Very simple question here, I almost feel bad for asking it..

Lets say we have a function bounded between $0$ and $1$. This function is high dimensional:
$0<f(X) \le1, ~~~ X \in \mathbb{R}^D$

Now, we calculate the limit for all elements of $X$ going to plus and minus infinity. We find out that they are zero.

Can we say that the integral of the function over the entire domain of $X$ is finite?

Can we say that if we get even non-zero limit?

Finally, if the zero limit is insufficient, is there some other condition that suffices?


2 Answers 2


Imagine a simple example in 1 dimension $$f(x)=\left\{\begin{array}{c}1/|x|,\quad x>1\\ 0,\quad \mbox{otherwise}\end{array}\right.$$

This function is bounded, and its limit at infinity is zero. However, $\int_{-\infty}^{\infty}f(x)dx=\infty$. the condition $\lim_{x\to\pm\infty}=0$ is a necessary condition for the converging of the integral, but it is not sufficient.

In order to get a convergence, you need an extra assymptotic behavior of $f(x)$: it has to go to zero faster than $1/x$. For example, the function

$$f(x)=\left\{\begin{array}{c}1/|x|^p,\quad x>1\\ 0,\quad \mbox{otherwise}\end{array}\right.$$ with converges $\forall p\in(1,\infty)$

  • $\begingroup$ That messes my brain up a lot. I changed the question slightly, is that ok? $\endgroup$ Apr 29, 2016 at 7:25
  • $\begingroup$ Well, I would assume (with great caution) that this is not very important. I mean this function does have limit zero and is defined over $R^D$, in a bounded way. $\endgroup$ Apr 29, 2016 at 7:28
  • $\begingroup$ @gebruiker Try $f(x)=\min\left(1,\frac{1}{||X||}\right)$ to extend to the multidimensions of the original question $\endgroup$
    – Henry
    Apr 29, 2016 at 7:28
  • 1
    $\begingroup$ Great example, I really like how you gave the marginal case as a counter example. $\endgroup$ Apr 29, 2016 at 7:35

No, you cannot conclude that the integral is finite.

As an example take a function like $$f(x,y)=\min\{1, 1/\sqrt{xy}\}$$ This will be bounded between $0$ and $1$, but for large $x$ and $y$ the function will be $1/\sqrt {xy}$ and its integral behaves like $\sqrt{xy}$ and thus is not finite.

  • $\begingroup$ Is there an additional condition that will suffice? $\endgroup$ Apr 29, 2016 at 7:33
  • $\begingroup$ See the edit in seoanes answer. $\endgroup$
    – TheAbelian
    Apr 29, 2016 at 7:35

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