On the null set of absolute continuous signed measure

Question

Folland states the absolute continuity of a signed measure as follows:

If $\nu \ll \mu$, then $ \mu(E)=0$ implies that $E$ is a null set for $\nu$ because $\forall A \subset E : \mu(A)=0$ and so $\nu(A)=0$. It is also obvious that $ \nu(E)=0$ does not imply $ \mu(E)=0$ since we can choose $E$ such that $\nu^+(E)=\nu^-(E) \ne 0$.

Is it possible that $\nu \ll \mu$ and $E$ is a null set for $\nu$ but $ \mu(E) \ne 0$? If it is possible, would you give me an example? If it is not, would you prove it?

Note: $E$ is null set for $\nu$, if $\forall A \subset E : \nu(A)=0$.

Answer 1

This is very easy. Let $\mu$ be Lebesgue measure on $\mathbb R$ and $\nu (E)=\mu (E\cap (0,1))$. Then $\nu <<\mu$ and $\mathbb R \setminus (0,1)$ is a $\nu-$ null set but $\mu (\mathbb R \setminus (0,1)) >0$

Answer 2

First example : Choose the measurable space $(\Bbb{N}, \scr{P}({\Bbb{N}}))$

$\nu (E) = \sum_{n\in E } \frac{1}{2^n}$

$\mu(E) = \begin{cases} |E| & E : finite \\ \infty & otherwise \end{cases}$

Claim : $\nu\ll\mu $ but not the converse.

Proof : $E\subset \Bbb{N} $ and $\mu(E) = 0$

Then $E=\emptyset$ implies $\nu(\emptyset) =0$

But converse is not true. i.e $\exists \epsilon >0 $ such that $ \forall \delta>0 $, $\exists E\subset \Bbb{N} $ such that $\mu(E) < \delta $ but $\nu(E) \ge \epsilon$

$\nu(\Bbb{N}) =\sum_{n\in \Bbb{N} } \frac{1}{2^n}$

Since the above series converges $\exists k\in \Bbb{N}$ such that

$ \sum_{n\ge k } \frac{1}{2^n}<\delta$

Choose, $E= \{n\in \Bbb{N} : n\ge k\}$

Then, $nu(E) <\delta $ but $\mu(E) =\infty$

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To find $E\subset \Bbb{N}$ such that $\nu(E) =0$ but $\mu(E) \neq 0$ ...

Choose, $E_k=\{n\in \Bbb{N} : n\ge k\}$

Let, $E=\cap_{k\in \Bbb{N}} E_k$

Since, $(E_k) $ is a decreasing sequence of measurable sets and decreases to $E$ and $\nu(\Bbb{N})<\infty$ , by continuity of measure

$\begin{align} \nu(E) &=\lim_{k\to \infty } \nu(E_k)\\&= \lim_{k\to \infty} \sum_{n\ge k } \frac{1}{2^k}\\&=0\end{align} $

But, $\mu(E) \neq 0 $ as $E\neq \emptyset$.

Because, $(E_k) $ decreasing sequence of non-empty closed sets as $(\Bbb{N}, d_std) $ is discrete space and since discrete space is complete, from Cantor intersection theorem it follows that $E=\cap_{k\in \Bbb{N}} E_k \neq \emptyset$

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Second example :

$X$ be any uncountable set.

$\begin{align}{\scr{B}}{(X)}& = \{E\subset X : |E|\le \aleph_0 \text { or } | X\setminus E |\le \aleph_0\}\end{align}$

Then $(X, {\scr{B}}{(X) })$ is a measurable space.

$\nu(E) = \begin{cases} 0 & E : Countable \\ \infty & otherwise \end{cases}$

$\mu(E) = \begin{cases} |E| & E : finite \\ \infty & otherwise \end{cases}$

Then $\nu\ll \mu $ is clear as only null set in the measure $\nu$ is empty set and $\nu $ - measure of the empty set is $0$.

Now, choose $E$ to be any non empty countable set .

Then $\nu(E) =0 $ but $\mu(E) =|E|\neq 0$

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Third Example :

Choose the lebesgue measure space $(\Bbb{R}, { \scr{L}} , m) $ and

and $\mu$ : Counting measure.

Then $m\ll \mu $ but choose $E=\scr{C}$ : Cantor set.

Then, $m(E) =0 $ but $\mu (E) \neq 0 $ as Cantor set is uncountable.

Note :

1. A measure itself a signed measure .

2. Monotonicity and non-negativity implies measure $0$ and null set are equivalent.