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This is example 8.9(a) in Rudin's Real and Complex Analysis, (alternatively, exercise 10.2 in Rudin's Principles of Mathematical Analysis).

Let $X$ and $Y$ be the closed unit interval $[0,1]$, let $\{\delta_n\}$ be an increasing sequence of distinct points in $[0,1]$ that converges to $1$, and to each positive integer $n$, let $g_n$ be a real continuous function on $[0,1]$ with support in $(\delta_n,\delta_{n+1})$, and such that $\int_{0}^{1} g_n(t)~dt=1$. Define $f$ over $X\times Y$ as follows: $$ f(x,y):=\sum_{n=1}^\infty[g_n(x)-g_{n+1}(x)]g_n(y). $$ If $x$ is held fixed, then $g_n(x)-g_{n+1}(x)$ is nonzero for at most two choices of $n$, and $f$ reduces to a finite sum. So, \begin{align*} \int_X~dx\int_Y f(x,y)~dy&=\int_0^1~dx\int_0^1\sum_{n=1}^\infty[g_n(x)-g_{n+1}(x)]g_n(y)~dy \\ &=\int_0^1~dx\sum_{n=1}^\infty\int_0^1[g_n(x)-g_{n+1}(x)]g_n(y)~dy \\ &=\int_0^1\sum_{n=1}^\infty g_n(x)-g_{n+1}(x)~dx \\ &=\int_0^1 g_1(x)~dx=1. \end{align*} On the other hand, if $y$ is held fixed, then $g_n(y)$ is nonzero for at most one choice of $n$, and $f$ reduces to a finite sum. So, \begin{align*} \int_Y~dy\int_X f(x,y)~dx&=\int_0^1~dy\int_0^1\sum_{n=1}^\infty[g_n(x)-g_{n+1}(x)]g_n(y)~dx \\ &=\int_0^1~dy\sum_{n=1}^\infty\int_0^1[g_n(x)-g_{n+1}(x)]g_n(y)~dx\\ &=\int_0^1 \sum_{n=1}^\infty g_n(y)-g_n(y)~dy\\ &=\int_0^1 0~dy = 0. \end{align*}

The double integrals are not interchangeable in this example, in part, because $$ \int_0^1~dx\int_0^1|f(x,y)|~dy=\infty. $$ Why is this true?

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You cannot simply interchange the sum and integral without justifying it.

Let me repeat the premise, i.e.

Let $X$ and $Y$ be the closed unit interval $[0,1]$, let $\{\delta_n\}$ be an increasing sequence of distinct points in $[0,1]$ that converges to $1$, and to each positive integer $n$, let $g_n$ be a real continuous function on $[0,1]$ with support in $(\delta_n,\delta_{n+1})$, and such that $\int_{0}^{1} g_n(t)~dt=1$. Define $f$ over $X\times Y$ as follows: $$ f(x,y):=\sum_{n=1}^\infty[g_n(x)-g_{n+1}(x)]g_n(y). $$


Set $$h(x)=\int_0^1 f(x,y)dy,$$ then $$ h(x)=\begin{cases}\displaystyle\int_0^1 g_1(x)g_1(y)dy=g_1(x),& x\in(\delta_1,\delta_2);\\ \displaystyle\int_0^1 [g_n(x)g_n(y)-g_n(x)g_{n-1}(y)]dy=0,& x\in(\delta_n,\delta_{n+1}),\ \forall n\in\mathbb Z_{> 1}.\end{cases} $$ Hence $$ \int_0^1dx\int_0^1f(x,y)dy=\int_0^1h(x)dx=\int_0^1g_1(x)dx=1. $$


However, if we set $$ h(y)=\int_0^1f(x,y)dx, $$ then we have $$ h(y)=\int_0^1 [g_n(x)-g_{n+1}(x)]g_n(y)dx=0,\ x\in(\delta_n,\delta_{n+1}),\ \forall n\in\mathbb Z_{\ge 1}. $$ Hence $$ \int_0^1dy\int_0^1f(x,y)dx=\int_0^1h(y)dy=0. $$


To see that $$ \int_0^1dy\int_0^1|f(x,y)|dx=\infty ,$$ set $$ h(y)=\int_0^1|f(x,y)|dx ,$$ then when $y\in(\delta_n,\delta_{n+1})$, $n\in\mathbb Z_{\ge 1}$, we have \begin{align} h(y)&=\int_0^1|g_n(x)-g_{n+1}(x)||g_n(y)|dx\\ &=|g_n(y)|\int_0^1|g_n(x)-g_{n+1}(x)|dx\\ &=|g_n(y)|\left(\int_0^1|g_n(x)|dx+\int_0^1|g_{n+1}(x)|dx\right)\\ &\ge 2|g_n(y)|. \end{align} Then we have \begin{align} \int_0^1dy\int_0^1|f(x,y)|dx&=\int_0^1h(y)dy\\ &\ge\sum_{n=1}^N\int_{\delta_n}^{\delta_{n+1}}2|g_n(y)|dy\\ &\ge 2N\to\infty\ \text{as}\ N\to\infty. \end{align} By Fubini Theorem, we know that $$ \int_0^1dx\int_0^1|f(x,y)|dy=\int_0^1dy\int_0^1|f(x,y)| dx=\infty.$$

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  • $\begingroup$ He has justified his exchange of sum and integral when he says the sum is finite. Hasn't he? $\endgroup$ – P-addict Nov 9 at 16:11
  • $\begingroup$ Also, can you explain how you go from $\int_{0}^{1} |g_{n}(x) - g_{n+1}(x)| dx$ to the below expression? $\endgroup$ – P-addict Nov 9 at 16:18

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