I have the (infinitely) discontinuous function defined by $$f(x)=\begin{cases}
\frac{1}{n} & x\in(\frac{1}{n+1},\frac{1}{n}]\\
0 & x=0
\end{cases}$$
Who's graph is as such: 
Now, by inspection the "area under the curve" as is the informal definition of an integral is clearly greater than $\frac{1}{2}$, however using Riemann Integrals to find the integral to 1.d.p. (between $0$ and $1$) I did the following:
Proof
Let $(\mathscr{D}_{n})=\{0,\frac{1}{n},\ldots,\frac{n-1}{n},1\}$ be the regular dissection of $[0,1]$ of size $n$, then $$l_{\mathscr{D}} =\sum_{j=1}^{n}\inf\{f(x):a_{j-1}\leq x\leq a_{j}\}(a_{j}-a_{j-1})$$ $$u_{\mathscr{D}} =\sum_{j=1}^{n}\sup\{f(x):a_{j-1}\leq x\leq a_{j}\}(a_{j}-a_{j-1})$$ Clearly, $f(x)$ is increasing, so: $$m_{j}=\inf\left\{ f(x):\frac{j-1}{n}\leq x\leq\frac{j}{n}\right\} \\=f\left(\frac{j-1}{n}\right)\\=\frac{j-1}{n}$$ $$M_{j}=\sup\left\{ f(x):\frac{j-1}{n}\leq x\leq\frac{j}{n}\right\} \\=f\left(\frac{j}{n}\right)\\=\frac{j}{n}$$ So our Riemann sums are: $$l_{\mathscr{D}_{n}}=\frac{1}{n}\sum_{j=1}^{n}\frac{(j-1)}{n}\\=\frac{1}{n^{2}}\sum_{k=1}^{n-1}(k)\\=\frac{1}{n^{2}}\left(\frac{n(n-1)}{2}\right)\\=\frac{1}{2}(1-\frac{1}{n})$$ and $$u_{\mathscr{D}_{n}}=\frac{1}{n}\sum_{j=1}^{n}\frac{j}{n}\\=\frac{1}{n^{2}}\left(\frac{n(n+1)}{2}\right)\\=\frac{1}{2}(1+\frac{1}{n})$$ Then, obviously $\lim u_{\mathscr{D}_{n}} = \lim l_{\mathscr{D}_{n}}=\frac{1}{2}$ so $\intop_{0}^{1} f(x) = \frac{1}{2}$. So my question is this, why is my answer counter to the "Intuitive" idea of area under a curve, is it just this is a poor way of estimating the integral? Do the infinite points of discontinuity make a difference to the idea of "area under a curve"? Is my method even correct? Thank you.