If $(X,\mathcal S ) $ is a measurable space and $f:X \to [-\infty ,\infty] $ is a function such that $f^{-1}((a ,\infty)) \in \mathcal S $ for every $a \in R$ . Then $f \ $ is $\ $ $ \mathcal S $ - measurable . I am trying to disprove this statement by searching counterexample .I am trying to construct a characteristic function for counterexample .Can any one give me some counter example if this statement is false and if true then please give me proof .
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1$\begingroup$ Do the sets $(a, +\infty)$, $a \in \Bbb R$ generate the Borel algebra on $\overline{\Bbb R}$? If so then $f$ is measurable, if not then maybe not. $\endgroup$– Henno BrandsmaCommented Mar 19, 2021 at 13:09
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$\begingroup$ So, is it the case that $(a, \infty ] = (a, \infty) \cup \{ \infty \} ?$ If yes, then can I concluded that the open sets in $\mathbb{R}$ generate the extended borel sets in $\mathbb{R}, \mathcal{B}(\mathbb{R}) ?$ $\endgroup$– MathRookie2204Commented Sep 8, 2023 at 4:40
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$\begingroup$ The whole idea here is this: Can I treat $\infty$ as just a symbol and adjoin it to open rays and go ahead or is there any structure related properties ? $\endgroup$– MathRookie2204Commented Sep 8, 2023 at 4:48
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2 Answers
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Consider $\mathcal{B}=\{A \subseteq \overline{\Bbb R}=[-\infty,+\infty]\mid f^{-1}[A] \in \mathcal{S}\}$. It's easy to check that $\mathcal{B}$ is a $\sigma$-algebra on $\overline{\Bbb R}$ and it contains all of $\mathcal{B}':=\{(a+\infty)\mid a \in \Bbb R\}$. If $\mathcal{B}'$ generates $\text{Bor}(\overline{\Bbb R})$ then $f$ will be measurable by definition, as then $\text{Bor}(\overline{\Bbb R}) \subseteq \mathcal{B}$ from the given $\mathcal{B'}\subseteq \mathcal{B}$ inclusion.
answered Mar 19, 2021 at 13:17
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Take Borel sigma algebra for simplicity. Define $f(x) = +\infty$ if $x \in V$ (V for Vitali or any non-measurable set you like) and $f(x) = -\infty$ otherwise. Then $f^{−1}((a,∞)) = \emptyset$ for all $a$, but $f$ is not measurable.
