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Let $(\Omega,\mathcal{F})$ be a measurable space. The following are equivalent:

  1. $\ X:\Omega \to \mathbb{R} $ is a Borel function.

  2. $\{\omega\in\Omega:X(\omega)>a\}\in\mathcal{F}$ for all $a\in\mathbb{R}$.

  3. $\{\omega\in\Omega:X(\omega)< a\}\in\mathcal{F}$ for all $a\in\mathbb{R}$.

  4. $\{\omega\in\Omega:X(\omega) \in B\}\in\mathcal{F}$ for all open subsets $B\subset\mathbb{R}$.

  5. $\{\omega\in\Omega:X(\omega) \in B\}\in\mathcal{F}$ for all closed subsets $B\subset\mathbb{R}$.

How on earth would I prove this? I have no idea where to start. Any help would be very much appreciated. Thanks

1 Answers 1

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1) $\Rightarrow$ 2) is clear since $(a;+\infty)$ is a Borel set.

2) $\Rightarrow$ 3) because $\{\omega, X(\Omega) and each set of the union is of the form of the sets involved in 2), after taking complement.

3)$\Rightarrow$ 4) can be shown as the following: first show that it works when $B$ is an interval, then use the fact that an open subset of the real line can be written as a countable union of open intervals.

4)$\Rightarrow$ 5) using complement.

5)$\Rightarrow$ 1) Let $\mathcal A:=\{A\subset R, X^{-1}(A)\mbox{ is a Borel set}\}$. Show that $\mathcal A$ is a $\sigma$-algebra.