# $[0,1]$ is not of measure zero

$[0,1]$ is not of measure zero ?!

Proof sketch :

I assumed that there is an open covering of length $<1/2$

Meaning $[0,1] \subset \bigcup \limits_{i=1}^{\infty} I_i$ such that $I_i$ is open interval and $\sum \limits_{i=1}^{\infty} len(I_i) < \frac{1}{2}$

By Heine-Borel theorem and because $[0,1]$ is compact then there is finite sub covering meaning there is $n\in \mathbb{N}$ such that $I_1, \cdots I_n$ are cover for $[0,1]$ and are open intervals and $len(I_1) + \cdots + len(I_n) < \frac{1}{2}$

I said by induction on $n$, if $n=1$ then $0 \not \in I_1$ then $[0,1] \not\subset I_1$ else $0 \in I_1$ and because $len(I_1) < 1/2$ and $I_1$ is open interval then $1 \not \in I_1$ so $[0,1] \not \subset I_1$

but i couldn't proceed in induction step proof, please help me, any hints or a sketch for how to proceed.

• Come on. When you have finitely many intervals, you can order them and add their lengths, $I_1$ is one that covers $0$. If $\sup I_1>1$ you are done $len(I_1)>1$. If not, define $I_2$ to be one interval that covers $\sup I_1$. If $\sup I_2>1$ then $len(I_1)+len(I_2)>1$. If not, define $I_3$ in the analogous way and continue. – user545963 Apr 4 '18 at 17:39

Here is a standard solution:

Let $\cup_{n} I_n$ of open intervals that cover $[0,1]$. Heineken Borel implies that there is a finite sub collection of $(a_i,b_i)$.

Let $I_1=(a_1,b_1)$ contain $0$. Then $b_1 \in [0,1]$, so do the same thing to obtain $I_2$ and continue in this fashion. This process must terminate at some $k$, in which case $b \in (a_k,b_k)$. In this case

$\sum l(I_n) \geq \sum_{i=1}^{k} l(b_i-a_i) =b_k-(a_k-b_{k-1})-\dots a_1 \geq b_k-a_1 >1-0=1.$

the penultimate inequality comes from the fact that $a_i<b_{i-1}$.

• can you please give a ref for this proof – Ahmad Apr 4 '18 at 18:10
• Heineken Borel. That sounds like a fun theorem. ;-) – tomasz Apr 4 '18 at 20:40

Here's a trick solution. Consider the set $$X_N=\{1/N,2/N,3/N,\ldots,N/N\}$$ for $N\in\Bbb N$. For a set $A\subseteq[0,1]$ define $$\mu_N(A)=|A\cap X_N|.$$ If $A=I$ is an interval $$\text{len}(I)=\lim_{N\to\infty}\frac{\mu_N(A)}N.$$

Here, the covering property gives $$\sum_{i=1}^n\mu_N(I_i)\ge\mu_N([0,1]).$$ Dividing by $N$ and letting $N\to\infty$ gives $$\sum_{i=1}^n\text{len}(I_i)\ge1.$$