# Thread: sets of measure zero

1. ## sets of measure zero

Prove that the union of two sets of measure zero has measure zero.

2. Originally Posted by friday616
Prove that the union of two sets of measure zero has measure zero.
Let $\displaystyle X_1, X_2$ be sets of measure zero. Let $\displaystyle \sum_{n=1}^{\infty}|I_n^{i}|,$ for i=1,2 be the sum of lengths of open intervals that cover for $\displaystyle X_1, X_2$, respectively. Since $\displaystyle X_1, X_2$ are measure zero, given $\displaystyle \epsilon >0$, we can have $\displaystyle \sum_{n=1}^{\infty}|I_n^{i}| < \frac{\epsilon}{2}$, for i=1,2, respectively.

Now, if $\displaystyle \epsilon >0$, then we have $\displaystyle \sum_{n=1}^{\infty}|I_n^{1}| + \sum_{n=1}^{\infty}|I_n^{2}| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ as a cover for $\displaystyle X_1 \cup X_2$. Thus, a union of two sets of measure zero is of measure zero.

3. Hello,

I don't understand why you talk about intervals... we don't know if we're in the real line, do we ?

Here is a way... The idea is to write $\displaystyle X_1\cup X_2$ as a union of disjoint sets of measures 0.

In particular, we have $\displaystyle X_1\cup X_2=(X_1\backslash X_2)\cup(X_2\backslash X_1)\cup(X_1\cap X_2)$

where $\displaystyle A\backslash B=A\cap B^c$

Since for any set A,B, $\displaystyle A\cap B\subset A$, it follows that $\displaystyle \mu(X_1\backslash X_2)\leq \mu(X_1)=0$
And since $\displaystyle \mu(X_1\backslash X_2)\geq 0$, it follows that $\displaystyle \mu(X_1\backslash X_2)=0$

The same is applied to $\displaystyle X_2\backslash X_1$ and $\displaystyle X_1\cap X_2$

And we have an axiom that says that the measure of the union of disjoint sets is the sum of their measures.

And you're done

4. Originally Posted by Moo
Hello,

I don't understand why you talk about intervals... we don't know if we're in the real line, do we ?

Here is a way... The idea is to write $\displaystyle X_1\cup X_2$ as a union of disjoint sets of measures 0.

In particular, we have $\displaystyle X_1\cup X_2=(X_1\backslash X_2)\cup(X_2\backslash X_1)\cup(X_1\cap X_2)$

where $\displaystyle A\backslash B=A\cap B^c$

Since for any set A,B, $\displaystyle A\cap B\subset A$, it follows that $\displaystyle \mu(X_1\backslash X_2)\leq \mu(X_1)=0$
And since $\displaystyle \mu(X_1\backslash X_2)\geq 0$, it follows that $\displaystyle \mu(X_1\backslash X_2)=0$

The same is applied to $\displaystyle X_2\backslash X_1$ and $\displaystyle X_1\cap X_2$

And we have an axiom that says that the measure of the union of disjoint sets is the sum of their measures.

And you're done
The notion of measure zero does not involve any Lebesgue measure theory in some situations (link). Some analysis books introduce "measure zero" before introducing Lebesgue measure theory.

The definition of measure zero I had used is

"A set of points capable of being enclosed in intervals whose total length is arbitrarily small" (link).

5. Yeah, I referred to the Lebesgue measure (more precisely to the real line) because you were talking about intervals. And I didn't know this definition, thanks

6. Wow, what a way to overprove something. Use the subaditivity anyone?
$\displaystyle \mu(A \cup B) \leq \mu(A) + \mu(B)=0$