Integration of a piecewise defined discontinuous function In a proof i posted recently on this site (link) i made the mistake of thinking that a function $f$, which is bounded on a closed interval $[a,b]$, would assume its infimum at a certain $x$ in the domain of $f$. I was presented with a counter example which, for my ability, seemed rather complicated. So i tried to find an easier counter example. I found the following function:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
Here, $ inf \{ f(x) : 0 \le x \le 1 \} = 0 $ but $f(x) \neq 0$ for all $x$.
I guess this is suitable as a counter example. Then i came up with the idea of trying to integrate this function. So came to the following proposition:
Proposition: Let $f$ be a function defined on $[a,b]$ as follows:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
This function is integrable with:
$$ \int_0^b = \frac{b^2}{2}$$
Proof: I use a partition $ P = \{t_0, ... , t_n \} $ of $[a,b]$ with
$$ t_i - t_{i-1} = \frac{b}{n} $$
$$ t_i = \frac{ib}{n} $$
$$ t_{i-1} = \frac{(i-1)b}{n} $$
$$ m_i = inf \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{(i-1)b}{n} $$
$$ m_1 = inf \{ f(x) : t_0 \le x \le t_1 \} = 0 $$
$$ M_i = sup \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{ib}{n} $$
$$ M_1 = sup \{ f(x) : t_0 \le x \le t_1 \} = 1 $$
Then we have
$$ L(f, P) = \sum_{i=2}^n m_i \cdot \frac{b}{n} + m_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{(i-1)b}{n} \cdot \frac{b}{n} + 0 \cdot \frac{b}{n} = \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j = \frac{b^2}{2} \cdot \frac{n-1}{n}$$
and
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{ib}{n} \cdot \frac{b}{n} + 1 \cdot \frac{b}{n} = \Biggl[ \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j+1 \Biggl] + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
For the difference of the upper and lower sums this results in
$$ U(f, P) - L(f, P) = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{2}{n} + \frac{b}{n} = \frac{b^2+b}{n}$$
So in order get $ U(f, P) - L(f, P) < \epsilon$ we can choose $ n > \frac{b^2+b}{\epsilon} $. Thus $f$ is integrable and since
$$ \frac{b^2}{2} \cdot \frac{n-1}{n} \le \frac{b^2}{2} \le \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
and the integral is unique, if it exists, we have
$$ \int_0^b = \frac{b^2}{2}$$
as required. $ \blacksquare $
All these equations and manipulations are quite complex regarding my ability. So might anyone tell me if this is correct or point me towards my mistakes? Thanks in advance.

In retrospection i think i found a mistake in my proof. I used $\sum_{j=1}^{n-1} j+1 = \frac{n(n+1)}{2}$ which is false, since $\sum_{j=1}^{n-1} j+1 = \frac{n^2 + n -2}{2}$.
With this discovery and the hint of @Andrea S. i found
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{ib}{n} \cdot \frac{b}{n} + 1 \cdot \frac{b}{n} = \Biggl[ \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j+1 \Biggl] + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{(n-1)(n+2)}{n^2} + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{(n-1)}{n} \cdot \frac{n+2}{n} + \frac{b}{n}$$
and
$$ L(f, P) = \frac{b^2}{2} \cdot \frac{n-1}{n}$$
Now, since $\lim_{n\to \infty} \frac{(n-1)}{n} = \lim_{n\to \infty} \frac{n+2}{n} = 1$ and $\lim_{n\to \infty} \frac{b}{n} = 0$ we have
$$\lim_{n\to \infty}L(f,P)=\lim_{n\to \infty}U(f,P_n)=\frac{b^2}{2}$$
as required. $\blacksquare$
