# Deciding whether function $f(x,y)$ is integrable.

First of all, I am sorry if this seems basic. I just don't know where to begin.

Let $X=[0,1]$. Let $\mu$ be the Lebesgue measure. Consider the functions $g_1=1_{[0,1/2]}~,g_2=1_{[1/2,3/4]}~,g_3=1_{[3/4,7/8]}~\ldots$ The question I want to ask is whether or not the function $f(x,y) = \sum_{n=1}^\infty [g_n(x)-g_{n+1}(x)]g_n(y)$ is integrable on $[0,1]\times [0,1].$

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should $g_n = 1_{[0, 1/n]}$? – user29743 May 5 '12 at 15:56
@countinghaus. Sorry I made a typo. – Josh May 5 '12 at 16:03
Calculate the integral of the partial sum, then take limit to see what happens. – leo May 5 '12 at 16:15

The function is clearly measurable, so it's enough to find a dominating integrable function. Notice $$|f(x,y)|\leq \sum_{n=1}^{\infty} g_n(x)+g_{n+1}(x)$$ for all $x,y \in [0,1]^2$. Now notice that, for each $x \in [0,1]$ at most four terms of the sum above can be non-zero at $x$ (the intervals defining the $g_n$ have disjoint interior), so we have that the sum is bounded above by (say) $4$, for all $x$. Since $4$ is integrable in $[0,1]^2$ we're done.
what happened to $g_n(y)$? – Josh May 5 '12 at 17:07
@Josh It's a characteristic function, so its absolute value is bounded above by $1$. – Jose27 May 5 '12 at 17:19
@Josh: Sure, pick any $x\in [0,1]$, the $g_n$ 's are characteristic functions of sets $A_n$ satisfying $A_n\cap A_m\neq \emptyset$ if and only if $m=n-1,n,n+1$. Since these $A_n$ cover the unit interval $x$ must belong to some $A_k$, by what's been said, $x$ is then either in the interior of $A_k$ (in which case only $g_k(x) \neq 0$), or it's an endpoint of the interval $A_k$ in which case it belong to either $A_{k-1}$ or $A_{k+1}$ (never both!). Since each $g_n$ appears at most two times in the series, the bound follows. – Jose27 May 5 '12 at 19:12