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Caleb Proctor 2022-07-07 Answered
Fix some probability space $\left(\mathrm{\Omega },\mathcal{F},\mathbb{P}\right)$, and let $\mathcal{A}\subset \mathcal{F}$ be a sub-$\sigma$-algebra. Suppose that $g,h:\mathrm{\Omega }\to \mathbb{R}$ are two $\mathbb{P}$-integrable and $\mathcal{A}$-measurable functions.
I need help understanding why $\left\{h>g\right\}\in \mathcal{A}$? I understand that since $g$ and $h$ both are $\mathcal{A}$-measurable, it holds that for all $B\in \mathcal{B}\left(\mathbb{R}\right)$:
${h}^{-1}\left(B\right)\in \mathcal{A},\phantom{\rule{1em}{0ex}}{g}^{-1}\left(B\right)\in \mathcal{A}.$
However, I can't seem to make the connection as to why $\left\{h>g\right\}\in \mathcal{A}$ also holds true.
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Keegan Barry
$\left\{h>g\right\}=\left\{h-g>0\right\}\in \mathcal{A}$ because if $h$ and $g$ are $\mathcal{A}$-measurable, then the difference $h-g$ is also $\mathcal{A}$-measurable.
To see this, note that for all $t\in \mathbb{R}$, $h-g for some rational $q$. So
$\left\{h-gq-t\right\}$
Hence $\left\{h-g as a countable union of elements of $\mathcal{A}$, and therefore $h-g$ is $\mathcal{A}$-measurable.

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