Cantor Set has Zero Lebesgue Measure

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Theorem

Let $\mathcal C$ be the Cantor set.

Let $\lambda$ be the Lebesgue measure on the Borel $\sigma$-algebra $\mathcal B \left({\R}\right)$ on $\R$.


Then $\mathcal C$ is $\mathcal B \left({\R}\right)$-measurable, and $\lambda \left({\mathcal C}\right) = 0$.

That is, $\mathcal C$ is a $\lambda$-null set.


Proof

Consider the definition of $\mathcal C$ as a limit of a decreasing sequence.

In the notation as introduced there, we see that each $S_n$ is a collection of disjoint closed intervals.

From Closed Set Measurable in Borel Sigma-Algebra, these are measurable sets.


Furthermore, each $S_n$ is finite.

Hence by Sigma-Algebra Closed under Union, it follows that $C_n := \displaystyle \bigcup S_n$ is measurable as well.


Then, as we have:

$\mathcal C = \displaystyle \bigcap_{n \mathop \in \N} C_n$

it follows from Sigma-Algebra Closed under Countable Intersection that $\mathcal C$ is measurable.


The $C_n$ also form a decreasing sequence of sets with limit $\mathcal C$.

Thus, from Characterization of Measures: $(3')$, it follows that:

$\lambda \left({\mathcal C}\right) = \displaystyle \lim_{n \to \infty} \lambda \left({C_n}\right)$


It is not too hard to show that, for all $n \in \N$:

$\lambda \left({C_n}\right) = \left({\dfrac 2 3}\right)^n$


Now we have by Sequence of Powers of Number less than One that:

$\displaystyle \lim_{n \mathop \to \infty} \left({\frac 2 3}\right)^n = 0$

and the result follows.

$\blacksquare$


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