Counterexample to Absolute Continuity condition (for measures)

Question

I'm taking an Analysis class and today we proved the following lemma:

Let $\nu$ be a finite measure. Then $\nu\ll\mu$ iff for all $\varepsilon>0$ there exists $\delta>0$ s.t $\mu\left(A\right)<\delta\Longrightarrow\nu\left(A\right)<\varepsilon$

Then the instructor gave the following counter-example showing this is not true if we only require $\nu$ to be $\sigma$-finite: Take $X=\left(0,1\right)$ , $\mu$ the Lebesgue measure and $\nu\left(A\right)=\int_{A}\frac{1}{x}d\mu$ . Then for $A_{n}=\left[\frac{1}{n},\frac{2}{n}\right],\ n\geq2$ one has $\nu\left(A_{n}\right)=\log2$ for all $n$ but $\mu\left(A_{n}\right)\overset{n\to\infty}{\longrightarrow}0$

I don't understand how this shows the claim is not true in this case. I would appreciate it if someone could clarify the instructor's intention.

Answer 1

First we show that $\nu$ is $\sigma-$finite. Define $$A_n=\left[\frac{1}{n},\frac{2}{n}\right),\quad n\geq 2$$ Then it is clear that $$X=(0,1)=\bigcup_{n=2}^\infty A_n$$ and $$\nu(A_n)=\int_{\frac{1}{n}}^{\frac{2}{n}}\frac{1}{x}dx=\log\frac{2}{n}-\log\frac{1}{n}=\log 2<\infty$$ This shows that $\nu$ is $\sigma-$finite. And it is clear that $\nu$ is absolutely continuous w.r.t $\mu$.

However for every $\delta>0$, we can always choose $N\in\mathbb{N}$ with $\frac{1}{N}\leq\delta$, thus $$\mu(A_n)\leq \mu(A_N)=\frac{1}{N}<\delta,\quad\forall n\geq N$$ Choose $\epsilon\in(0,\log 2)$, we have $\nu(A_n)=\log 2>\epsilon$ for all $n\geq N$. Thus in this case $\mu(A_n)<\delta$ does not imply $\nu(A_n)<\epsilon$.

Edit: Fixed incorrect inequality.

Answer 2

The measure $\nu$ has by definition a density with respect to the Lebesgue measure $\mu$; namely the density equals

$$f(x) = \frac{1}{x}.$$

Consequently, $\nu \ll \mu$.

Now consider $\epsilon := \frac{\log}{2}$. Since

$$\mu(A_n) = \frac{1}{n} \qquad \text{and} \qquad \nu(A_n) = \log 2 > \epsilon$$

for all $n \in \mathbb{N}$, we cannot choose $\delta>0$ such that the condition

$$\mu(A)<\delta \implies \nu(A)<\epsilon$$

is satisfied. (Just assume the contrary, i.e. such $\delta>0$ exists. For $n$ sufficiently large, we have $\mu(A_n)<\delta$, but $\nu(A_n)<\epsilon$ does not hold true.)