Let $f(x,t) : [0,1]^2 \to \mathbb{R}$ be a measurable function such that \begin{equation} x \to f(x, t) \text{ is integrable for a.e. } t \in [0,1] \end{equation} and \begin{equation} t \to \int_0^1 f(x,t) dx \text{ is a.e. finite and nonnegative } \end{equation} plus \begin{equation} \int_0^1\int_0^1 f(x,t) dx dt \text{ is finite}. \end{equation}
Under such conditions, is it still possible to have \begin{equation} \int_0^1\int_0^1 \lvert f(x,t)\rvert dx dt = \infty? \end{equation}
I believe so, but cannot easily come up with an example myself.. Could anyone please help me?
You obviously need to take a function $f$ not integrable, otherwise this would be impossible by Fubini's theorem. I suggest you take a classical counterexample to Fubini in Wikipedia. Namely consider, $$f(x,t)=\frac{t^2-x^2}{(t^2+x^2)^2}.$$
You can see that $\int_0^1 f(x,t)dx = \frac{1}{1+t^2}$ and $\int_0^1 f(x,t)dt = \frac{-1}{1+x^2}$. Since $\int_0^1 \frac{1}{1+x^2}dx= arctan(1)$, you get that integrating in a different order gives different values, so the function must not be integrable. In this case, you also have that $\int_0^1 f(x,t)dx \geq 0$ for all $t>0$.
Edit:
Following the discussion in the comments, we can also see that $$\int_0^1\vert f(x,t)\vert dx \geq \int_0^t \vert f(x,t)\vert dx=\int_0^t f(x,t) dx.$$
The last term is equal to $\frac{t}{2t^2}=\frac{1}{2t}$. We know that $\int_0^1 \frac{1}{2t}= \infty$.