroup$ – shahab Jun 17 '15 at 15:47

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