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If $f:I\subseteq \mathbb{R}\to \mathbb{R}$ is a Lebesgue-measurable function ($I$ is a closed interval), $\mu$ is the Lebesgue measure, my question is whether the limit
$\underset{u\to {u}_{0}^{-}}{lim}\mu \left({f}^{-1}\left(\left(u,{u}_{0}\right]\right)\right)=0.$
I don't really know how to justify the value of this limit, if it is 0 or possibly $\mu \left({f}^{-1}\left(\left\{{u}_{0}\right\}\right)\right)$. I would appreciate your help.
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Brendan Bush
wlog you can use a sequence ${u}_{n}↑{u}_{0}$. We have $\mu \left({f}^{-1}\left(\mathbb{R}\right)\right)=\mu \left(I\right)=\lambda \left(I\right)<\mathrm{\infty }$. Then you can see that
${f}^{-1}\left(\left({u}_{n},{u}_{0}\right]\right)=\left\{x:f\left(x\right)\in \left({u}_{n},{u}_{0}\right]\right\}\supseteq \left\{x:f\left(x\right)\in \left({u}_{n+1},{u}_{0}\right]\right\}={f}^{-1}\left(\left({u}_{n+1},{u}_{0}\right]\right)$
and therefore, since inverse images preserve set operations:
$\bigcap _{n\in \mathbb{N}}{f}^{-1}\left(\left({u}_{n},{u}_{0}\right]\right)={f}^{-1}\left(\bigcap _{n\in \mathbb{N}}\left({u}_{n},{u}_{0}\right]\right)={f}^{-1}\left(\left\{{u}_{0}\right\}\right)$
By continuity of measures (we have $\mu \left({f}^{-1}\left(\left({u}_{1},{u}_{0}\right]\right)\right)\le \lambda \left(I\right)$ we then have
$\underset{n\to \mathrm{\infty }}{lim}\mu \left({f}^{-1}\left(\left({u}_{n},{u}_{0}\right]\right)\right)=\mu \left({f}^{-1}\left(\left\{{u}_{0}\right\}\right)\right)$
To see that it is not always 0, consider $I=\left[-1,1\right]$, $f\left(x\right)={\mathbf{1}}_{\left[0,1\right]}\left(x\right)+\left(1/2\right){\mathbf{1}}_{\left[-1,0\right)}\left(x\right)$ and ${u}_{0}=1$. Then
$\underset{n\to \mathrm{\infty }}{lim}\mu \left({f}^{-1}\left(\left({u}_{n},1\right]\right)\right)=\mu \left({f}^{-1}\left(\left\{1\right\}\right)\right)=\mu \left(\left[0,1\right]\right)=1$