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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 $nu E = int_{E} f dmu$ for all $Einmathcal{B}.$
Solution. Suppose there is a function $f$ such that $nu(E) = int_{E} f dmu$ for all $Einmathcal{B}$. Since $nu$ es finite, $f$ is integrable with respect to $mu$. Thus $E_0 = left{x : f(x)neq 0right}$ is countable. Now $0=nu(E_0)=int_{E_0}fdmu$. Contradiction. Hence there is no such function
I have a doubts:

Why $E_0$ is countable?

Why $int_{E_0}fdmu=0$ is a contradiction?

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
$$int_{[0,1]} |f| dmu geq int_{F_n} |f| dmugeq frac{1}{n}mu(F_n)=frac{|F_n|}{n}.$$

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 $int_{E_0} f dmu = 0$ 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 $$nu(E_0) = int_{E_0} f dmu geq f(x) mu({ x })= f(x)> 0,$$
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. $nu (E) = int_E f dmu$ holds for all Lebesgue-measurable $E$, so in particular, it must hold when $E$ is a singleton set $ { x }$. Thus we have
$$ 0 = nu ({ x })=int_{{ x}} f dmu=f(x) mu({ x })=f(x)$$
for all $x in [0,1]$, i.e. $f$ is identically zero.

But then, considering the case $E = [0,1]$, we have
$$ 1 = nu([0,1]) = int_{[0,1]}f dmu = int_{[0,1]}0 dmu = 0,$$

which is a contradiction.

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 $int_{E_0} f , dmu > 0.$