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This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in A$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And Nate Eldredge beat me to the proof for the Lebesgue measurable case.

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in R_{A}$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And Nate Eldredge beat me to the proof for the Lebesgue measurable case.

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in A$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And someone beat me to the proof for the Lebesgue measurable case.

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in A$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And Nate Eldredge beat me to the proof for the Lebesgue measurable case.

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in A$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof).

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in A$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And someone beat me to the proof for the Lebesgue measurable case.