I have a question as I look at the example 8.9(a) in Rudin's Real and Complex Analysis:

Let $X$ and $Y$ be the closed unit interval $[0,1]$, let $\{\delta_n\}$ be an increasing sequence of distinct points in $[0,1]$ that converges to $1$, and to each positive integer $n$, let $g_n$ be a real continuous function on $[0,1]$ with support in $(\delta_n,\delta_{n+1})$, and such that $\int_{0}^{1} g_n(t)~dt=1$. Define $f$ over $X\times Y$ as follows: $$ f(x,y):=\sum_{n=1}^\infty[g_n(x)-g_{n+1}(x)]g_n(y). $$

It is easy to check that the Fubini Theorem does not apply for $f(x,y)$. And the book says that this is because the function of $f(x,y)$ is not integrable, i.e.


But I could not easily see why it is not integrable.

  • 1
    $\begingroup$ From FAQ about tags: Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag. It's not easy to keep balance between too specific tags and not having enough tags, but it is always good to search first and to ask yourself, whether newly created tag is not too specific. (Of course, you can disagree with the removal of the tag you've created, and there is possibility for further discussion, if needed.) $\endgroup$ Jun 6 '19 at 9:36
  • $\begingroup$ IMO, fubini is a good and important topic. There are so many non-integrable functions under integration tags, and how this function f is formulated is not that easily to be specified nor does this $f$ has a famous name. Thus I feel in this case, the fubini context will help for a better searching from google, because it gives a specific context to this non-integrable function. I do appreciate your input and help for a better math.SE community $\endgroup$
    – Jay Zha
    Jun 7 '19 at 1:30

You are probably familiar with the following example of a non-integrable function

enter image description here

If you integrate over $x$ and then over $y$ you get $0$, but if you integrate first over $y$ and then over $x$ you get $1$. Here it is easy to see that the integral of the absolute value diverges and therefore the function is not integrable.

If you think about it, your function captures essentially the same idea. Yes, in your case the horizontal and vertical lines would be the sequence $\delta_{n}$ (and would not extend to infinity), and the function won't necessarily be constant on those squares (rectangles in your case). The integral of your $f$ would still evaluate to $1$ on each cell

$$\int_{\rm cell}f\left(x,y\right){\rm d}A=1$$


$$\int_{\rm cell}\left|f\left(x,y\right)\right|{\rm d}A\geq\left|\int_{\rm cell}f\left(x,y\right){\rm d}A\right|=1$$

and it is clear that $f$ is not integrable, since there is an infinite number of such cells.

  • $\begingroup$ Thanks, this is really a good answer! $\endgroup$
    – Jay Zha
    Jun 5 '19 at 4:10
  • $\begingroup$ You're welcome! $\endgroup$
    – eranreches
    Jun 5 '19 at 4:16

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