Cauchy sequence in $L^1$ Space I am learning about the $L^1$ space (the complete Riemann integrable functions) and I am not used to using $\epsilon, \delta$ in these type of problems yet. Here is my attempt. Below I want to that $f_n$ is a Cauchy sequence.
Let $\epsilon > 0$ be given.
    $$
 \int |f_n - f_m| = \int_{1/n}^{1/m} x^{-1/2} dx = 2 \cdot \left[\left(\frac{1}{m}\right)^{1/2} - \left(\frac{1}{n}\right)^{1/2}\right] < \epsilon $$
I do not know how to find $N$ such that $n, m \geq N$ the above happens.
Also if I just say as $n, m$ approaches infinity, then we can easily see that $\int |f_n - f_m|$ approaches zero. Is this okay enough so that we don't really have to use $\epsilon, \delta$ here?

Let $f_n = x^{-1/2}\chi_{[1/n, 1]}$.
Let $\epsilon >0$. Choose $N>\frac{4}{\epsilon^2}$ such that when $n,m > N$, then
$$\begin{align}\int |f_n - f_m| &= \int |x^{-1/2}\chi_{[1/n, 1]}-x^{-1/2}\chi_{[1/m, 1]}| \\
&= \int_{1/n}^{1/m} x^{-1/2} dx \\
&= 2 \cdot \left[\left(\frac{1}{m}\right)^{1/2} - \left(\frac{1}{n}\right)^{1/2}\right] \\
&< 2 \cdot \frac{1}{m}^{1/2} < 2 \frac{1}{N^{1/2}} < \epsilon
\end{align}$$ 
