# Are all nondecreasing $f: \mathbb R^d \to \mathbb R$ Borel-measurable?

It is well-known that any nondecreasing function $$f: \mathbb R \to \mathbb R$$ is Borel-measurable.

Can this property be generalized to nondecreasing functions defined on $$\mathbb R^d$$?

To be precise, let's fix the following definitions:

• A function $$f: \mathbb R^d \to \mathbb R$$ is said to be Borel-measurable if $$f^{-1}(B) \in \mathcal B (\mathbb R^d)$$ holds for all $$B \in \mathcal B(\mathbb R)$$, where $$\mathcal B(\mathbb R^k)$$ denotes the Borel sigma algebra on $$\mathbb R^k$$.
• Given $$x, x' \in \mathbb R^d$$ write $$x \le x'$$ if $$x_k \le x'_k$$ holds for all $$k = 1,\dots, d$$.
• A function $$f: \mathbb R^d \to \mathbb R$$ is said to be nondecreasing if $$\forall x, x' \in \mathbb R^d$$, $$x \le x'$$ implies $$f(x) \le f(x')$$

Question: Is every nondecreasing function $$f: \mathbb R^d \to \mathbb R$$ Borel-measurable?

• What do you mean by $x \le x'$ when these are $d$-uples? Commented May 10, 2020 at 11:13
• @Crostul the comparison is meant coordinate-wise: Given $x, x' \in \mathbb R^d$ write $x \le x'$ if $x_k \le x'_k$ holds for all $k = 1,\dots, d$ Commented May 10, 2020 at 11:15
• HINT: set $d=2$ and use induction Commented May 10, 2020 at 11:32

Choose a subset $$N$$ of $$\mathbb{R}$$ which is not Borel-measurable and define $$f:\mathbb{R}^2\to\mathbb{R}$$ by
$$f(x, y) = \begin{cases} 3, & x + y > 0; \\ 2, & x + y = 0 \text{ and } x \in N; \\ 1, & x + y = 0 \text{ and } x \notin N; \\ 0, & x + y < 0. \end{cases}$$
Then it is not hard to check that $$f$$ is non-decreasing. On the other hand, $$f^{-1}(\{2\})$$ is not Borel-measurable.
Intuitively speaking, a non-decreasing function in dimension $$\geq 2$$ can have uncountable number of jumps, which allows to encode certain non-measurability into the set of jumps.