Counterexample in Radon-Nikodym Theorem. Problem 38 Royden 2ed. Use the following example to show that the hypothesis in the
Radon-Nikodym Theorem that $\mu$ is $\sigma$-finite cannot be omitted. Let $X=[0,1],\ \mathcal{B}$ the class of Lebesgue measurable subsets of $[0, 1]$, and take $\nu$, to be Lebesgue measure and $\mu$ to be the counting measure on $\mathcal{B}$. Then $\nu$  is
finite and absolutely continuous with respect to $\mu$, but there is no function $f$ such that $\nu E = \int_{E} f d\mu$ for all $E\in\mathcal{B}.$ 
Solution.  Suppose there is a function $f$ such that $\nu(E) = \int_{E} f d\mu$ for all $E\in\mathcal{B}$. Since $\nu$ es finite, $f$ is integrable with respect to $\mu$. Thus $E_0 = \left\{x : f(x)\neq 0\right\}$ is countable. Now $0=\nu(E_0)=\int_{E_0}fd\mu$. Contradiction. Hence there is no such function
I have a doubts:
Why $E_0$ is countable?
Why $\int_{E_0}fd\mu=0$ is a contradiction?
 A: Why is $E_0$ countable?
Observe that $ E_0 = \bigcup_{n \in \mathbb N} F_n,$ where $ F_n : = \{ x : | f(x) | > \tfrac 1 n \}. $ Also observe that
$$\int_{[0,1]} |f| \ d\mu \geq \int_{F_n} |f| \ d\mu\geq \frac{1}{n}\mu(F_n)=\frac{|F_n|}{n}.$$
For $f$ to be $\mu$-integrable, we therefore require that $|F_n|$ is finite for each $n$, and this implies that $E_0$ is countable.
Why is $\int_{E_0} f d\mu = 0$ a contradiction?
It is worth pointing out that $f$ must be non-negative everywhere. (Because if $f$ is negative for some $x \in [0,1]$, then we would have $\nu(\{ x\}) = f(x) \mu(\{ x\}) = f(x) < 0$.)
Since $f$ is non-negative and non-zero on $E_0$, $f$ must in fact be strictly positive on $E_0$. If $E_0$ is non-empty (say $x \in E_0$), then $$\nu(E_0) = \int_{E_0} f \ d\mu \geq f(x) \mu(\{ x \})= f(x)> 0,$$
which is false, since the Lebesgue measure of any countable set is zero.
So $E_0$ must be empty. But then $f$ would be zero everywhere, which would imply that the Lebesgue measure is the zero measure, and this is absurd too.
Is there a more direct way of  proving this?
Personally, I would argue as follows. $\nu (E) = \int_E f \ d\mu$ holds for all Lebesgue-measurable $E$, so in particular, it must hold when  $E$ is a singleton set $ \{ x \}$. Thus we have
$$ 0 = \nu (\{ x \})=\int_{\{ x\}} f \ d\mu=f(x) \mu(\{ x \})=f(x)$$
for all $x \in [0,1]$, i.e. $f$ is identically zero.
But then, considering the case $E = [0,1]$, we have
$$ 1 = \nu([0,1]) = \int_{[0,1]}f \ d\mu = \int_{[0,1]}0 \ d\mu = 0,$$
which is a contradiction.
A: For a sum of the elements of a set to be finite, the number of non-zero elements must be at most countable. This is the reason for $E_0$ being countable.
The contradiction occurs because $f>0$ on $E_0,$ and $E_0$ is non-empty since $\nu>0,$ so $\int_{E_0} f \, d\mu > 0.$
