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 [1]). 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 $\mu _{2}(T)=\int _{X}f\cdot \chi _{T}\,d\mu _{1}=0$ so that $\mu _{2}\ll \mu _{1}$ (see [3] 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 [2]). 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 [3]). Namely, if we have $\forall \epsilon >0$ that there exists $\delta >0$ so that $\mu _{1}(T)<\delta \implies \mu _{2}(T)<\epsilon$, then $\mu _{2}\ll \mu _{1}$. Indeed, for any such $\epsilon >0$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 $\mu _{1}$.

References

[1]: Taylor, M. "Measure Theory and Integration". 50-51.

[2]: Craig, K. "Math 201a: Homework 8". Refer to question 2.

[3]: Rana, I. K. "Introduction to Measure and Integration". Second Edition. 311-313.