# Improper integral and sequences

I've got this question on homework assignment and couldnt solve it hope you can help! The question is: function $f$ is integrable on any subinterval $[c, d]$, and the improper integral of $f$ on the interval $[0,\infty)$ converges. Prove that there is a sequence $x_n$ such that $f(x_n)$ converges to $0$.

Suppose the function f is zero at finitely many points on $\mathbb{R}$. Then there exists a positive measure set $A_1$ such that $f(x)\leqslant 1$ there. Otherwise the integral becomes infinite. Choose one point $x_1$ from there. Again get a positive measure set $A_2$ such that $f(x)\leqslant 1/2$ there. We will get this $A_2$ due to the same reason. and thus we get a sequence with the above required property.

This is false. Let $h_0 =0,$ and for $n=1,2,\dots,$ let $h_n = \sum_{k=1}^{n}\dfrac{1}{k}.$ Then $h_n \to \infty.$ Define

$$f=\sum_{n=0}^{\infty}(-1)^n \chi_{[h_n,h_{n+1})}.$$

The series

$$(h_1-h_0)-(h_2-h_1) + (h_3-h_2) - (h_4-h_3) + \cdots = 1-\frac{1}{2} +\frac{1}{3} - \frac{1}{4} + \cdots$$

converges by the alternating series test (in fact, to $\ln 2$). This implies $\int_0^\infty f$ converges. However $f$ takes on only the values $\pm 1.$

• could you explain a little more about the series f and it's elements? I didn't quite understand the notation. how does it's elements look like? – Yaron Scherf Jun 26 '17 at 17:43
• @YaronScher $\chi_A$ is the function that is $1$ on $A$ and $0$ everywhere else. I edited my answer a bit. – zhw. Jun 26 '17 at 18:00
• This answer shows that "$f$ is integrable on $[0, \infty )$" is not sufficient. What we need is that "$f$ is absolutely integrable on $[0, \infty )$": in such a case the other answer works fine. – Crostul Jun 26 '17 at 18:13
• @Crostul Or assume $f$ is continuous – zhw. Jun 26 '17 at 18:16