# The Unapologetic Mathematician

## Approximating Sets of Finite Measure

So, we’ve got a $\sigma$-finite measure $\mu$ on a ring $\mathcal{R}$, and we extend it to a measure on the $\sigma$-ring $\mathcal{S}(\mathcal{R})$. But often it’s a lot more convenient to work with $\mathcal{R}$ itself than the whole of $\mathcal{S}(\mathcal{R}$. So, to what extent can we do this efficiently?

As it turns out, if $E\in\mathcal{S}(\mathcal{R}$ has finite measure and $\epsilon>0$, then we can find a set $E_0\in\mathcal{R}$ so that $\mu(E\Delta E_0)\leq\epsilon$.

Any set $E\in\mathcal{S}(\mathcal{R})$ can be covered by a sequence of sets in $\mathcal{R}$, and we know that

$\displaystyle\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i, \{E_i\}\subseteq\mathcal{R}\right\}$

That is, we can find such a cover satisfying

$\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)\leq\mu(E)+\frac{\epsilon}{2}$

But since $\mu$ is continuous, we see that

$\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)=\mu\left(\lim\limits_{n\to\infty}\bigcup\limits_{i=1}^nE_i\right)=\lim\limits_{n\to\infty}\mu\left(\bigcup\limits_{i=1}^nE_i\right)$

The sequence of numbers increases until it’s within $\frac{\epsilon}{2}$ of its limit. That is, there is some $n$ so that if we define $E_0$ to be the union of the first $n$ sets in the sequence, we have

$\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)\leq\mu(E_0)+\frac{\epsilon}{2}$

But now we can find

\displaystyle\begin{aligned}\mu(E\setminus E_0)\leq\mu\left(\left(\bigcup\limits_{i=1}^\infty E_i\right)\setminus E_0\right)&=\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)-\mu(E_0)\leq\frac{\epsilon}{2}\\\mu(E_0\setminus E)\leq\mu\left(\left(\bigcup\limits_{i=1}^\infty E_i\right)\setminus E\right)&=\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)-\mu(E)\leq\frac{\epsilon}{2}\end{aligned}

And thus $\mu(E\Delta E_0)\leq\epsilon$.