# atom and measurable set

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 $\mu(F)=\mu(E)$.)

• This might help: math.stackexchange.com/questions/866312/… – Prahlad Vaidyanathan Jun 17 '15 at 8:35
• 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? – PhoemueX Jun 17 '15 at 9:43
• 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 – shahab Jun 17 '15 at 9:55
• 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. – DisintegratingByParts Jun 17 '15 at 15:43
• question is relation between $L^1(\Omega)$ and $\Omega$ – shahab Jun 17 '15 at 15:47