The Measure of a Countable Coll. of Mea. Zero Subsets of Real Numb.

The Measure of a Countable Collection of Measure Zero Subsets of Real Numbers

Recall from The Measure of Countable Subsets of Real Numbers page that if $S \subset \mathbb{R}$ is a countable subset of $\mathbb{R}$ then the measure of $S$ is equal to $0$, i.e., $m(S) = 0$.

We will now look at a stronger theorem which says that if we have an infinite countable collection of measure $0$ sets, $F_1, F_2, ..., F_n, ...$ then the union $F = \bigcup_{j=1}^{\infty} F_j$ is also of measure $0$.

Theorem 1: Let $\{ F_j \}_{j=1}^{\infty}$ be a infinitely countable collection of subsets of $\mathbb{R}$ such that $m(F_j) = 0$ for all $j \in \{ 1, 2, ... \}$. Let $F = \bigcup_{j=1}^{\infty} F_j$. Then $m(F) = 0$.

Theorem 1 is also true when we take only a finite countably collection of subsets of $\mathbb{R}$ with measure $0$.

  • Proof: Let $\epsilon > 0$ be given. For each $j \in \{1, 2, ... \}$ since $m(F_j) = 0$ there exists a countable open interval covering $\{ I_{j_k} = (a_{j_k}, b_{j_k}) : k \in J_k \}$ such that $\displaystyle{F_j \subseteq \bigcup_{k \in J} I_{j_k} = \bigcup_{k \in J} (a_{j_k}, b_{j_k})}$ for some countable indexing set $J$ and such that:
\begin{align} \quad \sum_{k \in J} l(I_{j_k}) < \frac{\epsilon}{2^j} \quad (*) \end{align}
  • Now consider the union $\displaystyle{\bigcup_{j=1}^{\infty} F_j}$. This set has an infinitely countable open interval covering which is the union of the infinitely countable open interval covers, i.e., $F \subseteq \bigcup_{j=1}^{\infty} \left ( \bigcup_{k \in J} I_{j_k} \right )$ and furthermore from $(*)$ we have that the lengths of the open intervals in this infinitely countable open interval covering satisfy:
\begin{align} \quad \sum_{j=1}^{\infty} \left ( \sum_{k \in J} l(I_{j_k}) \right ) < \sum_{j=1}^{\infty} \frac{\epsilon}{2^j} = \epsilon \sum_{j=1}^{\infty} \frac{1}{2^j} = \epsilon \end{align}
  • Therefore $m(F) = 0$.
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