Using Darboux sums to find the value of an integral

Question

I was given the following problem. Given $$ f:[0,1] \longrightarrow \mathbb{R}, f(x)= \begin{cases} 0 & 0\leq x\leq \frac{1}{2} \\ 1 & \frac{1}{2}< x \leq 1 \end{cases} $$

Prove that $f$ is integrable in $[0,1]$ and show that $$ \int_0^1 f\, = \frac{1}{2}. $$

So, proving $f$ is integrable I think I understand fairly well. I chose a partition $ P = \{x_0, x_1, ..., x_n \} $ such that $ \frac{1}{2} \in (x_{t-1},x_t)$ and $ |x_t - x_{t-1}| < \epsilon $ with $ \epsilon > 0 $ for some $t \in \{1, 2, ..., n\}$. It naturally follows that the upper and lower sums under partition $P$ are $ U(f,P) - L(f,P) < \epsilon $, and thus the function is integrable. However, I fail to produce a convincing argument as to why the integral is $\frac{1}{2}$. The textbook I'm using just states that "it's obvious that $\frac{1}{2} $ is the only number that is $ L(f,P) \leq \frac{1}{2} \leq U(f,P) $ for every partition $ P $", but I don't quite see it. Is there a way to make this argument more explicitly, choosing a particular partition?

Answer 1

Take any partition $P=\{ x_0, x_1, \dots, x_n \}$ of $[0,1]$. If $x_k = 1/2$ for some $0 \leq k \leq n$ then I believe you can easily show $\displaystyle L(f,P) \leq \frac{1}{2} \leq U(f,P)$. Otherwise, consider $P'$ refinement of $P$ which includes $1/2$. Then you have the following chain of inequalites $L(f,P) \leq L(f,P') \leq \dfrac{1}{2} \leq U(f,P') \leq U(f,P)$.

Answer 2

For every partition $ P = \{x_0, x_1, ..., x_n \} $, there is a $t\in\{1,2,\dots,n\}$ such that $ \frac{1}{2} \in (x_{t-1},x_t)$.

For every interval before this one, the maximum and minimum of $f$ is $0$. For every interval after, the maximum and minimum of $f$ is $1$. For this particular interval, the minimum of $f$ is $0$ and the maximum is $1$. Then $$L(f,P)=1\times(1-x_t)\leq1-\frac12=\frac12$$ since $x_t\geq\frac12$. And $$U(f,P)=1\times(1-x_{t-1})\geq1-\frac12=\frac12$$ since $x_{t-1}\leq\frac12$.

Conclusion $$L(f,P)=\leq\frac12\leq U(f,P)$$

Answer 3

Way 1: pick the partitions $\{ 0,1/2-1/n,1/2+1/n,1 \}$ for $n=3,4,\dots$. Explicitly compute the lower sum to be $1/2-1/n$ and the upper sum to be $1/2+1/n$, then you're done. (These follow because the only subinterval where $f$ isn't constant is $[x_k,x_{k+1}]$.) But this requires the central theorem of Darboux integration, which is that the existence of the Darboux integral is equivalent to the lower sums and upper sums converging to each other for any particular sequence of partitions.

You might think that a valid "way 2" would be to take any arbitrary partition, refine it to include 1/2, and then conclude that the lower and upper sums automatically agree once 1/2 is included. This is actually not quite right because the lower sum will still not be exactly 1/2 if 1/2 is in the partition, since the interval to the right of 1/2 will still have exactly one point in it where $f(x)=0$, and you have to take the sup/inf on closed subintervals.

An actual valid way 2, not relying on this theorem, is as follows. Given any partition $P$, there is some $k$ such that $x_k \leq 1/2 < x_{k+1}$, the lower sum is $1-x_{k+1}$ and the upper sum is $1-x_k$ (the argument is the same as in way 1). The difference is $x_{k+1}-x_k \leq \| P \|$, so you're done.