# The Unapologetic Mathematician

## Borel Sets and Lebesgue Measure

Let’s consider some of the easy properties of the Borel sets and Lebesgue measure we introduced yesterday.

First off, every countable set of real numbers is a Borel set of measure zero. In particular, every single point $\{a\}$ is a Borel set. Indeed, $\{a\}$ can be written as the countable intersection

$\displaystyle\{a\}=\bigcap\limits_{n=1}^\infty\left[a,a+\frac{1}{n}\right)$

so it’s a Borel set. Further, monotonicity tells us that

$\displaystyle\mu\left(\{a\}\right)=\lim\limits_{n\to\infty}\mu\left(\left[a,a+\frac{1}{n}\right)\right)=\lim\limits_{n\to\infty}\frac{1}{n}=0$

and so the singleton $\{a\}$ has measure zero. But $\mu$ is countably additive, so given any countable collection $A\subseteq\mathbb{R}$ the measure $\mu(A)$ is the sum of the measures of the individual points, each of which is zero.

Next, as I said when I introduced semiclosed intervals, we could have started with open intervals, but the details would have been messier. Now we can see that the $\sigma$-ring $\mathcal{S}$ generated by the collection $\mathcal{P}$ of semiclosed intervals is the same as that generated by the collection $\mathcal{U}$ of all open sets.

We can see, in particular, that each open interval $\left(a,b\right)$ is a Borel set. Indeed, the point $\{a\}$ is a Borel set, as is the semiclosed interval $\left[a,b\right)$, and we have the relation $\left(a,b\right)=\left[a,b\right)\setminus\{a\}$. Every other open set in $\mathbb{R}$ is a countable union of open intervals, and so they’re all Borel sets as well. Conversely, we could write

$\displaystyle\{a\}=\bigcap\limits_{n=1}^\infty\left(a-\frac{1}{n},a+\frac{1}{n}\right)$

and find the singleton $\{a\}$ in the $\sigma$-ring generated by $\mathcal{U}$. Then we can write $\left[a,b\right)=\left(a,b\right)\cup\{a\}$ and find every semiclosed interval in this $\sigma$-ring as well. And thus $\mathcal{S}=\mathcal{S}(\mathcal{P})\subseteq\mathcal{S}(\mathcal{U})$

We can also tie our current measure back to the concept of outer Lebesgue measure we introduced before. Back then, we defined the “volume” of a collection of open intervals to be the sum of the “volumes” of the intervals themselves. We defined the outer measure of a set to be the infimum of the volumes of finite open covers. And, indeed, this is exactly the outer measure $\mu^*$ corresponding to Lebesgue measure $\mu$.

Remember that the outer measure $\mu^*(E)$ is defined for a set $E\subseteq\mathbb{R}$ by

$\displaystyle\mu^*(E)=\inf\left\{\mu(F)\vert F\in\mathcal{S},E\subseteq F\right\}$

Since $\mathcal{U}\subseteq\mathcal{S}$, we have the inequality

$\displaystyle\mu^*(E)\leq\inf\left\{\mu(U)\vert U\in\mathcal{S},E\subseteq F\right\}$

On the other hand, if $\epsilon$ is any positive number, then by the definition of $\mu^*$ we can find a sequence $\left\{\left[a_n,b_n\right)\right\}$ of semiclosed intervals so that

$\displaystyle E\subseteq\bigcup\limits_{n=1}^\infty\left[a_n,b_n\right)$

and

$\displaystyle\sum\limits_{n=1}^\infty(b_n-a_n)\leq\mu^*(E)+\frac{\epsilon}{2}$

We can thus widen each of these semiclosed intervals just a bit to find

$\displaystyle E\subseteq\bigcup\limits_{n=1}^\infty\left(a_n-\frac{\epsilon}{2^{n+1}},b_n\right)=U\in\mathcal{U}$

and

$\displaystyle\mu(U)\leq\sum\limits_{i=1}^\infty(b_n-a_n)+\frac{\epsilon}{2}\leq\mu^*(E)+\epsilon$

Since $\epsilon$ was arbitrary, we find that $\mu(U)\leq\mu^*(E)$. And, thus, that

$\displaystyle\mu^*(E)=\inf\left\{\mu(U)\vert U\in\mathcal{S},E\subseteq F\right\}$

In effect, we’ve replaced the messily-defined “volume” of an open cover by the more precise Lebesgue measure $\mu$, but the result is the same. The “outer Lebesgue measure” from our investigations of multiple integrals is the same as the outer measure induced by our new Lebesgue measure.

April 20, 2010 Posted by | Analysis, Measure Theory | 5 Comments