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Math Help - proving the cantor set is integrable

  1. #1
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    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 C be the Cantor set. Let f:[0,1]\rightarrow \mathbb{R} be determined by \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{when $x\in C $}\\<br />
0, &\text{when $x\notin C $}\\<br />
\end{cases}<br />
\]
    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    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 \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{when $x\in C $}\\<br />
0, &\text{when $x\notin C $}\\<br />
\end{cases}<br />
\]
    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?
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  3. #3
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    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.
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  4. #4
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    Try proving that  [0,1] - C has measure 1, implying that C has measure 0.
    Last edited by JG89; February 4th 2011 at 03:48 PM.
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  5. #5
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    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.
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  6. #6
    Senior Member roninpro's Avatar
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    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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