5

Lebesgue's criterion for Riemann integrability for functions of one variable states that:

$f:[a,b]\rightarrow \mathbb{R}$, with $a,b\in\mathbb{R},a\leq b$, is Riemann integrable if and only if $f$ is bounded and continuous $m$-a.e.,where $m$ is the Lebesgue measure on $\left(\mathbb{R},{\cal M}(\mathbb{R})\right)$.

Is there such a statement (or similar) for Riemann integrability on $[a,b]\times[c,d]$, with $a,b,c,d\in\mathbb{R},a\leq b,c\leq d$, of a function $f:[a,b]\times[c,d]\rightarrow \mathbb{R}$, in terms of boundedness and $m^2$-a.e. continuity, where $m^2$ is standard Lebesgue measure on $\left(\mathbb{R}^2,{\cal M}(\mathbb{R}^2)\right)$?

Thank you very much.

Comment: Thank you, Prahlad. Following Integration and Modern Analysis (Benedetto and Czaja), the proof of "$\Leftarrow$" in one variable needs basically the primitive $F(x) = \overline{\int_a^x}f(u)du$ (upper Riemann integral) whose derivative is equal to $f$ $m$-a.e. One then uses it to build a uniformly bounded sequence of functions $F_n(x) = n(F(x+1/n)-F(x))$ convergent to $f$ $m$-a.e. (LDC and some calculations then imply that the upper Riemann integral and Lebesgue integral are equal.) For the moment, I'm having trouble imagining correctly both $F$ and $F_n$ in two variables.

ir7
  • 6,249
  • 4
    Yes, the statement is identical, and the proof is also essentially the same. Can you try it and explain where you are getting stuck? – Prahlad Vaidyanathan Dec 30 '13 at 07:07
  • Have a look at Tom Apostol's Mathematical Analysis. I just glanced through it, and the proof appears to translate verbatim to multiple variables (He does not refer to this function $F(x)$ that you are using) – Prahlad Vaidyanathan Dec 30 '13 at 09:31
  • @Prahlad Ok. I'll give it a try (I find this result's proof technical, and I think you are right it doesn't have to involve the Lebesgue integral itself, just the measure itself and Riemann integral characterizations). Best regards. – ir7 Dec 30 '13 at 21:35

0 Answers0