Thm Prove that a subset of a set of measure zero has measure zero.
I attempted the proof, corrections appreciated.
Pf
Let $A=\{x_1,….,x_N\}$ be a finite set, and let $\epsilon > 0$ be given.
Then $\cup^{\infty}_{i=1}U_N = \{ (x_1 – \frac{\epsilon}{4N} , x_1 + \frac{\epsilon}{4N}),…….,(x_N – \frac{\epsilon}{4N} , x_N + \frac{\epsilon}{4N})\}$
is an open cover of A.
Also, $U_{n_k} = 1,….j$ is a finite subcover of $\cup^{\infty}_{i=1}U_N$
Let $A' = \{x_1,….x_j\}$ be a subset of A.
Since there are N intervals each of measure $\frac{\epsilon}{2N}$ so that our open cover has measure $N \frac{\epsilon}{2N} = \frac{\epsilon}{2} < \epsilon$
Thus are subcover also has measure zero.
Best Answer
You have only shown this for finite sets. In general, use that $A \subset B$ implies $\mu (A) \leq \mu (B)$. This follows from the additivity axiom.