# Let T be a countable set, <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="script">A

Let $T$ be a countable set, and $\mu :\mathcal{A}\to \left[0,\mathrm{\infty }\right)$ definided for every $A\in \mathcal{A}$ by

Then every set $A\in \mathcal{A}$, such that ${A}^{c}$ is finite, is an atom with respect of $\mu$.
But I can't see how it is concluded that every set $A\in \mathcal{A}$, such that ${A}^{c}$ is finite, is an atom with respect of $\mu$, can someone help me?
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Maggie Bowman
If ${A}^{c}$ is finite, then A must be infinite. Otherwise, $T=A\cup {A}^{c}$ would be finite. So $\mu \left(A\right)=1$. Now, if $E\subseteq A$ and $E\in \mathcal{A}$, then, by the latter condition, E is finite and satisfies $\mu \left(E\right)=0$, or ${E}^{c}$ is finite, which , again. implies that E is infinite and $\mu \left(E\right)$. So either $\mu \left(E\right)=\mu \left(A\right)$ or $\mu \left(E\right)=0$ for all $E\subseteq A$ with $E\in \mathcal{A}$, which is exactly the definition of A being an atom.