I am trying to solve the following proof: Let $||$ be the Lebesgue outer measure. Show that if $|B|=0$ then $|A\cup B|=|A|$. If I consider $B=\{x_1,x_2,….\}$ this is easy enough. Consider an arbitrary open covering of $A \subset \cup_{i=1}^{\infty} I_i$. Let us consider the open covering for $B\subset\cup_{n=1}^{\infty}(x_n-\frac{\epsilon}{2^n},x_n+\frac{\epsilon}{2^n})$. Then
Consider the open covering of $A\cup B\subset \cup_{n=1}^{\infty}\bigg\{ I_n\cup (x_n-\frac{\epsilon}{2^n},x_n+\frac{\epsilon}{2^n})\bigg\}$. Then we get
$$|A\cup B|\le \sum_{n=1}^{\infty}l(I_n)+\epsilon\sum_{n=0}^{\infty}\frac{1}{2^n}= \sum_{n=1}^{\infty}l(I_n)+2\epsilon.$$
Since the covering for A was arbitrary:
$$|A\cup B|\le |A|+2\epsilon.$$
Letting $\epsilon \rightarrow 0$
$$|A\cup B|\le |A|.$$ The reverse inequality holds by monotonicity. I can't think of another set $B$ that would have measure zero that is not a countable set of singletons. If there is can anyone provide an example?
Best Answer
The middle thirds set of Cantor, i.e. the set of all numbers in the interval $[0,1]$ whose ternary expansion does not contain the digit $1$, is an uncountable set of Lebesgue measure zero.