# Mathematics Prelims

## July 15, 2008

### Null Sets

Filed under: Analysis,Measure Theory — cjohnson @ 3:56 pm

We say that if $I$ is a bounded interval with endpoints $a$ and $b$ (where $a \leq b$), then the length of $I$ is $b - a$; this is denoted $l(I) = b - a$.  In this sense, the length of an interval doesn’t depend on whether or not the interval is open or closed; $l([0, 1)) = l((0, 1)) = l ([0, 1]) = l((0, 1]) = 1$.  This idea of the length of a (bounded) interval is very intuitive, and the ideas from measure theory (specifically the Lebesgue measure on $\mathbb{R}$) will tell us how to measure the “length” of more general sets than just intervals.

Recall that a covering of a set $M$ is a collection of sets $U_n$ of sets such that $M \subseteq \bigcup_{n=1}^\infty U_n$.  If a subset of $\mathbb{R}$ can be covered by intervals whose total length can be made arbitrarily small, then we say that $M$ is a null set.  That is, $M$ is a null set if for any given $\epsilon$ greater than 0, there exists a sequence of intervals $(I_n)$ that covers $M$ (so $M \subseteq \bigcup_{n=1}^\infty I_n$) and $\sum_{n=1}^\infty l(I_n) < \epsilon$.

Note that all singleton sets, $\{x\}$ are null sets, since they can be covered by $(x - \frac{\epsilon}{2}, x + \frac{\epsilon}{2})$.  This result can easily be generalized to any finite, or even countably infinite, set of elements.  This result extends to countable collections of arbitrary null sets (not just singletons) as follows: If $(M_n)_{n \in \mathbb{N}}$ is a sequence of null sets, then their union, $M = \bigcup_{n=1}^\infty M_n$, is null too.

Proof: Suppose $(M_n)_{n \in \mathbb{N}}$ is a sequence of null sets.  This means for each $M_n$ and each $\epsilon > 0$, there exists a sequence $(I_k^{(n)})_{k \in \mathbb{N}}$ such that $M_n \subseteq \bigcup_{k=1}^\infty I_k^{(n)}$ and $\sum_{k=1}^\infty l(I_k^{(n)}) < \epsilon$.  So, in particular, we can choose $(I_k^{(n)})$ such that $\sum_{k=1}^\infty l(I_k^{(n)}) < \frac{\epsilon}{2^n}$.

Since a countable union of countable sets is countable, we rearrange the elements of our $(I_k^{(n)})$ sequences so that

$\bigcup_{n=1}^\infty M_n \subseteq \bigcup_{n=1}^\infty \bigcup_{k=1}^\infty I_k^{(n)} = \bigcup_{n=1}^\infty L_n$, then calculate the total length of our $(L_n)$ sequence.

$\displaystyle\sum_{n=1}^\infty l(L_n) = \sum_{n=1}^\infty \sum_{k=1}^\infty l(I_k^{(n)}) < \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon$

And so a union of countably many null sets is null.