Interior points and measure zero set.

Question

I have a question about my proof. I assume $A$ has measure zero, and I want to prove that $A$ has no interior points. I assume by contradiction that $A$ does have an interior point $x$. So, there exists $\delta>0$ s.t there exists an open ball $B_x(\delta)\subseteq A$. Now since $A$ is a measure zero set, I know that for $\varepsilon = \frac{\text{Vol}(B_x(\delta))}{3}$ there exist cuboids $\{R_k\}_{k=1}^{\infty}$ s.t $A\subseteq \bigcup_{k=1}^{\infty}R_k$, and $\text{Vol}(\bigcup_{k=1}^{\infty}R_k)\leq\sum_{k=1}^{\infty}\text{Vol}(R_k)<\varepsilon$, but thats a contradiction to the fact that $B_x(\delta)\subseteq A\subseteq \bigcup_{k=1}^{\infty}R_k$ (we found a subset of a set whose volume is bigger than the volume of the set). Is this proof correct?

Answer 1

That proof is correct, but can be cleaned up slightly.

In my mind, it is easier to say that $B_\delta(x) \subseteq A$ for some $x$ and some $\delta$. Moreover, we know that $B_\delta(x)$ has positive measure. Then by monotonicity of measure, we have $0 < m(B_\delta(x)) \leq m(A)$, and so $A$ cannot be null.

Obviously your proof uses the same idea, but an appeal to open covers of small measure seems a bit excessive in this case. I'll emphasize again that your proof absolutely works, though!

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I hope this helps ^_^