Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $[0,1]\times [0,1].$

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer
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
okay. Im a little bit confused about the "at most four terms of the sum above can be non-zero at x;" Could you please elaborate. – Josh May 5 '12 at 18:44
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.