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.
As you say the function is measurable. It is also bounded and supported in a bounded region. So it is integrable.