Let R[0,1] be the set of Riemann integrable functions between 0,1 by considering $f_{n}(x)=\frac{1}{\sqrt{x}}\chi _{[n^{-1},1]}(x)$ Where $\chi _{[n^{-1},1]}$ is the indicator function for if x is between $n^{-1},1$ with respect to the norm $\left \| f \right \|_1=\int_{a}^{b}\left | f(x ) \right |dx$
Edit: I have to show this using the function $f_{n}(x)=\frac{1}{\sqrt{x}}\chi _{[n^{-1},1]}(x)$ so I don't think it's a duplicate since the other question is considering different functions and my issue is showing this function doesn't converge.
Thoughts so far
First I want to show that $f_{n}$ is Cauchy so I consider for $n>m$ $$\left \| f_{n }-f_{m} \right \|_1=\int_{0}^{1/n} 0\, dx+\int_{1/n}^{1/m} \frac{1}{\sqrt{x}}\,dx +\int_{1/m}^{1} 0 \, dx $$
I think this is Cauchy since as $n,m\rightarrow \infty $ then $1/n,1/m \rightarrow 0$ so $\left \| f_{n}-f_{m} \right \|_1 \rightarrow 0$ which is less than $\epsilon$ $ \forall$ $\epsilon >0$, But does this not imply $f_n$ converges to $0\in R[0,1]$
Edit: New idea Assume $f_{n}$ converges to $f$ then $\left \| f_{n}-f \right \|_1 < \epsilon $ for some $n>N\in \mathbb{N} $ and $\epsilon >0$ so $$\left \| f_{n}-f \right \|_1 =\int_{0}^{1/n} 0\, dx+\int_{1/n}^{1} \frac{1}{\sqrt{x}}\,dx $$
$\Rightarrow f(x)=1/\sqrt{x}$ when x is between $1/n$ and $1$ and 0 otherwise which is $\notin$R[0,1]
