# Absolutely Continuous Measures

## Definitions

Let $(X,{\mathcal {M}},\mu _{1})$ be a measure space. The measure $\mu _{2}:{\mathcal {M}}\rightarrow [0,\infty ]$ is said to be absolutely continuous with respect to the measure $\mu _{1}$ if we have that $\mu _{2}(T)=0$ for $T\in {\mathcal {M}}$ such that $\mu _{1}(T)=0$ (see ). In this case, we denote that $\mu _{2}$ is absolutely continuous with respect to $\mu _{1}$ by writing $\mu _{2}\ll \mu _{1}$ .

## Examples

Recall that if $f:X\rightarrow [0,\infty ]$ is a measurable function, then the set function $\mu _{2}(T)=\int _{T}f\,d\mu _{1}$ for $T\in {\mathcal {M}}$ is a measure on $(X,{\mathcal {M}},\mu _{1})$ . Observe that if $\mu _{1}(T)=0$ , then $\mu _{2}(T)=\int _{X}f\cdot \mathbb {1} _{T}\,d\mu _{1}=0$ so that $\mu _{2}\ll \mu _{1}$ (see  for further details on this example and others).

## Properties

It was previously established on a homework problem that for some nonnegative measurable $f\in L^{1}(\mu _{1})$ defined on the measure space $(X,{\mathcal {M}},\mu _{1})$ and some arbitrarily chosen $\epsilon >0$ , there exists $\delta >0$ such that $\int _{T}f\,d\mu _{1}<\epsilon$ whenever $\mu _{1}(T)<\delta$ (see ). The method that was used to establish this result can also be used to show that, in a finite measure space, if $\mu _{2}\ll \mu _{1}$ , then for some arbitrarily chosen $\epsilon >0$ , there exists $\delta >0$ such that $\mu _{1}(T)<\delta \implies \mu _{2}(T)<\epsilon$ .

In particular, we proceed by contradiction and suppose there exists $\epsilon >0$ so that for any $\delta >0$ and $\mu _{1}(T)<\delta$ , we have $\mu _{2}(T)\geq \epsilon$ . Now, define a sequence of sets $\{T_{n}\}_{n\in \mathbb {N} }\subseteq {\mathcal {M}}$ such that $\mu _{1}(T_{n})<{\frac {\epsilon }{2^{n}}}$ and denote $T=\limsup {T_{n}}=\cap _{n=1}^{\infty }G_{n}\in {\mathcal {M}}$ where $G_{n}=\cup _{k=n}^{\infty }T_{k}$ . We have from countable subadditivity that $\mu _{1}(G_{n})\leq \sum _{k=n}^{\infty }\mu _{1}(T_{k})<\sum _{k=n}^{\infty }{\frac {\epsilon }{2^{k}}}={\frac {\epsilon }{2^{n+1}}}\,\forall n\in \mathbb {N}$ . We have from monotonicity that $\mu _{2}(G_{n})\geq \epsilon \,\,\forall n\in \mathbb {N}$ . The monotonicity of the measure $\mu _{1}$ implies that $\mu _{1}(\cap _{n=1}^{\infty }G_{n})=\mu _{1}(\limsup {T_{n}})=\mu _{1}(T)=0<\delta$ . Applying continuity from above to $\mu _{2}$ , we also have $\mu _{2}(T)\geq \epsilon$ . However, this contradicts the definition of $\mu _{2}\ll \mu _{1}$ .

In fact, the converse to the above result also holds (see ). Namely, if we have $\forall \epsilon >0$ that there exists $\delta >0$ so that $\mu _{1}(T)<\delta \implies \mu _{2}(T)<\epsilon$ , then $\mu _{2}\ll \mu _{1}$ . Suppose that $\mu _{1}(T)=0$ for some $T\in {\mathcal {M}}$ . Then, for any such $\epsilon >0$ as in the preceding claim, we have $\mu _{1}(T)=0<\delta \implies \mu _{2}(T)<\epsilon$ . Since $\epsilon >0$ can be taken to be arbitrarily small, we have that $\mu _{2}(T)=0$ , as required for the measure $\mu _{2}$ to be absolutely continuous with respect to $\mu _{1}$ .