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Show that the supremum of an uncountable family of $[-\infty, \infty]$- valued Borel measurable functions on $\mathbb{R}$ can fail to be Borel measurable.

Does this mean that I need to find some kind of family such that the intersection of the sets $$\{x\in A | f_n(x) \leq t \}, $$ for all $t$, fails to be an Borel set? Or is there some easier way?

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Take a non-measurable set $E$ and for each $e\in E$ let $\chi_{\{e\}}$ be the characteristic function of the set $\{e\}$. Then the supremum of the (uncountable) family $\{\chi_{\{e\}}:~e\in E\}$ is the characteristic function $\chi_E$ of the set $E$, which is not measurable and therefore nor Borel measurable.

Edit

$\chi_E$ is the supremum of the family since it is greater or equal than any function of the family, and is the smallest function with such property.

$\chi_E \geq \chi_{\{e\}}$ for all $e\in E$; indeed,

  • if $x\in E$, then $\chi_E(x) = 1 \geq \chi_{\{e\}}(x)$ for all $e\in E$;
  • if $x\notin E$, then $x\neq e$ for all $e\in E$ and hence $\chi_E(x) = 0 = \chi_{\{e\}}(x)$.

$\chi_E$ is the minimal such function, in that if $f(x)<\chi_E(x)$ for some $x$, then there exists $e\in E$ such that $f(x) < \chi_{\{e\}}(x)$; indeed,

  • if $x\in E$, then $\chi_x$ belongs to the family and $f(x) < \chi_E(x) = 1 = \chi_{\{x\}}(x)$;
  • if $x\notin E$, then $f(x) < \chi_E(x) = 0 = \chi_{\{e\}}(x)$ for all $e\in E$.
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  • $\begingroup$ Why is $\chi_E$ the supremum of the family? The family consists of characteristic functions on a singleton set right? $\endgroup$ – nomadicmathematician Nov 5 '15 at 19:13
  • $\begingroup$ @takecare you are right. Check my edited answer for more details. $\endgroup$ – AndreasT Nov 12 '15 at 14:27

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