A Cauchy sequence in $L^2$ 
Problem
Is $f_n = 1_{(n, \infty)} \frac{1}{x^2} $ Cauchy in $L^2(0, \infty ) $
??

$$ ||f||_2 = (\int\limits_E |f|^2 )^{1/2} $$
Atttempt
We start as follows:
$$ || f_n - f_m ||^2_2 = \int\limits_{(0, \infty)} \frac{1}{x^4}|f_n - f_m|\,dx = \int\limits_{(n, \infty)} \frac{1}{x^4} + \int\limits_{(m, \infty)}\frac{1}{x^4} $$
Since we have $p$-integral both integrals must converge to finite number. Is it enough to conclude it is Cauchy in $L^2$? thanks
 A: Solution 1
You are basically done: let $\epsilon > 0$. We want to show that there exists $N = N(\epsilon) \in \mathbb{N}$ such that for every $n \ge m \ge N$ we have $$\int|f_m - f_n|^2 < \epsilon^2$$


*

*Note that $|f_m - f_n|^2 = \frac{1}{x^4}\chi_{(m,n]} \le \frac{1}{x^4}\chi_{(m,\infty)}$.

*Note also that $\frac{1}{x^4}\chi_{(m,\infty)} \to 0$ pointwise as $m \to \infty$

*We also have that for every $m \in \mathbb{N}$ $\frac{1}{x^4}\chi_{(m,\infty)} \le \frac{1}{x^4}\chi_{(1,\infty)} \in L^1((0,\infty))$

*By the Lebesgue dominated convergence theorem we get $$\int_{(0,\infty)}\frac{1}{x^4}\chi_{(m,\infty)} \to 0$$

*We can rewrite this as follows: for every $\delta > 0$ there exists $M = M(\delta)$ such that if $m \ge M$ then $$\int_{(0,\infty)}\frac{1}{x^4}\chi_{(m,\infty)} < \delta$$

*This shows that $N(\epsilon) := M(\epsilon^2)$ does the trick.


Solution 2
Note that a subset of the previous proof is sufficient to conclude:


*

*We have that for every $m \in \mathbb{N}$ $\frac{1}{x^4}\chi_{(m,\infty)} \le \frac{1}{x^4}\chi_{(1,\infty)} \in L^1((0,\infty))$

*By the Lebesgue dominated convergence theorem we get $$\int_{(0,\infty)}\frac{1}{x^4}\chi_{(m,\infty)} \to 0$$

*Converging sequences in a Banach space are Cauchy.
I hope this helps :)
A: A small mistakes in your calculation. Assume $n>m$, then $|f_n-f_m|=1_{(m,n]}\frac{1}{x^2}$, since the Riemann integral is convergent, then by the relation between Riemann and Lebesgue.
It is Cauchy sequence.
