What is wrong about this proof?

Question

Statement: if $f(x)\ge0$ for all x, and $\int_a^\infty f(x)dx$ converges for some constant a>0, then $\lim_{x\to \infty} f(x)=0$.

The statement is in fact wrong, and I know it, but I mananged to give a proof and I don't know what is wrong about it.

My proof:

By contradiction,assume the limit is not 0. Then, there exsists some $\varepsilon_0>0$, and an $x_0$, such that for all $x>x_0$:

$|f(x)-0|=|f(x)|=f(x)\ge \varepsilon_0$.

Define $g(x)=\varepsilon_0$, we know that for all $x>x_0: g(x)\le f(x)$.

$\int_a^\infty g(x)dx=\lim_{M\to \infty} \varepsilon_0(M-a)=\infty$

Both functions are non-negative, therefore by the comparison test, the spoken integral of f(x) diverges, in contradiction to the fact that it converges.

Answer 1

The statemente is not true because you can pick a function such as follows.

$$ f(x)=\left\{ \begin{array}{} 1 \text{, if } x\in\mathbb{Q} \\ 0 \text{, if } x\in\mathbb{R}\setminus\mathbb{Q} \end{array} \right. $$

The function doesn't have a limit, still the integral converges.

In your proof instead you are assuming the function has a limit.

Answer 2

Let's take another example from the box of counter-examples, with

$$f(x)=\left\{ \begin{array}{cl} n &\text{ if } x\in \left[n,n+\frac1{n^3}\right) \text{ for positive integer } n \\ 0 &\text{ otherwise } \end{array} \right.$$

This satisfies your conditions as you have $\int\limits_0^\infty f(x)\,dx = \sum\limits_1^\infty \frac1{n^2} =\frac{\pi^2}{6}$

but your statement

there exists some $\varepsilon_0>0$, and an $x_0$, such that for all $x>x_0$: $|f(x)-0|=|f(x)|=f(x)\ge \varepsilon_0$

is not correct.

Answer 3

An example can be provided by a concise formula: $$f(x)=(1-|\sin x|)^{x^2/\pi^2},\quad x\ge \pi$$ Then the limit points of $f(x)$ at infinity coincide with the interval $[0,1]$ and the function is integrable. Indeed $$\int\limits_\pi^\infty f(x)\,dx =\sum_{n=1}^\infty \int\limits_{n\pi}^{(n+1)\pi} f(x)\,dx= \sum_{n=1}^\infty \int\limits_{0}^{\pi} f(x+n\pi)\,dx$$ Next $$\int\limits_{0}^{\pi} f(x+n\pi)\,dx\le \int\limits_0^\pi (1-\sin x)^{n^2}\,dx=2\int\limits_0^{\pi/2} (1-\sin x)^{n^2}\,dx\\ \le 2\int\limits_0^{\pi/2}\left (1-{2\over \pi}x\right )^{n^2}\,dx ={4\over \pi (n^2+1)}$$