# The Unapologetic Mathematician

## Associated Metric Spaces and Absolutely Continuous Measures II

Yesterday, we saw that an absolutely continuous finite signed measure $\nu$ on a measure space $(X,\mathcal{S},\mu)$ defines a continuous function on the associated metric space $\mathfrak{S}$, and that a sequence of such finite signed measures that converges pointwise is actually uniformly absolutely continuous with respect to $\mu$.

We’re going to need to assume that $\nu$ is nonnegative. We’d usually do this by breaking $\nu$ into its positive and negative parts, but it’s not so easy to get ahold of the positive and negative parts of $\nu$ in this case. However, we can break each $\nu_n$ into $\nu_n^+$ and $\nu_n^-$. Then we can take the limits $\nu^{\geq0}(E)=\lim_n\nu_v^+(E)$ and $\nu^{\leq0}(E)=\lim_n\nu_v^-(E)$, which will still satisfy $\nu(E)=\nu^{\geq0}(E)-\nu^{\leq0}$. The only difference between this decomposition and the positive and negative parts is that this pair of set functions might have some redundancy that gets cancelled off in this subtraction. And so, without loss of generality, we will assume that all the $\nu_n$ are nonnegative, and that their limit $\nu$ is as well.

Now, given such a sequence, define the limit function $\nu(E)=\lim_n\nu_n(E)$. I say that $\nu$ is itself a finite signed measure, and that $\nu\ll\mu$. Indeed, $\nu(E)$ is finite by assumption, and additivity is easy to check. As for absolute continuity, if $\mu(E)=0$, then each $\nu_n(E)=0$ since $\nu_n\ll\mu$, and so $\nu(E)=0$ as the limit of the constant zero sequence.

What we need to check is continuity. We know that it suffices to show that $\nu$ is continuous from above at $\emptyset$. So, let $\{E_m\}$ be a decreasing sequence of measurable sets whose limit is $\emptyset$. We must show that the limit of $\nu(E_m)$ is zero. But we know that the limit of $\mu(E_m)$ is zero, and thus for a large enough $m$ we can make $\mu(E_m)<\delta$ for any given $\delta$. And since $\nu\ll\mu$ we know that for any $\epsilon$ there is some $\delta$ so that if $\mu(E)<\delta$ then $\nu(E)<\epsilon$. Thus we can always find a large enough $m$ to guarantee that $\nu(E_m)<\epsilon$, and so the limit is zero, as asserted.

Finally, what happens if we remove the absolute continuity requirement from the $\nu_n$? That is: what can we say if $\{\nu_n\}$ is a sequence of finite signed measures on $X$ so that $\nu(E)=\lim_n\nu_n(E)$ exists and is finite for each $E\in\mathcal{S}$. I say that $\nu(E)$ is a signed measure. What we need is to find some measure $\mu$ so that all the $\nu_n\ll\mu$, and then we can use the above result.

Since $\nu_n$ is a finite signed measure, we can pick some upper bound $c_n\geq\lvert\nu_n(E)\rvert$. Then we define

$\displaystyle\mu(E)=\sum\limits_{n=1}^\infty\frac{1}{2^nc_n}\lvert\nu_n\rvert(E)$

If any $\lvert\nu\rvert(E)\neq0$, then $\mu(E)\neq0$, and so $\lvert\nu_n\rvert\ll\mu$. And thus $\nu_n\ll\mu$ for all $n$, as desired.

August 17, 2010 - Posted by | Analysis, Measure Theory