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This is a question from past comp exam,

Let ${(g_n)_n }$ be a real continuous funtion with support in $(\frac{1}{n+1}, \frac{1}{n})$, such that $\int_0^1 g_n(t)dt =1$, for $n=1,2,3....$ Now define $$ f(x,y)= \sum _n^\infty [g_n(x)-g_{n+1}(x)] g_n(y)$$ on $[0,1]*[0,1]$.

I am trying to prove this function is not integrable on $[0,1]*[0,1]$. What I know is I need to prove that the absolute value of the function $f(x,y)$ is not integrable. on $[0,1]*[0,1]$. showing that the sum actually goes to infinity. I would love to see if someone there who like to prove this fact rigorously. In addition why this fact does not contradict the hypothesis of the Tonelli's and Fubini's Theorem. Thank you in advance.

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up vote 3 down vote accepted

Let $I_k:=[(k+1)^{-1};k^{-1})$ and $a_{j,k}:=\int_{I_j\times I_k}|f(x,y)|dydx$.

We have \begin{align} a_{j,k}&=\int_{I_j\times I_k}|g_k(x)-g_{k+1}(x)||g_k(y)|dydx\\ &\geqslant \int_{I_j}|g_k(x)-g_{k+1}(x)|dx \end{align} and in particular, as $g_{k+1}$ vanishes on $I_k$, $$a_{k,k}\geqslant\int_{I_k}|g_k(x)|\geqslant 1.$$ This proves the divergence of the series $\sum_{j,k}a_{j,k}$, hence the divergence of the considered integral.

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How do you argue it is greater than or equal to part (both of them), I did not see, please explain. @Davide – Deepak Dec 30 '12 at 4:36
For $a_{j,k}$, I can use Fubini as we have an integrable function. I first integrate with respect to $y$, then I use the fact that the absolute value of the integral is greater than the integral without the absolute value. – Davide Giraudo Dec 30 '12 at 11:26
Excuse me for my dumbness, can you add one or two line in the answer please. I am still missing some points. I can see the $a_{k,k}$ part though. – Deepak Dec 30 '12 at 17:33
I mean, I was okay with the line you added, what I did not see was how that absolute value of the integral is greater than the integral without the absolute value in the context of our problem. Since the $a_{j,k}$ itself defined in form of the absolute integral. I am sorry, I am still missing those points. Thanks – Deepak Dec 30 '12 at 19:10

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