Triviality of the Empty Set

2024-04-29 · 1 min read

Prove that if there is a set $A$ in the collection $\mathcal{A}$ for which $m(A) < \infty$, then $m(\emptyset)=0$.

Proof:

Suppose on the contrary that $m(\emptyset) = \epsilon$ for $\epsilon > 0$. Let $A \in \mathcal{A}$ be any set. Note that

$$A = A \cup \left (\bigcup_{i=1}^n\emptyset \right)$$

Since $\emptyset \cap A = \emptyset$ and $\emptyset \cap \emptyset = \emptyset$, we have

$$ \begin{aligned} m(A) &= m(A) + \sum_{i=1}^{\infty}m(\emptyset) \\ &= m(A) + \sum_{i=1}^{\infty}\epsilon \\ &= \infty \end{aligned} $$

contradicting the original hypothesis. Hence, we if there exists an $A \in \mathcal{A}$ with finite measure, we must that $m(\emptyset) = 0$.