I have some difficulty in dealing with the absolute value under the integral. Let $X=Y=[0,1], \mu=\lambda=$ the Lebesgue measure on $[0,1].$ Let $0=\delta_1<\delta_2<\delta_3..., \delta_n \rightarrow 1.$ Let $f_n$ be real continuous functions with support in $(\delta_n, \delta_{n+1}),$ such that $\int_0^1 g_n(t)\,dt=1, n=1,2,3...$
Let $f(x,y)=\sum_{n=1}^{\infty} [g_n(x)-g_{n+1}(x)]g_n(y).$
I need to show that: $$ \int_0^1 dx\,\, (\int_0^1 |f(x,y)|\,dy) = \infty.$$
It is clear that for each point $(x,y),$ at most one term of the sum is non-zero, unless $x,y$ belong to different intervals $(\delta_n, \delta_{n+1})$. I have troubles to see if $g_n$ can take negative values.
To compute the integral in parethesis one can fix $x$. If $x\in (\delta_n, \delta_{n+1}),$ then this integral is non-zero if $y\in (\delta_n, \delta_{n+1}).$ Thus: $$\int_0^1 |f(x,y)|\,dy=\int_{\delta_n}^{\delta_{n+1}} |f(x,y)|\,dy.$$
Unfortnately I cant go further. So I need some support or a solution proposal. Many thanks.
Just notice
$$\int_0^1 \left(\int_0^1 |f(x,y)|\,dy\right)dx = \sum_{n}\int_{\delta_n}^{\delta_{n+1}} \left(\int_0^1 |f(x,y)|\,dy\right)dx.$$
If $x\in (\delta_n, \delta_{n+1})$ then $f(x,y) = g_n(x)g_n(y)$ and $|f(x,y)| = |g_n(x)||g_n(y)|$. Therefore $$\int_{\delta_n}^{\delta_{n+1}} \left(\int_0^1 |f(x,y)|\,dy\right)dx = \int_{\delta_n}^{\delta_{n+1}} |g_n(x)| \left(\int_0^1 |g_n(y)|\,dy\right)dx.$$
Since $$1 = \left|\int_0^1 g_n(y)\,dy\right| \leq \int_0^1 |g_n(y)|dy$$ we have $\int_0^1 |g_n(y)|dy \geq 1$ and then $$ \int_{\delta_n}^{\delta_{n+1}} |g_n(x)| \left(\int_0^1 |g_n(y)|\,dy\right)dx \geq \int_{\delta_n}^{\delta_{n+1}}|g_n(x)|dx = \int_0^1 |g_n(x)|dx \geq 1.$$
So, we have
$$\int_0^1 \left(\int_0^1 |f(x,y)|\,dy\right)dx = \sum_{n}\int_{\delta_n}^{\delta_{n+1}} \left(\int_0^1 |f(x,y)|\,dy\right)dx \geq \sum_n 1 = \infty.$$