# The Unapologetic Mathematician

## Absolute Continuity I

We’ve shown that indefinite integrals are absolutely continuous, but today we’re going to revise and extend this notion. But first, to review: we’ve said that a set function $\nu$ defined on the measurable sets of a measure space $(X,\mathcal{S},\mu)$ is absolutely continuous if for every $\epsilon>0$ there is a $\delta$ so that $\mu(E)<\delta$ implies that $\lvert\nu(E)\rvert<\epsilon$.

But now I want to change this definition. Given a measurable space $(X,\mathcal{S})$ and two signed measures $\mu$ and $\nu$ defined on $\mathcal{S}$ we say that $\nu$ is absolutely continuous with respect to $\mu$ — and write $\nu\ll\mu$ — if $\nu(E)=0$ for every measurable set $E$ for which $\lvert\mu\rvert(E)=0$. It still essentially says that $\nu$ is small whenever $\mu$ is small, but here we describe “smallness” of $\nu$ by $\nu$ itself, while we describe “smallness” of $\mu$ by its total variation $\lvert\mu\rvert$.

This situation is apparently asymmetric, but only apparently; If $\mu$ and $\nu$ are signed measures, then the conditions

\displaystyle\begin{aligned}&\nu\ll\mu\\&\nu^+\ll\mu\quad\mathrm{and}\quad\nu^-\ll\mu\\&\lvert\nu\rvert\ll\mu\end{aligned}

are equivalent. Indeed, if $X=A\uplus B$ is a Hahn decomposition with respect to $\nu$ then whenever $\lvert\mu\rvert(E)=0$ we have both

\displaystyle\begin{aligned}0\leq\lvert\mu\rvert(E\cap A)&\leq\lvert\mu\rvert(E)=0\\{0}\leq\lvert\mu\rvert(E\cap B)&\leq\lvert\mu\rvert(E)=0\end{aligned}

Thus if the first condition holds we find

\displaystyle\begin{aligned}\nu^+(E)=\nu(E\cap A)&=0\\\nu^-(E)=-\nu(E\cap B)&=0\end{aligned}

and the second condition must hold as well. If the second condition holds we use the definition

$\displaystyle\lvert\nu\rvert(E)=\nu^+(E)+\nu^-(E)$

to show that the third must hold. And if the third holds, then we use the inequality

$\displaystyle0\leq\lvert\nu(E)\rvert\leq\lvert\nu\rvert(E)$

to show that the first must hold.

Now, just because smallness in $\nu$ can be equivalently expressed in terms of its total variation does not mean that smallness in $\mu$ can be equivalently expressed in terms of the signed measure itself. Indeed, consider the following two functions on the unit interval $[0,1]\subseteq\mathbb{R}$ with Lebesgue measure $\mu$:

\displaystyle\begin{aligned}f_1(x)&=\chi_{\left[0,\frac{1}{2}\right]}-\chi_{\left(\frac{1}{2},1\right]}\\f_2(x)&=x\end{aligned}

and define $\nu_i$ to be the indefinite integral of $f_i$. We can tell that the total variation $\lvert\nu_1\rvert$ is the Lebegue measure $\mu$ itself, since $\lvert f_1\rvert=1$. Thus if $\lvert\nu\rvert(E)=0$ then we can easily calculate

$\displaystyle\nu_2(E)=\int\limits_Ex\,d\mu=\int x\chi_E\,d\mu=0$

and so $\nu_2\ll\nu_1$. However, it is not true that $\nu_2(E)=0$ for every measurable $E$ with $\nu_1(E)=0$. Indeed, $\nu_1(\left[0,1\right])=0$, and yet we calculate

$\displaystyle\nu_2(\left[0,1\right])=\int x\,d\mu\geq\int\frac{1}{2}\chi_{\left[\frac{1}{2},1\right]}\,d\mu=\frac{1}{4}$

By the way: it’s tempting to say that this integral is actually equal to $\frac{1}{2}$, but remember that we only really know how to calculate integrals by taking limits of integrals of simple functions, and that’s a bit more cumbersome than we really want to get into right now.

One first quick result about absolute continuity: if $\mu$ and $\nu$ are any two measures, then $\nu\ll\mu+\nu$. Indeed, if $\mu(E)+\nu(E)=0$ then by the positivity of measures we must have both $\mu(E)=0$ and $\nu(E)=0$, the latter of which shows the absolute continuity we’re after.

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

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