Subsets of Real Numbers with Measure Zero

Subsets of the Real Numbers with Measure Zero

We will soon look at a very important theorem which will tell us whether a function $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ or not, but first, we will need to take a brief look at some fundamentals of measure theory on the set of real numbers $\mathbb{R}$.

We begin by defining what it is meant for a subset of $\mathbb{R}$ to have measure $0$.

 Definition: Let $S \subseteq \mathbb{R}$. If for all $\epsilon > 0$ there exists a countable open interval covering of subintervals $\{ I_k = (a_k, b_k) \}_{k \in K}$ (where $K$ is some countable indexing set) that covers $S$, i.e., $S \subseteq \bigcup_{k \in K} I_k$ and for $l(I_k) = b_k - a_k$ we also have that $\sum_{k \in K} l(I_k) < \epsilon$ then $S$ is said to have Measure Zero written $m(S) = 0$.

Let's now investigate some examples of subsets of $\mathbb{R}$ that have measure $0$.

First consider perhaps one of the simplest types of subsets of $\mathbb{R}$ - singleton sets. Let $x \in \mathbb{R}$. Then the set that contains only $x$ is $S = \{ x \}$. To show that $m(S) = 0$, let $\epsilon > 0$ and consider the open interval:

(1)
\begin{align} \quad I = \left (x - \frac{\epsilon}{2}, x + \frac{\epsilon}{3} \right ) = \left \{ y \in \mathbb{R} : x - \frac{\epsilon}{3} < y < x + \frac{\epsilon}{3} \right \} \end{align}

Then $\{ I \}$ is a countable open interval covering of $S = \{ x \}$ and $l(I) = \frac{2\epsilon}{3} < \epsilon$ and so $m(S) = 0$.

From this, it is not hard to see that any finite subset containing $n$ real numbers, say $S = \{x_1, ..., x_n\}$, is of measure $0$. For each $x_k \in S$, $k \in \{ 1, ..., n \}$ let:

(2)
\begin{align} \quad I_k = \left ( x_k - \frac{\epsilon}{2(n + 1)}, x_k + \frac{\epsilon}{2(n + 1)} \right ) \end{align}

The clearly $\{ I_k : k \in \{ 1, ..., n \} \}$ covers $S$, i.e., $S \subset \bigcup_{k=1}^{n} I_k$. Furthermore we see that for each $k$ that:

(3)
\begin{align} \quad l(I_k) = \frac{\epsilon}{n+1} \end{align}

Hence we have that:

(4)
\begin{align} \quad \sum_{k=1}^{n} l(I_k) = \sum_{k=1}^{n} \frac{\epsilon}{n+1} = \frac{n \epsilon}{n+1} < \epsilon \end{align}

Therefore $m(S) = 0$.