Let $f:[0,2] \rightarrow \mathbb{R}$ be given by
$$f(x) = \left\{ \begin{array}\ 10,\quad \ 0\leq x < 1, \\ 100,\quad \ x = 1, \\ -5,\quad \ 1 < x \leq 2. \\ \end{array} \right. $$
Prove that $f$ is Darboux integrable and compute $\int_{0}^{2}f$.
Attempt
That the function is Darboux integrable means for all $\epsilon > 0$, there exists a partition $P$ of $[0,2]$ such that $U(f,P) - L(f,P) < \epsilon$.
Suppose $P = \{t_{0}, \dots, t_{n}\}$ is a partition of $[0,2]$ with $t_{j-1} < 1 < t_{j}$.
For, $t_{0} < \dots < t_{j-1}$: $m_{i} = M_{i} = 10$
Also for $t_{j} < \dots < t_{n}$: $m_{i} = M_{i} = -5$
Now I'm having trouble with managing $x = 1$ which is where the discontinuity is and obviously the challenge of the problem. At first I was going to say that $m_{i} = M_{i} = 100$ whichever interval the number 1 is in and that would give me:
$$L(f,P) = \sum_{i = 1}^{n}m_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}-5(t_{i} - t_{i-1})$$
and
$$U(f,P) = \sum_{i = 1}^{n}M_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}-5(t_{i} - t_{i-1})$$
Then subtracting these I would get $U(f,P) - L(f,P) = 0 < \epsilon$.
But I feel this isn't the right answer and I need to express the partition a bit more explicitly. Also whatever $U(f,P) - L(f,P)$ is, it is going to eventually converge to what the integral is. And when calculating the integral (as a check using earlier calc techinques) I get $5$ which is not the value of the difference in the upper and lower sums. Where am I going wrong?
Fixed $\varepsilon$ and taken $\delta$ small enough, you can take $$ t_{j-1}=1-\delta,\qquad t_j=1+\delta, $$
so that \begin{align} L(f,P) &= \sum_{i = 1}^{n}m_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + \mathbf{(-5)}(t_{j} - t_{j-1}) + \sum_{j+1}^{n}(-5)(t_{i} - t_{i-1})=\\ &=10(1-\delta)-5\cdot2\delta-5(1-\delta)=5-15\delta\\ U(f,P) &= \sum_{i = 1}^{n}M_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}(-5)(t_{i} - t_{i-1})=\\ &=10(1-\delta)+100\cdot2\delta-5(1-\delta)=5+195\delta \end{align} so that to have $$ U(f,P)-L(f,P)=210\delta<\varepsilon $$ you have to choose $$ \delta<\frac{\varepsilon}{180}. $$