Absolute continuity of measures

A measure $\nu$ is absolutely continuous with respect to $\mu$, written $\nu \ll \mu$, when every $\mu$-null measurable set is also $\nu$-null. This is distinct from the regularity property of functions called absolute continuity.

A measure $\nu$ is absolutely continuous with respect to $\mu$ when every $\mu$-null set is also $\nu$-null.

A measure $\rho$ dominates a family of measures when every member is absolutely continuous with respect to $\rho$. Any two finite positive measures $\mu$ and $\nu$ have the common dominating measure $\mu +\nu$.

A function $f$ has Lusin's condition $(N)$ when it maps every Lebesgue-null set to a Lebesgue-null set.

Let $h:\mathbb{R}\to \mathbb{R}$ be continuous, strictly increasing, and satisfy Lusin condition N. Thendefines a measure on the Lebesgue measurable sets, and $\nu \ll \lambda$. Moreover,for $a\le b$.

If $\mu (\Omega )<\infty$, then $\mu \ll \nu$ exactly when for every $\epsilon >0$ there is $\delta >0$ such that $\nu (A)<\delta$ implies $\mu (A)<\epsilon$. For necessity, a contrary sequence with $\nu (A_n)<2^{-n}$ and $\mu (A_n)\ge \epsilon$ has tail unions decreasing to a $\nu$-null limsup; continuity from above for finite $\mu$ contradicts absolute continuity.

Two measures are mutually absolutely continuous when they have exactly the same null sets.

Two probability measures are equivalent when each is absolutely continuous with respect to the other, or equivalently when they have the same null events.

For sigma-finite measures, $\nu \ll \mu$ implies $\nu (A)={\int }_A(d\nu /d\mu )d\mu$ for a nonnegative measurable density unique almost everywhere.

For finite positive measures $\mu ,\nu$, put $\rho =\mu +\nu$. The linear functional $g\mapsto \int gd\nu$ is bounded on real $L^2(\rho )$ by the Cauchy-Schwarz inequality. The Riesz representation theorem gives a measurable $h$ with $\nu (A)={\int }_{A}hd\rho$. Testing indicators of its negative and greater-than-one level sets shows $0\le h\le 1$ almost everywhere. Subtraction gives $\mu (A)={\int }_A(1-h)d\rho$. This constructs the densities without assuming the Radon-Nikodym theorem as a prior result.

When $\nu \ll \mu$, the Radon-Nikodym derivative is the almost-everywhere unique measurable function satisfying $\nu (A)={\int }_A(d\nu /d\mu )d\mu$.

For an absolutely continuous change of probability measure, restricting its Radon-Nikodym derivative to the current sigma-algebra gives this nonnegative uniformly integrable martingale. If $Z_T>0$ almost surely, the measures are equivalent on ${\mathcal{F}}_T$. This finite-horizon equivalence need not imply equivalence on a larger sigma-algebra.

Mutual absolute continuity is equivalent to the Radon--Nikodym derivative being finite and strictly positive almost everywhere.

For sigma-finite measures $\nu$ and $\mu$, there is a unique decomposition $\nu ={\nu }_{\mathrm{ac}}+{\nu }_{\mathrm{s}}$ in which ${\nu }_{\mathrm{ac}}\ll \mu$ and ${\nu }_{\mathrm{s}}\perp \mu$.

For a finite positive measure on $[0,1)$, average its mass over each half-open dyadic interval. The resulting nonnegative martingale has constant expectation $\mu ([0,1))$: a parent interval's mass is the sum of its children's masses. Its almost-sure limit is integrable but the martingale need not be uniformly integrable. The limit recovers the density of the absolutely continuous component in the Lebesgue decomposition theorem.

The Conditional Fatou lemma gives $X_n\ge \mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ for the dyadic density martingale. Integration first on dyadic sets, then extension by the Monotone class theorem, makes $\nu =\mu -X_{\infty }\lambda$ a nonnegative measure. Its finite-level density is $Y_n=X_n-\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$, which tends to zero almost everywhere for Lebesgue measure. The density level sets separate a singular measure argument shows that this zero-limit set has $\nu$-measure zero. Thus $\nu$ and $\lambda$ are mutually singular measures, obtaining the decomposition without assuming it to identify the martingale limit.

Use the Hilbert-space construction of dominated measure densities with $\rho =\mu +\nu$. The set $B=\{h=1\}$ is a null set for $\mu$. Put $f=0$ on $B$ and $f=h/(1-h)$ elsewhere. Then $\int fd\mu =\nu (B^c)<\infty$ and $\nu (A)={\int }_{A}fd\mu +\nu (A\cap B)$. This is the Lebesgue decomposition into a density part and a part concentrated on a $\mu$-null set.

A countable family of probability measures is dominated by any mixture ${\sum }_n{w}_n{\nu }_n$ with every $w_n>0$ and ${\sum }_n{w}_n=1$.