The Unapologetic Mathematician

Mathematics for the interested outsider

Measurable Covers

Again, let \mathcal{R} be a ring, let \mathcal{S}(\mathcal{R}) be the smallest \sigma-ring containing \mathcal{R}, and let \mathcal{H}(\mathcal{R}) be the smallest hereditary \sigma-ring containing \mathcal{R}. Given a measure \mu on \mathcal{R}, it induces an outer measure \mu^* on \mathcal{H}(\mathcal{R}), which restricts to a measure \bar{\mu} on \mathcal{S}(\mathcal{R}).

The sets in E\in\mathcal{H}(\mathcal{R}) are easily described — they’re everything that can be countably covered by sets in \mathcal{R} — but we don’t have a measure on this collection. Instead, we need good approximations of these sets by sets in \mathcal{S}(\mathcal{R}), where we do have a measure. And so we say that a set F\in\mathcal{S}(\mathcal{R}) is a “measurable cover” of E if E\subseteq F, and if for every G\in\mathcal{S}(\mathcal{R}) with G\subseteq F\setminus E we have \bar{\mu}(G)=0. That is, F may be larger than E, but only negligibly.

If E\in\mathcal{H}(\mathcal{R}) has \sigma-finite outer measure, then it has a measurable cover F with \bar{\mu}(F)=\mu^*(E). Indeed, our hypothesis is that E can be written as the countable disjoint union of sets of finite outer measure, so if we can show this for sets of finite measure the \sigma-finite case will follow.

By the comparisons we showed last time, for every n there is some set F_n\in\mathcal{S}(\mathcal{R}) so that E\subseteq F_n and \bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}. Taking F to be the intersection of this sequence, we must have

\displaystyle \mu^*(E)\leq\bar{\mu}(F)\leq\bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}

Since n is arbitrary, we must have \bar{\mu}(F)=\mu^*(E). If G\in\mathcal{S}(\mathcal{R}) fits into F\setminus E, then E\subseteq F\setminus G. And so:

\bar{\mu}(F)=\mu^*(E)\leq\bar{\mu}(F\setminus G)=\bar{\mu}(F)-\bar{\mu}(G)\leq\bar{\mu}(F)

Since \bar{\mu}(F)<\infty we can subtract it off and find \bar{\mu}(G)=0.

In fact, any measurable cover F must have \bar{\mu}(F)=\mu^*(E). Further, any two of them have a negligible difference.

Indeed, if we have two measurable covers, F_1 and F_2, then E\subseteq F_1\cap F_2\supseteq F_1, which tells us that F_1\setminus(F_1\cap F_2)\subseteq F_1\setminus E. But since F_1 is a measurable cover, we conclude that \bar{\mu}(F_1\setminus(F_1\cap F_2))=0, and similarly that \bar{\mu}(F_2\setminus(F_1\cap F_2))=0. But F_1\Delta F_2 is the union of these two differences, and so \bar{\mu}(F_1\Delta F_2)=0.

If \mu^*(E)=\infty, then any measurable cover must have \bar{\mu}(F)=\infty as well. On the other hand, if \mu^*(E) is finite, then there exists at least one measurable cover with \bar{\mu}(F)=\mu^*(E). Then any other measurable cover differs negligibly, and so has the same measure.

And so if the measure \mu on \mathcal{R} is \sigma-finite, then so are the measures \bar{\mu} on \mathcal{S}(\mathcal{R}) and \bar{\mu} on \overline{\mathcal{S}}. We’ve already seen that \mu^* must be \sigma-finite in this situation, and so any \mu^*-measurable set can be covered by a countable sequence of \mu^*-finite sets. Applying the above results to each of the sets in this sequence gives our result.

April 1, 2010 - Posted by | Analysis, Measure Theory


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