# Cantor Set has Zero Lebesgue Measure

## Theorem

Let $\CC$ be the Cantor set.

Let $\lambda$ denote the Lebesgue measure on the Borel $\sigma$-algebra $\map \BB \R$ on $\R$.

Then $\CC$ is $\map \BB \R$-measurable, and $\map \lambda \CC = 0$.

That is, $\CC$ is a $\lambda$-null set.

## Proof

Consider the definition of $\CC$ 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 $C_n := \ds \bigcup S_n$ is measurable as well.

Then, as we have:

$\CC = \ds \bigcap_{n \mathop \in \N} C_n$

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

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

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

$\map \lambda \CC = \ds \lim_{n \mathop \to \infty} \map \lambda {C_n}$

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

$\map \lambda {C_n} = \paren {\dfrac 2 3}^n$

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

$\ds \lim_{n \mathop \to \infty} \paren {\frac 2 3}^n = 0$

and the result follows.

$\blacksquare$