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Let $\int$ denote Lebesgue integral . If $$\int_R \left(\int_R f(x,y) \, dx \right) \, dy\label{1}\tag{1}$$ exist , what can we say about $$ \int_{R^2} f(x,y) \, dxdy \label{2}\tag{2}$$ and $$\int_R \left(\int_R f(x,y) \, dy \right) \, dx\label{3}\tag{3}\:\:?$$

My attempt:

  1. First, if \eqref{2} exists, then by Fubini's theorem, both \eqref{1} and \eqref{3} exist.
  2. Second , if we only assume \eqref{1} exists, then \eqref{2} or \eqref{3} might not exist.
    Indeed, let $\varphi \in \mathscr{S}(\Bbb R)$ and assume $\int \varphi(x) \,dx \neq 0$, then by fourier transform we have $$\int \int \varphi(x)e^{-2\pi iyx}\, dx \, dy=\varphi(0)$$ However , with the simple observation, \eqref{2} and \eqref{3} do not exist.

My question:

a) If we want the integral \eqref{2} exist , then we must have \eqref{1}=\eqref{2}=\eqref{3}. However , can we found a function $f(x,y)$ , both \eqref{1} and \eqref{3} exist but \eqref{1}$\neq$\eqref{3}.

b) Can we found a function $f(x,y)$ such that \eqref{1}=\eqref{3} but \eqref{2} do not exist ?

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Let $(n,m)$ denote the open cube $x\in (n,n+1)$ , $y\in(m,m+1)$ for some integer $m,n$ .
First we show that we can find a function $f$ which satisfied condition $(b)$ . We can construct $f$ as follow : $$f(z)= \left\{ \begin{array}{lcc} 1 & z\in (n,n) \,\,\text{for all nonegative } n \\ \\ -1 & z\in (n,n+1) \,\,\text{or }\,(n+1,n) \,\,\text{for all nonegative } \, n \\ \\ 0 & \text{otherwise} \end{array} \right.$$ Then $$\int_R\,f(z)\, dx=\int_R \, f(z) \,dy=0$$ So we have $(1)=(3)=0$ . However , $(2)$ do not exist since $\int_{R^2} \, |f(z)| \, dxdy=+ \infty$ .

To construct a function $f$ with satisfied condition $(a)$ , we need a slight modification of $f$ defined above . Let $\{b_n \}_{n=0}^{\infty}$ denote a sequence of nonegative number with $b_0=0$ and $\sum_{n=0}^{\infty} b_n=s$ exist . Then let $a_N=\sum_{n=0}^N b_n$ . Next we can construct funtion $g$ which satisfied condition $(a)$ . $$g(z)= \left\{ \begin{array}{lcc} a_n & z\in (n,n) \,\,\text{for all nonegative } n \\ \\ -a_n & z\in (n+1,n) \,\,\text{for all nonegative } \, n \\ \\ 0 & \text{otherwise} \end{array} \right.$$
For $g$ defined above , we see $\int_R \,f(z)\, dx =0 $ and $\int_R \, f(x,y) \,dy =b_n$ whenever $n\lt x \lt n+1$ .So we have $\int_R (\int_R f(x,y) \, dy ) \, dx=s $ while $\int_R (\int_R f(x,y) \, dx ) \, dy=0 $

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