# 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?

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.

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 $$\int_{E_0} f \, d\mu > 0.$$