1

Let $f(\cdot,\cdot):R^2\rightarrow R$. Suppose that
(a) Fix any $y\in R$, $f(\cdot,y)$ is continuous almost everywhere.
(b) Fix any $x\in R$, $f(x,\cdot)$ is continuous almost everywhere.

Is it true that $f$ is separately continuous almost everywhere?

i.e. For almost all point $(x,y)\in R^2$,
holding $y$ fixed, $f(x',y)$ is continuous at $x'=x$;
holding $x$ fixed, $f(x,y')$ is continuous at $y'=y$

Yi-Hsuan Lin
  • 795
  • It seems like your questions are the same as your assumptions. Don't you want to ask if $f(\cdot,\cdot):R^2\rightarrow R$ is continuous almost everywhere? – Craig Hicks Mar 15 '17 at 21:33
  • @CraigHicks I think my assumptions are different to the question. The assumption says continuous a.e. as a single-variable function, holding another variable fixed. But the question is about separately continuity a.e. on $R^2$. It is possible that, for example, $f(x',y_0)$, as a function of $x'$, is continuous almost everywhere but discontinuous at $x'=x_0$. Then $f$ is not separately continuous at $(x_0,y_0)$. – Yi-Hsuan Lin Mar 15 '17 at 21:41
  • Is it the same as proving there is not a set of measure > 0 on which $f$ is not separately continuous at $(x0,y0)$？ – Craig Hicks Mar 15 '17 at 23:54
  • @CraigHicks Yes, I am asking if the set ${(x_0,y_0)\in R^2: f\ not\ separately\ continuous\ at}$ has measure 0 on $R^2$? – Yi-Hsuan Lin Mar 16 '17 at 15:29
  • If you any point $(x,y)$ randomly, then by (a) and (b) it will a.s. (almost surely) be separately continuous. – Craig Hicks Mar 17 '17 at 06:19
  • Assume a worst-case/upper-limit scenario where every discontinuity in either x or y is a discontinuity in both x and y. (I assert) that wouldn't change the conditions (a) or (b). Then you prove it using this problem/solution found online in a rutgers problem set: Prove the following assertion: Every measurable function is the limit a.e. of a sequence of continuous functions. [Hint: Let /psi be a step function, and let /epsilon be positive. Show there is a continuous function which is equal to /psi outside a set of measure $0$.] ( https://andromeda.rutgers.edu/~loftin/ra1/sol2.pdf ) – Craig Hicks Mar 17 '17 at 16:01
  • @CraigHicks Thanks for your reply, but I don't get it. What do you mean by every discontinuity in either x or y is a discontinuity in both x and y? How can you assume it? – Yi-Hsuan Lin Mar 18 '17 at 02:30
  • @CraigHicks I don't think the claim follows straightforward from (a) and (b). Because, for example, consider $x=1$ and $x=2$. Although $f(1,y)$ and $f(2,y)$ (view them as functions of $y$ only) are both continuous almost everywhere, they can have completely different discontinuous points. More generally, for each $x$, consider ${r\in R: f(x,y)\ discontinuous\ at\ y=r }$. This set has measure 0 by (b). But the union (over $x$) may have positive measure. – Yi-Hsuan Lin Mar 18 '17 at 02:35
  • Maybe this is helpful: https://books.google.com/books?id=zy3jBwAAQBAJ&pg=PA28&dq=extension+of+continuous+almost+everywhere+to+2+dimensions&hl=ja&sa=X&ved=0ahUKEwiv3_2Yjd_SAhXETSYKHYfDCMMQ6AEIXjAI#v=onepage&q=extension%20of%20continuous%20almost%20everywhere%20to%202%20dimensions&f=false

    pages 28 and 29

     – Craig Hicks Mar 18 '17 at 04:13

0 Answers0