The Unapologetic Mathematician

Mathematics for the interested outsider

Absolute Continuity II

Now that we’ve redefined absolute continuity, we should tie it back to the original one. That definition makes precise the idea of “smallness” as being bounded in size below some \epsilon or \delta, but the new one draws a sharp condition that “small” sets are those of measure zero. As it turns out, in the presence of a finiteness condition the two are the same: if \nu is a finite signed measure and if \mu is any signed measure such that \nu\ll\mu, then to every \epsilon>0 there is a \delta so that \lvert\nu\rvert(E)<\epsilon for every measurable E with \lvert\mu\rvert(E)<\delta.

So, let’s say that the conclusion fails, and there’s some \epsilon for which we can find a sequence of measurable E_n with \lvert\mu\rvert(E_n)<\frac{1}{2^n}, and yet \lvert\nu\rvert(E_n)>\epsilon for each n. Then we can define E=\limsup\limits_{n\to\infty}E_n, and find

\displaystyle\lvert\mu\rvert(E)\leq\sum\limits_{i=n}^\infty\lvert\mu\rvert(E_i)<\frac{1}{2^{n-1}}

for each n, and thus \lvert\mu\rvert(E)=0. But we also find, since \nu is finite,

\displaystyle\lvert\nu\rvert(E)=\lim\limits_{n\to\infty}\lvert\nu\rvert\left(\bigcup\limits_{i=n}^\infty E_i\right)\geq\limsup\limits_{n\to\infty}\lvert\nu\rvert(E_n)\geq\epsilon

But this contradicts the assertion that \nu\ll\mu.

Let’s make the connection even further by proving the following proposition: if \mu is a signed measure and if f is integrable with respect to \lvert\mu\rvert then we can integrate f with respect to \mu and define

\displaystyle\nu(E)=\int\limits_Ef\,d\mu

for every measurable E. It’s easy to see that \nu is a finite signed measure, and I say that \nu\ll\mu. Indeed, if \lvert\mu\rvert(E)=0 then \mu^+(E)=0=\mu^-(E). Thus we see that

\displaystyle\nu(E)=\int\limits_Ef\,d\mu=\int\limits_Ef\,d\mu^+-\int\limits_Ef\,d\mu^-=0-0=0

and so \nu\ll\mu as asserted. This extends our old result that indefinite integrals with respect to measures are absolutely continuous in our old sense.

It’s easy to verify that the relation \ll is reflexive — \mu\ll\mu — and transitive — \mu_1\ll\mu_2 and \mu_2\ll\mu_3 together imply \mu_1\ll\mu_3 — and so it forms a preorder on the collection of signed measures. Two measures \mu and \nu so that \nu\ll\mu and \mu\ll\nu are said to be equivalent, and we write \mu\equiv\nu.

For example, we can verify that \mu\equiv\lvert\mu\rvert. Indeed, \mu\ll\mu, and we know this implies that \lvert\mu\rvert\ll\mu. Similarly, \lvert\mu\rvert\ll\lvert\mu\rvert, which implies that \mu\ll\lvert\mu\rvert. This is useful because it allows us to show that \lvert\mu\rvert(E)=0 for a measurable set E if and only if \mu(F)=0 for all measurable subsets F\subseteq E. If \lvert\mu\rvert(E)=0, then \lvert\mu\rvert(F)=0 since \lvert\mu\rvert is a measure, and then \mu(F)=0 since \mu\ll\lvert\mu\rvert. Conversely, if \mu(F)=0 for all measurable subsets F\subseteq E, then in particular \mu(E)=0, and thus \lvert\mu\rvert(E)=0 since \lvert\mu\rvert\ll\mu.

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July 2, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. [...] is finite, we know that for every there is a so that if then . Using this , our assertion of continuity [...]

    Pingback by Associated Metric Spaces and Absolutely Continuous Measures I « The Unapologetic Mathematician | August 16, 2010 | Reply


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