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Any function $\mathbb{R} \to \mathbb{R}$ that is monotone is measurable. How far does this generalize?

Is a monotone function $\mathbb{R}^n \to \mathbb{R}$ measurable? A function from one totally ordered metric space to another?

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What do you mean by a monotone function from $\mathbb R^n$ into $\mathbb R$? are you using the Borel $\sigma$-algebra on the linearly ordered metric spaces? – azarel Apr 14 '12 at 23:47
- Measurable with respect to the Borel $\sigma$ algebra. - A function is monotone if it is monotone in each component (I am not set on the definition of monotone) – Nahpetz Apr 15 '12 at 0:00
up vote 1 down vote accepted

A monotone function $\mathbb{R}^n \to \mathbb{R}$ is Lebesgue measurable.

(Here, we equip $\mathbb{R}^n$ with the partial order $\le$ where $(x_1, \dots, x_n) \le (y_1, \dots, y_n)$ iff $x_i \le y_1$ for each $i$. A function $\mathbb{R}^n \to \mathbb{R}$ is monotone with respect to this partial order iff it is monotone in each coordinate, which is the OP's definition.)

Proof. We proceed by induction on $n$. The $n=0$ case is trivial (and $n=1$ is well known). Suppose, then, that every monotone function $\mathbb{R}^{n-1} \to \mathbb{R}$ is Lebesgue measurable. Let $f : \mathbb{R}^n \to \mathbb{R}$ be monotone increasing. Fix $y \in \mathbb{R}$; we will prove that $f^{-1}([y, \infty))$ is a Lebesgue measurable subset of $\mathbb{R}^n$.

Define $g : \mathbb{R}^{n-1} \to \mathbb{R}$ by $$g(x) = \inf\{t \in \mathbb{R} : f(x,t) \ge y\}.$$ I claim $g$ is monotone decreasing. Suppose $x \le x' \in \mathbb{R}^{n-1}$. If $f(x, t) \ge y$ then also $f(x', t) \ge y$, hence the infimum for $g(x)$ is taken over a smaller set than that for $g(x')$, and so $g(x) \ge g(x')$. Thus by the induction hypothesis, $g$ is Lebesgue measurable.

Now we observe that if $t > g(x)$ then $f(x,t) \ge y$, and if $t < g(x)$ then $f(x,t) < y$. So if we set $$A_1 = \{(x,t) : g(x) < t\}, \quad A_2 = \{(x,t) : g(x) \le t\}$$ then we have $$A_1 \subset f^{-1}([y, \infty)) \subset A_2.$$ Now $A_1, A_2$ are Lebesgue measurable: if we set $G(x,t) = g(x) - t$ then $G$ is Lebesgue measurable and $A_1 = G^{-1}((-\infty, 0))$, $A_2 = G^{-1}((-\infty, 0])$. On the other hand, $A_0 = A_2 \backslash A_1 = \{(x,t) : g(x) = t\}$ is just the graph of $g$, and it follows from Tonelli's theorem that $m(A_0) = 0$, since $$m(A_0) = \int_{\mathbb{R}^{n-1}} \int_\mathbb{R} 1_{A_0}(x,t)\,dt\,dx = \int_{\mathbb{R}^{n-1}} m(\{g(x)\})\,dx = 0.$$ Thus every subset of $A_0$ is Lebesgue measurable, and since we have $f^{-1}([y, \infty)) = A_2 \cup (A_0 \cap f^{-1}([y, \infty)))$ we are done.

I first encountered this question (in a slightly different form) as an exercise in Geoffrey Grimmett's Probability on Graphs, where it appears as Exercise 4.10. He gives a reference to

Graham, B. T., Grimmett, G. R. Influence and sharp-threshold theorems for monotonic measures, Annals of Probability 34 (2006), 1726–1745

where the statement appears as Theorem 4.4, with a proof very similar to mine.

Generalizing this further could be tricky. I would probably want to restrict my attention to Polish spaces with Borel total orderings, and ask some question about the definability of monotone functions. But that would be a question for a descriptive set theorist.

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Great! Thank you also for the reference, that is helpful. – Nahpetz Apr 16 '12 at 20:11

The answer is "No", as soon as $n\geqslant2$.

For a counterexample, consider any non measurable $S\subset\mathbb R$ and the set $A\subset\mathbb R^n$ of points $(x_k)_k$ such that either (i) $x_1+x_2\gt0$, or (ii) $x_1+x_2=0$ and $x_1$ is in $S$.

Then $A$ is not measurable. (If $A$ was measurable, its intersection $A\cap L$ with the line $L$ of equations $x_1+x_2=0$ and $x_k=0$ for every $k\geqslant3$, would be measurable, as well as the preimage of $A\cap L$ by the continuous map $g:x\mapsto (x,-x,0,\ldots,0)$. But $g^{-1}(A\cap L)=S$.)

Now, $f=\mathbf 1_A$ is a nondecreasing function which is not measurable since the set $f^{-1}(\{1\})=A$ is not measurable.

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It's worth noting that the projection of a measurable set is not, in general, measurable. What saves us in this case is that the projection is bijective when restricted to $A \cap L$. It might be better to say instead that $S$ is the preimage of $A \cap L$ under the continuous map $x \mapsto (x, -x, 0, \dots, 0)$, hence if $A \cap L$ is measurable then so is $S$. – Nate Eldredge Apr 15 '12 at 15:07
Thank you. Would the answer still be "No" if one asks for Lebesgue-measurable? $L$ has zero Lebesgue-measure, so its subset $S$ is Lebesgue-measurable. – Nahpetz Apr 15 '12 at 15:56
@Stephan: For Lebesgue-measurable, I believe the answer is yes. I think I once proved something similar as an exercise. I will try to dig it up and add an answer when I have some more time. – Nate Eldredge Apr 15 '12 at 16:33
@Nate: Many Thanks! – Nahpetz Apr 15 '12 at 16:41
@Nate: Post modified. Thanks. – Did Apr 15 '12 at 16:48

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