Measure Theory :Lemma on absolute continuity One class I covered Radon Nikodym theorem but we have not gone through the following lemma, I tried to study it by myself on the book Introduction to Measure and Integration from Taylor. However I do not get an intuition of what is going on the proof. Does thE proof means $v(E)$ is larger than $\mu(E)$? What is the intuition behind the proof? Could someone explain me like in class? Because I have been noticing books omit lots of steps. Thanks for reading!
Lemma: If $(\Omega,\mathcal{F},\mu)$ is a measure space and $v:\mathcal{F}\to\mathbb{R}$ is finite valued, $\sigma-\text{additive}$ and absolutely continuous with respect to $\mu$, then $v$ satisfies condition(6.4.1): $\mu(E)<\delta\Rightarrow|v(E)|<\epsilon$. Proof: Then if (6.4.1) is false, tehere is an $\epsilon>0$ and a sequence $\{E_n\}$ of sets of $\mathcal{F}$ such that $v(E_n)>\epsilon$ and $\mu(E_n)<2^{-n}$. Put $E=\lim \sup E_n$.
Then
$\mu(E)\leqslant\mu(\bigcup\limits_{r=n+1}^{\infty} E_{r})\leqslant\sum\limits_{r=n+1}^{\infty}\mu( E_{r})<2^{-n}$
$v(E)=\lim v(\bigcup\limits_{r=n+1}^{\infty} E_{r})\geqslant\lim\sup v(E_r)$
so that $v(E)\geqslant\epsilon$. This contradicts $v«u$.$\blacksquare$
 A: Remember, saying that $\nu$ is absolutely continuous w.r.t $\mu$ is the same as saying that $\mu(E) = 0$ implies that $\nu(E) = 0$ for all measurable subsets $E$.
You are missing some very important "for all"'s and "there exists"'s in your statement of the theorem. It should say:
"If $\nu$ is absolutely continuous w.r.t $\mu$, then, for all $\epsilon > 0$, there exists a $\delta > 0$ such that $\mu (E) < \delta $ implies $\nu (E) < \epsilon$ for all measurable subsets $E$."
I could make a feable attempt at putting this in plain English: If every set of zero $\mu$-size  is also of zero $\nu$-size, then is always possible guarantee that a subset has a $\nu$-size smaller than given value by imposing the condition that the $\mu$-size of the subset is suitably small.
Now, I'll write out the proof in a more wordy way...
Suppose it is not the case that, for all $\epsilon > 0$, there exists a $\delta > 0$ such that $\mu (E) < \delta$ implies $\nu(E) < \epsilon$. Then there exists some value of $\epsilon > 0$ such that it is possible to find a subset $E_1$ with $$\mu(E_1) < 1/2, \ \ \ \  \nu(E_1) \geq \epsilon,$$
and such that it is also possible to find a subset $E_2$ with $$\mu (E_2) < 1/4, \ \ \ \ \nu(E_2) \geq \epsilon,$$
and such that it is even possible to find a subset $E_3$ with $$\mu(E_3) < 1/8, \ \ \ \ \nu(E_3) \geq \epsilon,$$
and so on.
In other words, I'm able to generate a sequence of subsets $E_n$ whose $\mu$-measures tend to zero, and yet, whose $\nu$-measures always remain above $\epsilon$.
Now, I'm going to use these subsets $E_1, E_2, E_3 \dots $ to define a new collection of subsets:
$$ F_1  = E_1 \cup E_2 \cup E_3 \cup E_4 \cup \dots $$
$$ F_2  = E_2 \cup E_3 \cup E_4 \cup \dots $$
$$ F_3  = E_3 \cup E_4 \dots $$
These newly-defined subsets $F_1, F_2, F_3 \dots $ also have the property that their $\mu$-measures tend to zero, but their $\nu$-measures always stay above $\epsilon$:
$$ \mu(F_1) < 1 , \ \ \ \nu(F_1) \geq \epsilon,$$
$$ \mu(F_2) < 1/2, \ \ \ \nu(F_2) \geq \epsilon,$$
$$ \mu(F_3) < 1/4, \ \ \ \nu(F_3) \geq \epsilon, $$
and so on. (The proof is by countable additivity and by summing geometric series.)
But $F_1, F_2, F_3, \dots$ also satisfy a really nice property that our original sets $E_1, E_2, E_3, \dots$ didn't satisfy: the sequence of sets  $F_1, F_2, F_3, \dots$ is a nested sequence. This means that
$$ F_1 \supset F_2 \supset F_3 \supset F_4 \dots$$
Finally, define $F$ to be the intersection of the $F_i$'s:
$$ F = F_1 \cap F_2 \cap F_3 \cap \dots$$
Since $F_1, F_2, F_3, \dots$ is a nested sequence, it is easy to work out the $\mu$-measure of $F$ using countable additivity:
$$ \mu(F) = \lim_{n \to \infty} \mu (F_n) < \lim_{n \to \infty} \frac 1 {2^{n-1}} = 0.$$
Meanwhile, we have
$$ \nu(F) = \lim_{n \to \infty} \nu(F_n) \geq \epsilon$$
So we have succeeding in constructing a measurable subset, $F$, whose $\mu$-measure is zero, but whose $\nu$-measure is strictly positive. This contradicts our original assumption that $\nu$ is absolutely continuous w.r.t. $\mu$.
