TITLE: Measure without sets of measure zero
QUESTION [5 upvotes]: Let $(X,\Sigma,\mu)$ be a finite $\sigma$-additive measure space such that 
$$
E\in\Sigma,\quad\mu(E)=0\implies E=\varnothing
$$
Is it true that $(X,\Sigma,\mu)$ is purely atomic? You may assume that $\mu$ is complete.
I think I know how to prove this via Maharam theorem, but that looks like an overkill. I wonder if there is a simpler proof or maybe my guess is even wrong?

REPLY [3 votes]: For each $x\in X$ it's easy to show that $x\in E_x\in\Sigma$ with $$\mu(E_x)=\inf\{\mu(E): x\in E\in\Sigma\}.$$Now $x\in E_x$ shows that $\mu(E_x)>0$.
And $E_x$ is an atom: Suppose $A\subset E_x$ and $\mu(A)>0$. If $x\in E_x\setminus A$ then $\mu(E_x\setminus A)<\mu(E_x)$ contradicts the minimality of $\mu(E_x)$. So $x\in A$; now again the choice of $E_x$ shows that $\mu(A)=\mu(E_x)$; so $\mu(E_x\setminus A)=0$, hence $E_x\setminus A=\emptyset$, or $A=E_x$.
So $X$ is the union of atoms, and it's clear that the set of atoms is countable.
Quiz Question: Where did we use the fact that $\mu$ is finite?