# 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 $[0,\infty)$ but $f(x)$ does not tend to zero as $x\rightarrow\infty$.

First of all, let $g_{n}$ be the function $\displaystyle\sin(2^{n}x)\chi_{[0,\frac{\pi}{2^{n}}]}$, 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

 $\int_{\mathbb{R}}g_{n}\,d\mu=2^{1-n},$ (1)

and $\displaystyle g_{n}\left(\frac{\pi}{2^{n+1}}\right)=1$. 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 $\displaystyle f=\sum_{r=0}^{\infty}h_{r}$. There are no convergence problems with this sum since for a given $x\in\mathbb{R}$, at most one $h_{r}$ takes a non-zero value at $x$. Also $f(x)$ does not tend to 0 as $x\rightarrow\infty$ 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 $\displaystyle f_{n}=\sum_{r=0}^{n}h_{r}$. We must check the hypotheses of the MCT. Clearly $f_{n}\rightarrow f$ as $n\rightarrow\infty$, 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 $\displaystyle\int_{\mathbb{R}}f_{n}\,d\mu\leq 4$ for all $n$. Therefore, the MCT applies and $f$ is integrable.

Title an integrable function that does not tend to zero AnIntegrableFunctionThatDoesNotTendToZero 2013-03-22 16:56:09 2013-03-22 16:56:09 silverfish (6603) silverfish (6603) 9 silverfish (6603) Example msc 28-01