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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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