An example of $\int_{X \times Y} f d(\mu \times \lambda )$ is finite but such that $\int_{Y} f_{x} d\lambda $ is infinite for some $x \in (0,1).$

Question

Let $X = Y = (0,1)$ and let $ \mu = \lambda)$ be the Lebesgue measure on $(0,1).$ Find an example of a positive continuous function $f: (0,1) \times (0,1) \rightarrow \mathbb{R},$ such that $$\int_{X \times Y} f d(\mu \times \lambda )$$ is finite but such that $\int_{Y} f_{x} d\lambda $ is infinite for some $x \in (0,1).$

My trial:

I was thinking about the function in this integration:

$$\int_{E} \frac{y}{x} e^{-x} \sin x d\mu$$ where $\mu$ is the product of Lebesgue measure on $\mathbb{R}$ with itself, and $E = \{(x,y) : 0 \leq y \leq \sqrt{x}\}.$ But I am not sure.

Also, will this question help me Showing that $\int_{[0,1]}\frac{1}{x}\, d\lambda(x)$ isn't finite? I do not know if anyone can help me I will appreciate this too much.

Answer 1

Consider $$f(x,y) = \frac{1}{|y|^{1-\sqrt{|x-1/2|}}}.$$ Clearly, $f$ is positive and continuous on $(0,1) \times (0,1)$. For $x=1/2$ we have $$\int_{(0,1)} f\left(\frac{1}{2},y\right) \, dy = \int_{(0,1)} \frac{1}{|y|} \, dy = \infty.$$ On the other hand, $$\int_0^1 \frac{1}{|y|^{1-\sqrt{|x-1/2|}}} \, dy = \frac{1}{\sqrt{|x-1/2|}}$$ for each $x \neq 1/2$, and so $$\int_0^1 \int_0^1 f(x,y) \, dy \, dx = \int_0^1 \frac{1}{\sqrt{|x-1/2|}} \, dx < \infty.$$