Riemann integration of a discontinuous function

Question

I'm asked to show that $\int_{0}^{1}f=1$ where $f$ is defined with piece-wise: $$ f=\begin{cases} 0 & x=0 \\ 1 & x\ne0.\\ \end{cases} $$ To show this I am using the upper/lower sums definition of a Riemann integral. This is what I have so far.

Let $H$ be the set of all refinements of the partition $P \in [0,1]$ of $f$. Take the upper integral of $f \in [0,1]$ to be $U(f) = \inf\{U(P,f) \colon P \in H\}$. Then,

\begin{align*} U(f) &= M_{1}(x_{1}-x_{0})+M_{2}(x_{2}-x_{1})+ \dots +M_{n}(x_{n}-x_{n-1})\\ &= 0+M_{2}(x_{2}-x_{1})+ \dots +M_{n}(x_{n}-x_{n-1})\\ &=0+1((x_{2}-x_{1})+ \dots +(x_{n}-x_{n-1})). \end{align*}

This is where I am confused, I know $M_{i}$, the supremum of each partition is $1$ once $x>0$. In my proof I assume there is one case where an area is $0$ since $0 \in I$ but if this is the case I'm confused as to how $1(x_{i}-x_{i-1})$ sums up to $1$.

Sorry about this choppy question I can't quite seem to word it the way I want to. After I finish this part of the proof I intend to show it for the lower sums, then since both upper/lower sums are equal to $1$ the integral from $0$ to $1$ is $1$.

Answer 1

The supremum in the first subinterval should be $1$ as well, not $0$. This is because the subinterval goes from $x_0=0$ to $x_1>0$, so the function takes both the values $0$ (at $x=0$) and $1$ (elsewhere) in this subinterval.

...and even though you didn't ask yet... for the lower sum you will have the first infimum being $0$ so your sum will be $1-x_1$. But when you refine the partitions, $x_1$ will go to $0$ so the sum goes to $1$.

Answer 2

Your calculation is not quite correct. For one thing, you should calculate $U(f, P)$ with respect to a partition $P$, then try to deduce $U(f)$. I'm not sure why you are referring to refinements of $P$ before you fix a partition, this is meaningless and does not give the correct definition of $U(f)$. Instead, $H$ should be the set of partitions of $[0, 1]$.

If $P = \{x_0, x_1, \dots, x_n\}$ is a partition of $[0, 1]$ with $0 = x_0 < x_1 < \dots < x_n = 1$. Then

$$U(f, P) = M_1(x_1 - x_0) + \dots + M_n(x_n - x_{n-1}).$$

where $M_i = \sup_{[x_{i-1}, x_i]}f(x)$. On each subinterval, including the first one, the supremum is equal to one, i.e. $M_1 = \dots = M_n = 1$.

It seems that, for you, the confusion lies in the fact that $M_1 = 1$. To see that this is the case, note that $f(x) \leq 1$ for all $x \in [x_0, x_1]$, so $M_1 = \sup_{[x_0, x_1]}f(x) \leq 1$, but as $x_1 > x_0 = 0$, $f(x_1) = 1$ so $M_1 = \sup_{[x_0, x_1]}f(x) \geq f(x_1) = 1$, and therefore $M_1 = 1$.

With this in mind, the upper sum becomes

$$U(f, P) = (x_1 - x_0) + \dots + (x_n - x_{n-1}) = x_n - x_0 = 1 - 0 = 1.$$

As $U(f, P) = 1$ for every partition $P$, $U(f) = 1$.