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I'm an undergraduate student looking for help on this homework question.

I've seen this answer to a similar question, which seems to use Littlewood's First Principle mentioned here in another related question.

I tried reading a proof of the Lebesgue Density Theorem (which doesn't use characteristic functions) linked here but couldn't understand all of it. Also we haven't covered this theorem in class yet, so I would prefer a more elementary approach which doesn't use this theorem.

I have read about the "regularity" of the Lebesgue outer measure (Theorem 1.17 from page 9 of these notes) that any subset $E \subset \mathbb R^n$ can be approximated by an open set $U$ such that $m^*(E) \le m^*(U) \le m^*(E) + \epsilon$ for any $\epsilon > 0$.

I don't think Littlewood's First Principle applies here, since $E$ is not necessarily Lebesgue measurable.

Does the solution involve the fact that since a ball $B$ is Lebesgue measurable, then for any $E \subset \mathbb R^n$

$$m^*(E) = m^*(E \cap B) + m^*(E-B)$$

and that the outer measure is countably subadditive, so

$$m^*(B) \le m^*(E \cap B) + m^*(B-E)$$

and with some algebra it would give the desired result?

Or is there something else I should focus on?

edit: Actually after reading the proof of Littlewood's First Principle, I guess it applies for $E$ as well. Then the only difference from the first question linked is that I am looking for a ball rather than an interval...

spenceryue
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2 Answers2

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The Lebesgue outer measure satisfies the following property:

For every $\epsilon>0$, there exists some open set $G\supseteq E$ such that $\mu^{\ast}(E)\leq\mu^{\ast}(G)\leq\mu^{\ast}(E)+\epsilon$.

Now find some open set $G\supseteq E$ such that $\mu^{\ast}(E)\leq\mu^{\ast}(G)\leq(1/\alpha)\mu^{\ast}(E)$.

Now find a sequence of disjoint balls $\{B_{i}\}$ such that $G=\displaystyle\bigcup_{i}B_{i}$, then \begin{align*} \mu^{\ast}(E)\leq\sum_{i}\mu^{\ast}(E\cap B_{i}), \end{align*} and \begin{align*} \mu^{\ast}(G)=\sum_{i}\mu^{\ast}(B_{i}), \end{align*} then \begin{align*} \sum_{i}\mu^{\ast}(B_{i})\leq\dfrac{1}{\alpha}\sum_{i}\mu^{\ast}(E\cap B_{i}). \end{align*} So there exists some $i$ such that $\mu^{\ast}(B_{i})\leq\dfrac{1}{\alpha}\mu^{\ast}(E\cap B_{i})$.

user284331
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  • Thank you for you help. I have one more question about your answer. Could you explain why the sum $\sum_i \mu^(E\cap B_i)$ converges? I know there might be an obvious reason, but I can't see why exactly. Is it because the sum $\sum_i \mu^(B_i)$ converges and all the terms are larger in this second sum? Or is there a different way to think about it? – spenceryue Feb 08 '18 at 06:21
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    $\displaystyle\sum_{i}\mu^{\ast}(E\cap B_{i})\leq\displaystyle\sum_{i}\mu^{\ast}(B_{i})=\mu^{\ast}(G)<\infty$. – user284331 Feb 08 '18 at 06:24
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The accepted answer is flawed: why can one always write $G=\bigcup_i B_i$ for disjoint open balls $B_i$ and open set $G\subseteq\mathbb{R}^d$? In particular, if $G$ is connected but not a ball, then by definition it cannot be a union of disjoint open sets. However, one can modify the proof by noting that for any open set $G\subseteq\mathbb{R}^d$, $G=(\bigcup_i B_i)\bigcup N$ where $B_i$ are disjoint open balls and $m^*(N)=0$.

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