# Measure theory - Lebesgue measure

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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You should post your two problems as two separate questions. –  Chris Taylor Oct 18 '12 at 8:54

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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Thanks Matt N..:) –  ccc Oct 18 '12 at 8:58
@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? –  Matt N. Oct 18 '12 at 10:34
I'm sorry Matt N,that should be m*. –  ccc Oct 18 '12 at 12:55
@cccjay I see. Is $m^\ast$ an outer measure on $\mathbb R$? –  Matt N. Oct 18 '12 at 12:59
@MattN.: So you suggest we should remove the tag from all resolved questions? –  tomasz Apr 25 at 15:45

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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