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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?

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  • $\begingroup$ What do you mean by $x \le x'$ when these are $d$-uples? $\endgroup$
    – Crostul
    Commented May 10, 2020 at 11:13
  • $\begingroup$ @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 $k = 1,\dots, d$ $\endgroup$
    – Epiousios
    Commented May 10, 2020 at 11:15
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    $\begingroup$ HINT: set $d=2$ and use induction $\endgroup$
    – Masacroso
    Commented May 10, 2020 at 11:32

1 Answer 1

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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.

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