How to compute $\lim_{u\to u_{0}^{-}} \mu\left( f^{-1}(u,u_{0}] \right)$?

Question

If $f\colon I\subseteq\mathbb{R} \rightarrow \mathbb{R}$ is a Lebesgue-measurable function ($I$ is a closed interval), $\mu$ is the Lebesgue measure, my question is whether the limit $$\lim_{u\to u_{0}^{-}} \mu\left( f^{-1}((u,u_{0}]) \right) = 0.$$ I don't really know how to justify the value of this limit, if it is $0$ or possibly $\mu(f^{-1}(\{u_{0}\}))$. I would appreciate your help.

Answer 1

wlog you can use a sequence $u_n \uparrow u_0$. We have $\mu(f^{-1}(\mathbb{R}))=\mu(I)=\lambda(I)<\infty$. Then you can see that $$f^{-1}((u_n,u_0])=\{x:f(x)\in (u_n,u_0]\}\supseteq \{x:f(x)\in (u_{n+1},u_0]\}=f^{-1}((u_{n+1},u_0])$$ and therefore, since inverse images preserve set operations: $$\bigcap_{n \in \mathbb{N}}f^{-1}((u_n,u_0])=f^{-1}\bigg(\bigcap_{n \in \mathbb{N}}(u_n,u_0]\bigg)=f^{-1}(\{u_0\})$$ By continuity of measures (we have $\mu(f^{-1}((u_1,u_0]))\leq \lambda(I)$) we then have $$\lim_{n \to \infty }\mu(f^{-1}((u_n,u_0]))=\mu(f^{-1}(\{u_0\}))$$ To see that it is not always $0$, consider $I=[-1,1]$, $f(x)=\mathbf{1}_{[0,1]}(x)+(1/2)\mathbf{1}_{[-1,0)}(x)$ and $u_0=1$. Then $$\lim_{n \to \infty }\mu(f^{-1}((u_n,1]))=\mu(f^{-1}(\{1\}))=\mu([0,1])=1$$