# Thread: proving the cantor set is integrable

1. ## proving the cantor set is integrable

I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.

Let $\displaystyle C$ be the Cantor set. Let $\displaystyle f:[0,1]\rightarrow \mathbb{R}$ be determined by $\displaystyle $f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $\displaystyle f$ is Riemann integrable on $\displaystyle [0,1]$ and $\displaystyle \int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]

I want to approach it like this but I'm not sure if it is correct:
We want to show that $\displaystyle f$ is integrable we need to show that $\displaystyle \{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\displaystyle \epsilon >0$.
We can make $\displaystyle \sum s=0$ when $\displaystyle x\notin C$. So i think all we have to do is to show that $\displaystyle \sum S< \epsilon$. However, I'm not sure how to do this.

2. Originally Posted by zebra2147
I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.

Let $\displaystyle C$ be the Cantor set. Let $\displaystyle f:[0,1]\rightarrow \mathbb{R}$ be determined by $\displaystyle $f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $\displaystyle f$ is Riemann integrable on $\displaystyle [0,1]$ and $\displaystyle \int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]

I want to approach it like this but I'm not sure if it is correct:
We want to show that $\displaystyle f$ is integrable we need to show that $\displaystyle \{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\displaystyle \epsilon >0$.
We can make $\displaystyle \sum s=0$ when $\displaystyle x\notin C$. So i think all we have to do is to show that $\displaystyle \sum S< \epsilon$. However, I'm not sure how to do this.
So start with the definition of the Cantor set. There are certain "removed" intervals, what are they?

3. The middle third of each line segment is removed. For example, for [0,1] we would have [0,1/3]U[2/3,1]. So, (1/3,2/3) would be removed.

4. Try proving that $\displaystyle [0,1] - C$ has measure 1, implying that C has measure 0.

5. Ok, here is what I want to say but I'm not sure how to write it correctly of if its even in the right direction...

If we let $\displaystyle s$ be a lower step function of $\displaystyle f$, then we see that $\displaystyle \sum s=0$. Then if we choose some upper step function, $\displaystyle S$, such that $\displaystyle \sum S=\sum A_{i}(x_{i}-x_{i-1})$ where $\displaystyle A_{i}=\text{height of the rectangle}$ then for any interval contained in $\displaystyle [0,1]$, we can find an element, $\displaystyle x_i$ that is contained in the Cantor set and is infinitely close to another element $\displaystyle x_{i-1}$ that is not contained in the Cantor set. Thus, we have $\displaystyle x_{i}-x_{i-1}\approx 0$. Thus, $\displaystyle \sum A_{i}(x_{i}-x_{i-1})\approx 0=\sum S$.
Thus, $\displaystyle \sum s+\sum S\approx 0$ which is less then any $\displaystyle \epsilon >0$. Thus, $\displaystyle \int_{0}^{1}f=0$.

I know that is not exactly the direction that you were leading me to but I think it is similar.

6. I think that it would be better to show that the upper sum and lower sum are equal to each other. Recall that

$\displaystyle \displaystyle U(f)=\inf_{P} U(f,P)$ and $\displaystyle \displaystyle L(f)=\sup_{P} L(f,P)$

where $\displaystyle P$ refers to some partition of $\displaystyle [0,1]$. We say that $\displaystyle f$ is Riemann-integrable if $\displaystyle U(f)=L(f)$.

Now I propose that we consider a sequence of partitions based on the cuts performed in the construction of the Cantor set. Let $\displaystyle P_1=\{0,1/3,2/3,1\}$ (corresponding to the removal of the middle third of the interval). Then, $\displaystyle U(f,P_1)=L(f,P_1)=2/3$. Let $\displaystyle P_2=\{0,1/9,2/9,1/3,2/3,7/9,8/9,1\}$ (again, corresponding to the middle thirds of the remaining intervals). Then, $\displaystyle U(f,P_2)=L(f,P_2)=4/9$. Perform this same procedure to get $\displaystyle P_3, P_4, \ldots, P_n,\ldots$. Now taking $\displaystyle n\to \infty$, we have $\displaystyle U(f,P_n)\to 0$ and $\displaystyle L(f,P_n)\to 0$. This shows that

$\displaystyle U(f)\leq U(f,P_n)\to 0$ and $\displaystyle L(f)\geq L(f,P_n)\to 0$

or in other words,

$\displaystyle U(f)\leq 0\leq L(f)$

But at the same time, from the definition of the upper and lower sums,

$\displaystyle U(f)\geq L(f)$

Therefore $\displaystyle U(f)=L(f)=0$, so this function $\displaystyle f$ is Riemann-integrable with integral value 0.