Thm Prove that a subset of a set of measure zero has measure zero.

I attempted the proof, corrections appreciated.


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.

  • 1
    $\begingroup$ I have no idea what you are trying to do with your proof. The entire thing is extremely puzzling. $\endgroup$ Feb 14 '16 at 1:01
  • 1
    $\begingroup$ First of all your measure should be complete, meaning that each subset of a zero measure set is measurable. I assume you are working with Lebesgue measure in $\mathbb R$ so this is OK. You can use that every subset has an outer measure less than or equal to the measure of the set.. $\endgroup$
    – Svetoslav
    Feb 14 '16 at 1:08

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.