The Unapologetic Mathematician

Mathematics for the interested outsider

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.

April 7, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. […] ring generated by the is itself countable, and so we may assume that is itself a ring. But then we know that for every and for every positive we can find some ring element so that . Thus is a […]

    Pingback by The Metric Space of a Measure Ring « The Unapologetic Mathematician | August 6, 2010 | Reply

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