Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(x,y)$ be a real valued function on $[0,1]\times [0,1]$. My question is that, if $f(x,y)$ is measurable in $x$ when $y$ is fixed, and monotone in $y$ when $x$ is fixed, is it true that $f$ is measurable?

share|cite|improve this question
up vote 2 down vote accepted

Consider $E = \{(x,y): f(x,y) \ge \alpha\}$. Let $A = \{x: f(x,0) \le f(x,1)\}$ and $B = \{x: f(x,0) \ge f(x,1)\}$. Since $f(\cdot,0)$ and $f(\cdot,1)$ are measurable, $A$ and $B$ are measurable sets, and $[0,1] = A \cup B$. For $x \in A$, $f(x,\cdot)$ is nondecreasing. Let $g(x) = \inf \{y \in [0,1]: f(x,y) \ge \alpha\}$. Since $g(x) \le t$ with $0 < t < 1$ iff $f(x,t-1/n) < \alpha \le f(x,t+1/n)$ for all integers $n$ large enough that $0 < t-1/n < t+1/n < 1$. and $f(\cdot, t\pm 1/n)$ are measurable, $g$ is measurable.

EDIT: For $(x,y) \in A \times [0,1]$, $f(x,y) < \alpha$ if $y < g(x)$ and $f(x,y) \ge \alpha$ if $y > g(x)$. As Byron pointed out, we don't know about $f(x,g(x))$. However, the graph $G = \{(x,g(x)): x \in A\}$ has Lebesgue measure $0$. Since the difference between $E \cap(A \times [0,1])$ and $\{(x,y): x \in A, y \ge g(x)\}$ has measure $0$, $E \cap (A \times [0,1])$ is (Lebesgue) measurable.

Similarly $E \cap (B \times [0,1])$ is measurable, and so is their union $E$, so $f$ is measurable.

share|cite|improve this answer
I think there may be a problem with $E \cap (A \times [0,1]) = \{(x,y): x \in A, y \ge g(x)\}$. From $y\geq g(x)$ we can only deduce $f(x,y+)\geq \alpha$, not $f(x,y)\geq \alpha$. Didier's answer here gives a counterexample:… – Byron Schmuland Apr 15 '12 at 13:48
Hmm... you're right. For example, you could have $f(x,y) = 1$ for $x < y$, $f(x,y) = 0$ for $x > y$, and $f(x,x)$ could be an arbitrary function from $[0,1]$ to $[0,1]$. In particular, $f$ doesn't have to be Borel measurable. On the other hand, I think I can still rescue Lebesgue measurable. – Robert Israel Apr 15 '12 at 15:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.