Let $(X,\Sigma,\mu)$ be a measurable space with $\mu(X)<\infty$.
Prove that $\mu(\Sigma)$ is closed.
I've been stumped with this for quite a while. I've tried every usual way of showing a set is closed, to no avail. The elements in the preimage of $\mu$ are sets, meaning I can't extract subsequences like I would do with real numbers.
Here's one approach you can take. Since $\mu(X)<\infty$, there can only be countably many atoms in $X$ (modulo null sets), so let $A_1,A_2,\dots$ be all the atoms, let $A=\bigcup A_n$, and let $B=X\setminus A$. Then the restriction of $\mu$ to $B$ is atomless, and thus takes every value in the interval $[0,\mu(B)]$. Note that $\mu(\Sigma)$ is just the image of $\mu(\Sigma\cap\mathcal{P}(A))\times\mu(\Sigma\cap\mathcal{P}(B))$ under the addition map $\mathbb{R}\times\mathbb{R}\to\mathbb{R}$. To show that $\mu(\Sigma)$ is closed, it thus sufffices to show $\mu(\Sigma\cap\mathcal{P}(A))\times\mu(\Sigma\cap\mathcal{P}(B))$ is compact. Since $\mu(\Sigma\cap\mathcal{P}(B))=[0,\mu(B)]$ is compact, it suffices to show $\mu(\Sigma\cap\mathcal{P}(A))$ is compact as well.
But the restriction of $\mu$ to $A$ is atomic: every measurable subset of $A$ has the form $\bigcup_{n\in S} A_n$ for some $S\subseteq\mathbb{N}$, up to null sets. So $\mu(\Sigma\cap\mathcal{P}(A))$ is just the image of the map $m:\{0,1\}^\mathbb{N}\to\mathbb{R}$ defined by $$m(f)=\mu\left(\bigcup_{f(n)=1} A_n\right)=\sum_{f(n)=1}\mu(A_n).$$ But $m$ is continuous with respect to the product topology on $\{0,1\}^\mathbb{N}$ (for instance, by the dominated convergence theorem--this is where we use the hypothesis $\mu(X)<\infty$). Since $\{0,1\}^\mathbb{N}$ is compact, the image of $m$ is thus compact, and so $\mu(\Sigma\cap\mathcal{P}(A))$ is compact.
