How to prove an interval $[0, 1]$ is not a null set?!

Question

I am studying Lebesgue's criterion for Riemann integrability. By the way, an example of this criterion says that the Dirichlet function is not integrable because " $[0, 1]$ is not a null set ".

How can I show that $[0, 1]$ is not a null set?

I tried to find explanations about this but all of them were related to measure theory, which I don't study. (I am just reading the introduction to real analysis by Bartle) Are there any proofs without measure theory?

Answer 1

I suppose your definition of a "null set" is a set $A\subseteq\mathbb R$ such that, given any $\varepsilon\gt0,$ we can find a sequence $\{I_n\}$ of open intervals such that $A\subseteq\bigcup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty|I_n|\lt\varepsilon$ where $|I_n|$ is the length of $I_n.$

Assume for a contradiction that $[0,1]$ is a null set. Then there is a sequence $\{I_n\}$ of open intervals such that $[0,1]\subseteq\bigcup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty|I_n|\lt1.$ Since $[0,1]$ is compact, there is some finite $k$ such that $[0,1]\subseteq\bigcup_{n=1}^k I_n.$

Let $f_n$ be the characteristic function of $I_n, $ i.e., $f_n(x)=1$ if $x\in I_n,$ and $f_n(x)=0$ if $x\notin I_n.$ Then $\sum_{n=1}^{k}f_n(x)\ge1$ for all $x\in[0,1],$ so $$1=\int_0^11dx\le\int_0^1\sum_{n=1}^kf_n(x)dx=\sum_{n=1}^k\int_0^1f_n(x)dx\le\sum_{n=1}^k|I_n|\lt1,$$ which is absurd.

Answer 2

Given a subset $X$ of $\mathbb{R}^n$, we can say that $X$ isn't a null set if $int(X) \neq \emptyset$ (it's not a definition). So, since $int([0,1]) = (0,1) \neq \emptyset$, we have that $[0,1]\subset \mathbb{R}$ isn't a null set.