an integrable function that does not tend to zero

In this entry, we give an example of a function $f$ such that $f$ is Lebesgue integrable on $[0,\mathrm{\infty })$ but $f(x)$ does not tend to zero as $x\to \mathrm{\infty }$.

First of all, let $g_n$ be the function $\sin (2^{n}x){\chi }_{[0,\frac{\pi }{2^n}]}$, where ${\chi }_I$ denotes the characteristic function of the interval $I$. In other words, $\chi$ takes the value $1$ on $I$ and 0 everywhere else.

Let $\mu$ denote Lebesgue measure. An easy computation shows

$${\int }_{ℝ}{g}_n𝑑\mu =2^{1-n},$$ (1)

and $g_n\left({\displaystyle \frac{\pi }{2^{n+1}}}\right)=1$. Let $h_n(x)=g_n(x-n\pi )$, so $h_n$ is just a “shifted” version of $g_n$. Note that

$$h_n\left(n\pi +\frac{\pi }{2^{n+1}}\right)=1.$$ (2)

We now construct our function $f$ by defining $f={\displaystyle \sum _{r=0}^{\mathrm{\infty }}}{h}_r$. There are no convergence problems with this sum since for a given $x\in ℝ$, at most one $h_r$ takes a non-zero value at $x$. Also $f(x)$ does not tend to 0 as $x\to \mathrm{\infty }$ as there are arbitrarily large values of $x$ for which $f$ takes the value $1$, by (2).

All that is left is to show that $f$ is Lebesgue integrable. To do this rigorously, we apply the monotone convergence theorem (MCT) with $f_n={\displaystyle \sum _{r=0}^n}{h}_r$. We must check the hypotheses of the MCT. Clearly $f_n\to f$ as $n\to \mathrm{\infty }$, and the sequence $(f_n)$ is monotone increasing, positive, and integrable. Furthermore, each $f_n$ is continuous and zero except on a compact interval, so is integrable. Finally, from (1) we see that ${\displaystyle {\int }_{ℝ}}{f}_n𝑑\mu \le 4$ for all $n$. Therefore, the MCT applies and $f$ is integrable.

Title	an integrable function that does not tend to zero
Canonical name	AnIntegrableFunctionThatDoesNotTendToZero
Date of creation	2013-03-22 16:56:09
Last modified on	2013-03-22 16:56:09
Owner	silverfish (6603)
Last modified by	silverfish (6603)
Numerical id	9
Author	silverfish (6603)
Entry type	Example
Classification	msc 28-01