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I'm looking for a function $g(x, y)$ that is well defined on $(0, 1)^2$.

Denote

$$ M(x) = \int_x^1 g(x, y) dy\\ N(y) = \int_0^y g(x, y) dx$$

$g(x, y)$ must be non-negative, continuous and real on its domain. $M(x)$ and $N(y)$ must be strictly positive, and that should also hold for the limiting cases $x\to 0$, $x\to 1$, $y\to 0$, $y \to 1$

In particular, this means (by my understanding) that at the limiting cases, $g(x, y)$ must explode:

$$\lim_{x\to 1} \int_x^1 g(x, y) dy > 0 \Rightarrow \lim_{x\to 1} g(x, y) \to \infty$$

And similarly for $\lim_{y\to 0} g(x, y)$.

I have tried dozens of different functions such as $\frac{1}{1+x}\frac{1}{1+i}$, $\frac{1}{1+xi}$, but all of them just failed on at least one of the corner cases. Typically, when I fix $\lim_{x\to 1} M(x) > 0$, I get that $\lim_{x\to 0} M(x) \to \infty$, and similar issues.

Does such a function even exist? How do I attack this problem? If none exists, I'd be happy to concede with a function where $M(x)$ is strictly positive and $N(y)$ is just positive.

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    Is $M(1)$ defined via $\lim_{x\to1}M(x)$? Otherwise $M(1)$ will always be zero. I assume you want $g$ to be continuous on $(0,1)^2$? – 2017-02-12
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    @s.harp I think you're right, I do want to have $(0, 1)^2$ as domain, but I want the limits to be well-defined. – 2017-02-12
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    I deleted my answer, since I had missed that introducing a $1/(1-x)$ term makes $N(1)$ explode. – 2017-02-13

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