How to show that $\mu$ is $\sigma$-finite This is part of a problem from an old prelim exam in analysis. I am studying to prepare for my own prelim.
Let $\{q_n\}=\mathbb Q$, and let $f_n : \mathbb R \to [0,\infty)$ be a Borel measurable function with $\int f_n d\lambda=1$ and with support $[q_n-2^{-n-1},q_n+2^{-n-1}]$. $\lambda$ is the Lebesgue measure.
We furthermore define a Borel measure $\mu$ by $\mu(A) := \int_A \sum_{n=1}^\infty f_n d\lambda$.
For the first part of this problem, I have already proven that $\sum_{n=1}^\infty f_n(x) < \infty\ \mathrm{a.e}$, that $\mu <\!\!<\lambda$, and that every open subset of $\mathbb R$ has infinite $\mu$-measure. Now we are asked to show that $\mu$ is $\sigma$-finite, and a I find myself fully at a loss to see how to do this. It seems very difficult to see how this could even be true, since again, every open subset of $\mathbb R$ has infinite $\mu$-measure.

If you've shown $f = \sum f_n <\infty$ on a set $A$ with $\lambda (\mathbb R\setminus A)=0,$ we can do this: For $j,k=1,2, \dots,$ let $B_{jk} = \{x\in [-j,j]\cap A: f(x)\le k\}.$ Then $A\cap [-j,j] = \cup_k B_{jk},$ and $\mu(B_{jk}) = \int_{B_{jk}} f \,d\lambda\le k\cdot 2j<\infty$ for all $j,k.$ We then have $A= \cup _{j,k} B_{j,k},$ which shows $\mu$ is $\sigma$-finite.
