proving the cantor set is integrable

**zebra2147** · February 2nd 2011, 03:02 PM

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 $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $ f(x)= \begin{cases} 1, &\text{when $x\in C $}\\ 0, &\text{when $x\notin C $}\\ \end{cases} $
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$ . [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 $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$ .
We can make $\sum s=0$ when $x\notin C$ . So i think all we have to do is to show that $\sum S< \epsilon$ . However, I'm not sure how to do this.

**Drexel28** · February 2nd 2011, 05:10 PM

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 $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $ f(x)= \begin{cases} 1, &\text{when $x\in C $}\\ 0, &\text{when $x\notin C $}\\ \end{cases} $
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$ . [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 $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$ .
We can make $\sum s=0$ when $x\notin C$ . So i think all we have to do is to show that $\sum S< \epsilon$ . 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?

**zebra2147** · February 2nd 2011, 07:14 PM

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.

**JG89** · February 3rd 2011, 11:47 PM

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

**zebra2147** · February 7th 2011, 08:11 AM

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

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

**roninpro** · February 7th 2011, 05:46 PM

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

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

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

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

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

or in other words,

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

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

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

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

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