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 \begin{equation} \label{g} \int_{\mathbb{R}} g_n \, d\mu= 2^{1-n},\end{equation}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 \begin{equation} \label{h} h_n \left( n \pi + \frac{\pi}{2^{n+1}} \right) = 1. \end{equation} 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 ().

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 () we see that $\displaystyle \int_{\mathbb{R}} f_n \, d\mu \leq 4$ for all $n$ . Therefore, the MCT applies and $f$ is integrable.