Characterization of Measures that Share Null Sets

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$.

The following statements are equivalent:

$(1) \quad$ $\nu \ll \mu$ and $\mu \ll \nu$, where $\ll$ denotes absolute continuity

$(2) \quad$ $A \in \Sigma$ is a $\mu$-null set if and only if it is a $\nu$-null set

$(3) \quad$ there exists a positive $\Sigma$-measurable function $g : X \to \R$ such that:

$\ds \map \nu A = \int_A g \rd \mu$

for each $A \in \Sigma$.

Proof

$(1)$ implies $(2)$

Suppose that $\nu \ll \mu$ and $\mu \ll \nu$.

Let $A \in \Sigma$ have $\map \mu A = 0$.

Then, since $\nu \ll \mu$, we have $\map \nu A = 0$ from the definition of absolute continuity.

So if a $\Sigma$-measurable set is $\mu$-null, it is $\nu$-null.

Similarly, let $B \in \Sigma$ have $\map \nu B = 0$.

Then, since $\mu \ll \nu$, we have $\map \mu B = 0$ from the definition of absolute continuity.

So if a $\Sigma$-measurable set is $\nu$-null, it is $\mu$-null.

So a $\Sigma$-measurable set is $\nu$-null if and only if it is $\mu$-null.

Hence $(2)$.

$\Box$

$(2)$ implies $(1)$

Suppose $A \in \Sigma$ is a $\mu$-null set if and only if it is a $\nu$-null set.

Then $A \in \Sigma$ has $\map \nu A = 0$ if and only if it has $\map \mu A = 0$.

In particular, if $A \in \Sigma$ has $\map \nu A = 0$, then it has $\map \mu A = 0$, so $\nu \ll \mu$.

Similarly, if $A \in \Sigma$ has $\map \mu A = 0$ then it has $\map \nu A = 0$ so $\mu \ll \nu$.

So $\nu \ll \mu$ and $\mu \ll \nu$, hence $(1)$.

$\Box$

$(3)$ implies $(2)$

Suppose there existed a positive $\Sigma$-measurable function $g : X \to \R$ such that:

$\ds \map \nu A = \int_A g \rd \mu$

for each $A \in \Sigma$.

Let $A \in \Sigma$ be a $\mu$-null set.

Then, from Integral of Integrable Function over Null Set, we have:

$\ds \int_N g \rd \mu = 0$

so:

$\map \nu A = 0$

So $\nu \ll \mu$.

Now let $A \in \Sigma$ be a $\nu$-null set.

Suppose that $A$ is not a $\mu$-null set, then:

$\map \mu A > 0$

We have:

$\ds \int_A g \rd \mu = 0$

That is:

$\ds \int g \chi_A \rd \mu = 0$

From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have:

$\map g x \map {\chi_A} x = 0$ for $\mu$-almost every $x \in X$.

But $\map {\chi_A} x = 1$ for $x \in A$, so we must have $\map g x = 0$ for $x \in A$.

We have that $\map \mu \O = 0$ from Empty Set is Null Set, while $\map \mu A > 0$.

So we cannot have that $A = \O$.

This contradicts that $\map g x > 0$ for all $x \in X$.

So we must have $\map \mu A = 0$.

$\Box$

$(2)$ implies $(3)$

Suppose that $\mu \ll \nu$ and $\nu \ll \mu$.

From the Radon-Nikodym Theorem, we have that there exists a $\Sigma$-measurable function $g : X \to \hointr 0 \infty$ such that:

$\ds \map \nu A = \int_A g \rd \mu$

for each $A \in \Sigma$.

Note that this is not sufficient, since we may have $\map g x = 0$ for some $x \in X$.

We show that we have $\map g x > 0$ for $\mu$-almost every $x \in X$ so that we can modify the values of $g$ unproblematically.

Suppose that there exists $A \in \Sigma$ such that $\map \mu A > 0$ but $\map g x = 0$ for each $x \in A$.

Then we have that $\map g x \map {\chi_A} x = 0$ for all $x \in X$, and so:

$\ds \int g \chi_A \rd \mu = 0$

But then:

$\ds \int_A g \rd \mu = 0$

giving:

$\map \nu A = 0$

So $A$ is a $\nu$-null set that is not $\mu$-null.

This contradicts the hypothesis that $\mu \ll \nu$.

So there cannot exist an $A \in \Sigma$ such that $\map \mu A > 0$ but $\map g x = 0$ for each $x \in A$.

So, if $A \in \Sigma$ has $\map g x = 0$ for all $A \in \Sigma$, then we have $\map \mu A = 0$.

From Measurable Functions Determine Measurable Sets, we have:

$\set {x \in X : \map g x = 0} \in \Sigma$

From what we have just shown we therefore have:

$\map \mu {\set {x \in X : \map g x = 0} } = 0$

Write:

$N = \set {x \in X : \map g x = 0}$

Note that $N$ may be empty.

Define $g^\ast : X \to \openint 0 \infty$ by:

$\ds \map {g^\ast} x = \map g x + \map {\chi_N} x = \begin{cases}\map g x & x \not \in N \\ 1 & x \in N\end{cases}$

for each $x \in X$.

From Characteristic Function Measurable iff Set Measurable:

$\chi_N$ is $\Sigma$-measurable.

So from Pointwise Sum of Measurable Functions is Measurable:

$g^\ast$ is $\Sigma$-measurable.

From Characteristic Function of Null Set is A.E. Equal to Zero, we have:

$g^\ast = g$ $\mu$-almost everywhere.

So, from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 2, we have:

$\ds \int_A g^\ast \rd \mu = \int_A g \rd \mu = \map \nu A$

for each $A \in \Sigma$.

So $g^\ast$ is the $\Sigma$-measurable function required.

$\blacksquare$