0
$\begingroup$

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 $1$.

$\endgroup$
2
$\begingroup$

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 $1$.

$\endgroup$
1
$\begingroup$

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 $P$, $U(f) = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.