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I have two problems that I would like some help with.

  1. Show that every countable subset of $\mathbb{R}$ has Lebesgue measure zero.

  2. For two arbitrary sets $A$ and $B$ show that $$\lvert m^*(A)-m^*(B)\rvert \leq m^*(A \triangle B)$$ where $\triangle$ is the symmetric difference operator.

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    $\begingroup$ You should post your two problems as two separate questions. $\endgroup$ – Chris Taylor Oct 18 '12 at 8:54
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For the first question:

Let $A = \bigcup_{n \in \mathbb N} \{a_n \} \subset \mathbb R$. The Lebesgue measure of a point is zero: by construction of the Lebesgue measure, $\lambda [a,b] = b - a$. For the one element set $\{ a \} = [a,a]$ we have $\lambda [a,a] = a-a =0$.

Since the Lebesgue measure is additive, we have $$ \lambda A = \sum \lambda \{a_n\} = \sum 0 = 0$$

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  • $\begingroup$ @cccjay Glad to help : ) In question 2 you write "arbitrary" set but then you apply $m$ to them (which I assume to be the Lebesgue measure). So shouldn't $A,B$ be measurable sets? $\endgroup$ – Rudy the Reindeer Oct 18 '12 at 10:34
  • $\begingroup$ I'm sorry Matt N,that should be m*. $\endgroup$ – ccc Oct 18 '12 at 12:55
  • $\begingroup$ @cccjay I see. Is $m^\ast$ an outer measure on $\mathbb R$? $\endgroup$ – Rudy the Reindeer Oct 18 '12 at 12:59
  • $\begingroup$ m* is defined for subsets of R,isn't it? $\endgroup$ – ccc Oct 18 '12 at 13:02
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For Question 1, first write the set into $ \left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },\cdots \right\} $, we can do so since it is countable. Then consider that for every $ \epsilon > 0 $, there is a interval $\left( { a }_{ i }-\frac { \epsilon }{ { 2 }^{ i } } , { a }_{ i }+\frac { \epsilon }{ { 2 }^{ i } } \right) $ containing $ {a}_{i} $ point. Sum them up and you will get the result.

For Question 2, if we assume A and B are measurable, by Cratheodory's Theorem, $ { m }^{ \ast }(A) = { m }^{ \ast }(A\cap B) - { m }^{ \ast }(A\setminus B) $, right hand side is equal to ${ m }^{ \ast }(A\cup B)-{ m }^{ \ast }(A\cap B)$. Meanwhile, without loss of generality, we assume $ { m }^{ \ast }(A) > { m }^{ \ast }(B) $. Then the left hand side is reduced to $ { m }^{ \ast }(A)-{ m }^{ \ast }(B)$. This is automatically true because ${ m }^{ \ast }(A)\le{ m }^{ \ast }(A\cup B)$ and ${ m }^{ \ast }(A\cap B) \le { m }^{ \ast }(B) $.

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1-The Lebesgue measure $\nu$ satisfies by definition $\nu([a,b])=b-a$. In particular $\nu(\{a\})=\nu([a,a])=a-a=0$. Thus $\nu$ is zero on all unitary set. And now using the aditivity of $\nu$ we have for all countable set $A=\{a_n\}_{n=1}^\infty=\bigcup_{n=1}^\infty\{a_n\}$: $$\nu(A)=\nu\left(\bigcup_{n=1}^\infty\{a_n\}\right)=\sum_{n=1}^\infty\nu(\{a_n\})=\sum_{n=1}^\infty 0=0$$

2-$A \cup B=A\triangle B+A\cap B$ where $+$ denotes the disjoint union of sets. Then $m^*(A\triangle B)=m^*(A\cup B)-m^*(A\cap B)\geq \begin{cases} m^*(A)-m^*(B)\\ m^*(B)-m^*(A)\end{cases} $. Hence $m^*(A\triangle B)\geq |m^*(A)-m^*(B)|$

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