Caleb Proctor

2022-07-07

Fix some probability space $(\mathrm{\Omega},\mathcal{F},\mathbb{P})$, 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 $\{h>g\}\in \mathcal{A}$? I understand that since $g$ and $h$ both are $\mathcal{A}$-measurable, it holds that for all $B\in \mathcal{B}(\mathbb{R})$:

${h}^{-1}(B)\in \mathcal{A},\phantom{\rule{1em}{0ex}}{g}^{-1}(B)\in \mathcal{A}.$

However, I can't seem to make the connection as to why $\{h>g\}\in \mathcal{A}$ also holds true.

I need help understanding why $\{h>g\}\in \mathcal{A}$? I understand that since $g$ and $h$ both are $\mathcal{A}$-measurable, it holds that for all $B\in \mathcal{B}(\mathbb{R})$:

${h}^{-1}(B)\in \mathcal{A},\phantom{\rule{1em}{0ex}}{g}^{-1}(B)\in \mathcal{A}.$

However, I can't seem to make the connection as to why $\{h>g\}\in \mathcal{A}$ also holds true.

Keegan Barry

Beginner2022-07-08Added 18 answers

$\{h>g\}=\{h-g>0\}\in \mathcal{A}$ 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 $t\in \mathbb{R}$, $h-g<t\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h<t+g\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h<q<t+g$ for some rational $q$. So

$\{h-g<t\}=\bigcup _{q\in \mathbb{Q}}\{h<q<t+g\}=\bigcup _{q\in \mathbb{Q}}\{h<q\}\cap \{g>q-t\}$

Hence $\{h-g<t\}\in \mathcal{A}$ as a countable union of elements of $\mathcal{A}$, and therefore $h-g$ is $\mathcal{A}$-measurable.

To see this, note that for all $t\in \mathbb{R}$, $h-g<t\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h<t+g\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h<q<t+g$ for some rational $q$. So

$\{h-g<t\}=\bigcup _{q\in \mathbb{Q}}\{h<q<t+g\}=\bigcup _{q\in \mathbb{Q}}\{h<q\}\cap \{g>q-t\}$

Hence $\{h-g<t\}\in \mathcal{A}$ as a countable union of elements of $\mathcal{A}$, and therefore $h-g$ is $\mathcal{A}$-measurable.

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