The Measure of Countable Subsets of Real Numbers

# The Measure of Countable Subsets of Real Numbers

Recall from the Subsets of Real Numbers with Measure Zero that a subset $S \subset \mathbb{R}$ is said to have measure $0$ denoted $m(S) = 0$ if there exists a countable open interval covering $\{ I_k = (a_k, b_k) \}_{k \in K}$ (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 have that:

(1)\begin{align} \quad \sum_{k \in K} l(I_k) < \epsilon \end{align}

We noted that any singleton and any finite subset $S$ of $\mathbb{R}$ has measure $0$. We will now see further than any countable set also has measure $0$.

Theorem 1: Let $S \subset \mathbb{R}$ be countable. Then $m(S) = 0$. |

**Proof:**Let $\epsilon > 0$ be given. If $S$ is finite then we're done since we've already seen that a finite subset of $\mathbb{R}$ has measure $0$ so assume that $S$ is countably infinite, say:

\begin{align} \quad S = \{ x_1, x_2, ..., x_n, ... \} \end{align}

- For each $x_k \in S$ let:

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

- Then $\{ I_k \}_{k=1}^{\infty}$ is countable and clearly covers $S$. We also see that:

\begin{align} \quad l(I_k) = \frac{\epsilon}{2^{k+1}} \end{align}

- Taking the sum of $l(I_k)$ from $k = 1$ to $\infty$ gives us:

\begin{align} \quad \sum_{k=1}^{\infty} l(I_k) = \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k+1}} = \epsilon \sum_{k=1}^{\infty} \frac{1}{2^{k+1}} < \epsilon \sum_{k=1}^{\infty} \frac{1}{2^k} = \epsilon \end{align}

- (We use the fact that the series $\sum_{k=1}^{\infty} \frac{1}{2^k}$ converges to $1$). Therefore $m(S) = 0$. $\blacksquare$