Continuous Measures: Range Let $\Omega$ be a sigma-finite measure space with no atoms.
(Reminder: A subset $A\in\Sigma$ is an atom if $\mu(E)<\mu(A)$ implies $\mu(E)=0$ for all $E\subseteq A$.)
Then the measure attains every number $\mu(F)=c$ for $\mu(\varnothing)\leq c\leq\mu(\Omega)$.
By sigma-finiteness there are subsets of arbitrarily large but finite mass and since it is atomless there are subsets of arbitrarily small but nonvanishing mass.
How can I combine these two properties to reach any mass up to any error?
 A: For a finite atomless measure $\mu$, one can show that $\mu$ attains every value in $[0, \mu(\Omega)]$. A sketch of the proof, taken from this Wikipedia page (http://en.wikipedia.org/wiki/Atom_%28measure_theory%29#Non-atomic_measures) is as follows
Let
$$
\Gamma := \{ \varphi : M \to \mathcal{A} \mid \{0, \mu(\Omega)\} \subset M \text{ and } \varphi \text{ has property } \ast\},
$$
Here we say that a map $\varphi : M \to \mathcal{A}$, where $\mu : \mathcal{A} \to [0,\mu(\Omega)]$, has property $\ast$, if


*

*$\mu(\varphi (s)) = s$ for all $s \in M$,

*$\varphi(s) \subset \varphi(t)$ for $s,t \in M$ with $s \leq t$.


It is easy to see that $\Gamma \neq \emptyset$ and that $\Gamma$ is inductively ordered by "$\subset$" (i.e. $\varphi \leq \psi$ if $\psi$ is an extension of $\varphi$).
By Zorn's Lemma, $\Gamma$ has a maximal element $\varphi_0 : M_0 \to \mathcal{A}$. Using the property that $\mu$ is atomless, one can show that $M_0 = [0,\mu(\Omega)]$.
Now, if $\mu$ is $\sigma$-finite, and atomless, $\mu$ still attains every value of $[0, \mu(\Omega)]$. In the case $\mu(\Omega) < \infty$, this follows from the above.
Otherwise, let $\alpha < \infty$ be arbitrary. Using $\sigma$-finiteness of $\mu$, it is easy to construct $\Omega ' \subset \Omega$ with $\alpha < \mu(\Omega') < \infty$. Now apply the above to $\mu$ restricted to (the measurable subsets of) $\Omega'$, to conclude the proof.
