The Unapologetic Mathematician

Mathematics for the interested outsider

Associated Metric Spaces and Absolutely Continuous Measures I

If \mathfrak{S} is the metric space associated to a measure space (X,\mathcal{S},\mu), and if \nu is a finite signed measure that is absolutely continuous with respect to \mu. Then \nu defines a continuous function on \mathfrak{S}.

Indeed, if E\in\mathcal{S} is any set with \mu(E)<\infty, then E represents a point of \mathfrak{S}, and \nu(E) defines the value of our function at this point. If F\subseteq\mathcal{S} is another set representing the same point, then \mu(E\Delta F)=0. By absolute continuity, \nu(E\Delta F)=0 as well, and so \nu(F)=\nu(E). Thus our function doesn’t depend on the representative we use.

As for continuity at a point E, given an \epsilon>0, we want to find a \delta>0 so that if \mu(E\Delta F)<\delta then \lvert\nu(E)-\nu(F)\rvert<\epsilon. We calculate

\displaystyle\begin{aligned}\lvert\nu(E)-\nu(F)\rvert&=\lvert\nu(E\setminus F)-\nu(F\setminus E)\rvert\\&\leq\lvert\nu(E\setminus F)\rvert+\lvert\nu(F\setminus E)\rvert\\&\leq\lvert\nu(E\Delta F)\rvert+\lvert\nu(E\Delta F)\rvert\\&=2\lvert\nu(E\Delta F)\rvert\\&\leq2\lvert\nu\rvert(E\Delta F)\end{aligned}

Since \nu is finite, we know that for every \epsilon>0 there is a \delta>0 so that if \lvert\mu\rvert(E\Delta F)=\mu(E\Delta F)<\delta then \lvert\nu\rvert(E\Delta F)<\epsilon. Using this \delta, our assertion of continuity follows.

Now, if \{\nu_n\} is a sequence of finite signed measures on X that are all absolutely continuous with respect to \mu, and if the limit \lim_n\nu_n(E) exists and is finite for each E\in\mathcal{S}, then the sequence is uniformly absolutely continuous with respect to \mu. That is,

For any \epsilon>0 we can define the set


Since each \nu_n is continuous as a function on \mathfrak{S}, each of these \mathfrak{E}_k is a closed set. Since the sequence \{\nu_n(E)\} always converges to a finite limit, it must be Cauchy for each E, and so the union of all the \mathfrak{E}_k is all of \mathfrak{S}. Thus the countable union of these closed subsets has an interior point. But since \mathfrak{S} is a complete metric space, it is a Baire space as well. And thus one of the \mathfrak{E}_k must have an interior point as well.

Thus there is some k_0, some radius r_0, and some set E_0 so that the ball \{E\in\mathfrak{S}\vert\rho(E,E_0)<r_0\} is contained in \mathfrak{E}_{k_0}. Let \delta be a positive number with \delta<r_0, and so that \lvert\nu_n(E)\rvert<\frac{\epsilon}{3} whenever \mu(E)<\delta and 1\leq n\leq k_0 This \delta will suffice (by definition) for all n up to k_0. We will show that it works for higher n as well. Note that if \mu(E)<\delta, then

\displaystyle\begin{aligned}\rho(E_0\setminus E,E_0)=\mu\left((E_0\setminus E)\Delta E_0\right)&=\mu(E_0\cap E)\leq\mu(E)<\delta<r_0\\\rho(E_0\cup E,E_0)=\mu\left((E_0\cup E)\Delta E_0\right)&=\mu(E\setminus E_0)\leq\mu(E)<\delta<r_0\end{aligned}

so E_0\setminus E and E_0\cup E are both inside \mathfrak{E}_{k_0}. And so we calculate

\displaystyle\lvert\nu_n(E)\rvert\leq\lvert\nu_{k_0}(E)\rvert+\lvert\nu_n(E_0\cup E)-\nu_{k_0}(E_0\cup E)\rvert+\lvert\nu_n(E_0\setminus E)-\nu_{k_0}(E_0\setminus E)\rvert

The first term is less than \frac{\epsilon}{3} by the definition of \delta. The second and third terms are less than or equal to \frac{\epsilon}{3} because E_0\cup E and E_0\setminus E are in \mathfrak{E}_{k_0}. Since the same \delta works for all n, the absolute continuity is uniform.

August 16, 2010 Posted by | Analysis, Measure Theory | 1 Comment



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