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$\DeclareMathOperator{\essinf}{ess inf}$ $\DeclareMathOperator{\esssup}{ess sup}$ Let $f:[0,1]\to\mathbb{R}$ be Lebesgue-Borel measurable and essentially bounded. If $\essinf{f}<\esssup{f}$, is it possible to find $t\in(0,1)$ such that it holds either $$ \essinf{f_{|[0,t]}}=\essinf{f}\quad\text{and}\quad\esssup{f_{|[t,1]}}=\esssup{f} $$ or $$ \esssup{f_{|[0,t]}}=\esssup{f}\quad\text{and}\quad\essinf{f_{|[t,1]}}=\essinf{f}\text{ ?} $$

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Define $f$ as

$$f=\sum_{n=1}^\infty (-1)^n(1-1/n) \chi_{(1/(n+1),1/n]} \quad \text{ and } \quad f(0)=0.$$

Then $f$ is measurable, since it's a limit of simple functions. We have that the essential supremum is $1$ and the essential infimum is $-1$. But these both are only achieved on $[0,t]$ for $t \in (0,1)$.

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    And the scales fell from my eyes... Thank you! To be technically correct, could you please let sum start at $n=2$. :)2012-12-31
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    @precarious Yes, I wrote it a bit quickly. I've made two small changes.2012-12-31
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    Thank you for providing a true fix! I wish you all the best for 2013!2012-12-31