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Consider the function $f: \mathbb{R}^2\to \mathbb{R}$ defined by $$f(x,y):= \begin{cases} 1 \qquad x \geq 0,x \leq y < x + 1\\ -1 \>\quad x \geq 0,x + 1 \leq y < x + 2\\ 0 \qquad \text{else}\end{cases}$$

Then it is easy to verify that $$1 = \int_\mathbb{R}\left[ \int_\mathbb{R} f(x,y) d\lambda(x) \right]d\lambda(y) \neq \int_\mathbb{R}\left[ \int_\mathbb{R} f(x,y) d\lambda(y) \right]d\lambda(x) = 0$$ Now the question is why this does not contradict Fubini's theorem. My guess is that $f \notin \mathcal{L}^1(\mathbb{R}^2)$. Is this right? How do I prove this?

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    I don't think this is the kind of answer you are looking for, so I'll just post it as a comment, but the fact that you can prove Fubini's theorem and that these two integrals are not the same (I didn't check this but I'll believe you for now) already implies that the function does not satisfy the premises required for Fubini's theorem. You could also just calculate $\int \int |f(x,y)| d\lambda(x,y)$ to check if it is in $L^1$ directly. – 2017-01-25
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    How did you get 1 for the LHS of the integral? – 2018-03-21

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You're correct. This function isn't from $L^1(\mathbb R^2)$. To prove that just take its absolute value $$ |f(x,y)|=\begin{cases} 1,&x\ge0,x\le y< x+2,\\ 0,&\text{elsewhere}, \end{cases} $$ and find that $$ \iint_{\mathbb R}|f(x,y)|d\lambda(x,y)=\lambda(\{x\ge0,x\le y< x+2\})=+\infty. $$

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    What would be the justification that the specified set has measure $\infty$? I mean it is quite clear but I do not see yet the formal justification. Thanks. – 2017-01-25
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    For example, you can put arbitrary number of pairwise disjoint rectangles $1\times1$ into this set. – 2017-01-25
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    Nice! That idea came to me right after I wrote the comment. – 2017-01-25