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I am still not sure about this problem and I was hoping someone can help me put it to rest once and for all.

Suppose $X=[0,1]$ and $m$ is the $\sigma$ algebra of Lebesgue measurable sets on $X$ and $\mu$ is Lebesgue measure on $X$, consider the function $g_1=2\chi_{[0,1/2)}, g_2=4\chi_{[1/2,3/4)}, g_3=8\chi_{[3/4,7/8)}$ and so on. Is the function $$f(x,y)=\sum_{n=1}^{\infty}(g_n(x)-g_{n+1}(x))g_n(y)$$ integrable on $[0,1]x[0,1]$?

I am currently working on Fubini and Tonelli theorems and I was given this problem but I'm not sure how to go about it.

Here are a few ideas I'm considering,

Since the simple functions $g_n$ are measurable and integrable over the space $[0,1]$, the function $f(x,y)$ is measurable .

Now I wish to show integrability. and As I understand, I need to show that $\int_{X \times Y}| f(x,y)|d\mu (x,y)<\infty$

For each $x$ , I notices that $\int_0^1g_n(x)dx=1$.

So I go like, for a fixed $x$,

$$ \int_0^1\int_0^1 f(x,y)=\int_0^1\int_0^1\sum_{n=1}^{\infty}(g_n(x)-g_{n+1}(x))g_n(y)dydx$$ And since $g_n(y)\geq 0$ I can switch the integral and sum. $$= \int_0^1\sum_{n=1}^{\infty}\int_0^1 (g_n(x)-g_{n+1}(x))g_n(y)dydx= \int_0^1\sum_{n=1}^{\infty} (g_n(x)-g_{n+1}(x))dx$$ That is because $\int_0^1g_n(y)dy=1$. So that yields $$=\int_0^1\sum_{n=1}^{\infty} g_n(x)dx-\int_0^1\sum_{n=1}^{\infty} g_{n+1}(x)=\int_0^1\sum_{n=1}^{\infty}2^n\chi_{E_n}(x)dx-\int_0^1\sum_{n=1}^{\infty}2^{n+1}\chi_{E_n+1}(x)dx$$ But since the $E_n's$ are Disjoint I have $$\sum_{n=1}^{\infty}2^n\chi_{E_n}=2^n\chi_{\cup_nE_n}=2^n\chi_{[0,1]}$$ So I get $$\int_0^12^n\chi_{[0,1]}(x)dx-\int_0^12^{n+1}\chi_{[0,1]}(x)dx$$ $$=1-1=0$$ On the other hand

Keeping $y$ fixed. Can I write $$\int_0^1\int_0^1\sum_{n=1}^{\infty}(g_n(x)-g_{n+1}(x))g_n(y)dxdy$$ $$=\int_Y\int_0^1\sum_{n=1}^{\infty} g_n(x)g_n(y)dxdy-\int_Y\int_0^1\sum_{n=1}^{\infty} g_n(x)g_{n+1}g_n(y)dxdy$$

$$=\int_0^1g_n(y)dy-\int_0^1g_n(y)dy=0$$

can someone pls check my idea , or show me show me how to view or tackle suck problems, thanks.

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    $\begingroup$ "As I understand, I need to show that $\int_X f(x,y)d\mu (x,y)<\infty.$" No, that's not right. You need to show $$\int_{X\times X} |f(x,y)| d\mu_2(x,y) < \infty,$$ where $\mu_2$ is Lebesgue measure on $\mathbb R^2.$ $\endgroup$ – zhw. May 3 '17 at 16:53
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    $\begingroup$ Yes, my mistake. thanks $\endgroup$ – J. Kyei May 10 '17 at 19:37
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As you say the function is measurable. It is also bounded and supported in a bounded region. So it is integrable.

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    $\begingroup$ okay, The measrability was relatively easier to see but I cant tell the boundedness by looking at the function, how can I tell. thank you $\endgroup$ – J. Kyei May 3 '17 at 5:53
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    $\begingroup$ The supports of the $g_i$ are disjoint. So for each $(x,y)$ at most one $(g_i(x)-g_{i+1}(x))g_i(y)$ is nonzero. $\endgroup$ – Lord Shark the Unknown May 3 '17 at 5:56
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    $\begingroup$ oh okay, how about my computation? does it make sense? $\endgroup$ – J. Kyei May 3 '17 at 6:00

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