Calculating U(f,$P_0$) and L(f,$P_0$) Consider f(x) = {5 if x=1, 2 if x$\neq$1
Let $\epsilon$ > 0. Consider the partition of [0, 10] given by $P_0$ = {0, 1-$\epsilon\over{12}$, 1+$\epsilon\over{12}$,10}. How much is U(f,$P_0$) and L(f,$P_0$)?
For L(f,$P_0$), I have the following, but am not sure if I am doing it correctly:
2($\epsilon\over{12}$) +5 (10 - (0+$\epsilon\over{12}$)) = 50 - $\epsilon\over{4}$.
 A: Here, the partition is fixed, right? So $L(f,P_0)$ depends on $P_0$.
The definition of $L(f,P_0)$ is the following: Suppose that $P_0 = [x_0,x_1,\ldots,x_n]$, where $x_0=0$ and $x_1=10$ (in this case), and $x_i \neq x_j$. Then, the lower sum $L(f,P_0)$ is defined by:
$$
L(f,P_0) = \sum_{i=1}^{n} (x_i-x_{i-1}) \inf_{[x_{i-1},x_{i}]} f(x)
$$
And the upper limit is defined as:
$$
U(f,P_0) = \sum_{i=1}^{n} (x_i-x_{i-1}) \sup_{[x_{i-1},x_{i}]} f(x)
$$
Hence, since $P_0 = [0,1-\frac{\epsilon}{12},1+\frac{\epsilon}{12},10]$, we get that:
$$
L(f,P_0) = 2\left(1-\frac{\epsilon}{12}\right) + 2\left(\frac{2 \epsilon}{12}\right) + 2\left(9-\frac{\epsilon}{12}\right)
$$
(Take the differences, and note that the value $5$ is taken at $x=1$, so the infimum of $f$ is $2$ in each interval).
This simplifies to:$$
L(f,P_0) = 20
$$
as you can check.
Similarly, in the upper sum, only the supremum of the middle interval changes to $5$.
So:
$$
U(f,P_0) = 2\left(1-\frac{\epsilon}{12}\right) + 5\left(\frac{2 \epsilon}{12}\right) + 2\left(9-\frac{\epsilon}{12}\right)
$$
Hence the upper sum evaluates to:
$$
L(f,P_0) = 20 + \frac{\epsilon}{2}
$$
As you can check.
To see that the function is integrable, see that when $\epsilon \to 0$, then $U(f)$ and $L(f)$ coincide, hence the function is integrable, and it's integral is $20$.
