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For this question, label:
Part A: Is $f(x,y)$ integrable? Question $3$-$7$ from Spivak's Calculus on Manifolds
Part B: Prove $f:[0,1]\times[0,1]→\mathbb{R}$ is integrable.

For part A and B question, I believe that the set of points of discontinuity is exactly equal to $[0,1]\times [0,1]\cap \mathbb{Q}\times \mathbb{Q}$ because part A is the extension of Thomae's function to $\mathbb{R}^2$ and part B is produced by taking a cross product of Thomae's function.

But when I use the sequential continuity definition, I get a larger set.
For Eg: For this part A; Let $p_n$ be a sequence of rational points converging to $\frac{1}{\sqrt{2}}$ but the sequence $f\left(p_n,\frac{1}{2}\right)=\frac{1}{2}$ doesn't converge to zero i.e. $f\left(\frac{1}{\sqrt{2}},\frac{1}{2}\right)=0$

Is there something wrong with the method? Else if the set is larger, what does the set look like?

I have asked this question on MSE but didn't get any answer and couldn't start a bounty due to reputation restrictions. Any Help will be appreciated. Thanks in advance.

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  • $\begingroup$ Your idea seems to be based on the fallacious assumption that if $f_1, f_2$ have discontinuity sets $D_1, D_2$ respectively, then the discontinuity set of $f(x,y) = f_1(x) + f_2(y)$ will be the product $D_1 \times D_2$, i.e. that $f$ is discontinuous at $(x,y)$ iff both the functions $f_1, f_2$ are discontinuous at $x,y$ respectively. But in fact the "both" should be "either". $\endgroup$
    – Nate Eldredge
    Commented Oct 30, 2020 at 15:25
  • $\begingroup$ @NateEldredge As the user has shown that for part A, $f(x,y)$ is discontinuous at $(\frac{1}{\sqrt{2}},\frac{1}{2})$ but in this case, $f_1(x)$ is continuous at $x\in [0,1]$. So, I suppose there is something more to it other than just the replacement of "both" with "either". $\endgroup$
    – Kumar
    Commented Oct 31, 2020 at 4:52

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