Preimage of Open Ball has Finite Measure

Question

I am working through example 3.2.16 in Ringrose and Kadison's book on Operator Algebras. Let $(X, \mu)$ be a $\sigma$-finite measure space, $f \in L^{\infty}(X, \mu)$, and let $\lambda \in \Bbb{C}$ be such that $\mu (f^{-1}(U)) > 0$ for each open subset $U$ of $\Bbb{C}$ containing $\lambda$. Why is it the case that $f^{-1}(B(\lambda, \frac{1}{n})$ will have finite positive measure? Why couldn't it be infinite?

Answer 1

As stated, it's not true. Consider $L^\infty(\mathbb R,m)$, $f=\lambda=1$. Then $f^{-1}(B(1,\tfrac1n))=\mathbb R$.

That said, the assertion in your question is nowhere to be seen in Kadison Ringrose's Example 3.1.16. The only thing that I see that is similar, is when they consider a subset of finite positive measure of $f^{-1}(\mathscr O_n)$.