1
$\begingroup$

Suppose $f(x,y)$ is a function mapping from $R^2$ to $R$ and it is continuous in each variable separately (separable continuity), then why $f(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n})$ is Lebesgue measurable where $m,n$ are positive integers? How to show the measurability? $\left\lfloor {mx} \right\rfloor $ denotes the largest integer no larger than $mx$.

Actually similar question has been asked here Showing a function of two variables is measurable and here Separate continuity implies measurability. I have examined the answers to the two posts but still don't get it. Hope someone can help. Thank you!

$\endgroup$
1
$\begingroup$

For any function $f:\mathbb{R}^2\to\mathbb{R}$ the composition $f_{m,n}(x,y) = f(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n})$ is Lebesgue measurable. Indeed, $f_{m,n}$ is constant on half-open rectangles of size $(1/m)$ by $(1/n)$. Therefore, for any set $A\subset\mathbb{R}$ the preimage $f_{m,n}^{-1}(A)$ is a countable union of rectangles and is therefore measurable.

As Davide Giraudo wrote, for a fixed $m$, $f_{m,n}(x,y)\to f\left(\frac{\lfloor mx\rfloor}m,y\right)$ as $n\to\infty$, because $f$ is continuous in the second variable. Similarly, $f\left(\frac{\lfloor mx\rfloor}m,y\right)\to f(x,y)$ as $m\to\infty$, because $f$ is continuous in the first variable. It follows that

$$f = \lim_{m\to\infty}\lim_{n\to\infty} f_{m,n}$$ is measurable, being the pointwise limit of measurable functions.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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