# The Unapologetic Mathematician

## 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 $\displaystyle\mathfrak{E}_k=\bigcap\limits_{m=k}^\infty\bigcap\limits_{n=k}^\infty\left\{E\in\mathfrak{S}\bigg\vert\lvert\nu_n(E)-\nu_m(E)\rvert\leq\frac{\epsilon}{3}\right\}$

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) is contained in $\mathfrak{E}_{k_0}$. Let $\delta$ be a positive number with $\delta, 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

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