Let $\Omega$ be a measurable space and $X$ a separable dual Banach space. $X$ is known to have the Radon-Nikodym property: If $\mu$ is a $\sigma$-additive measure with range in $X$ and $\nu$ a $\sigma$-additive measure with range in $\mathbb{R}_+$ such that $\mu<<\nu$, then there exists some $\frac{d\mu}{d\nu}:\Omega\rightarrow X$ such that $\mu(E)=\int_E\frac{d\mu}{d\nu}d\nu$ (see theorem III.3.1 in "vector measures" by Diestel and Uhl).
I have an example that seem to contradict this result and I do not understand how:
Define $\Omega=[0,1],\ X=X^*=L^2[0,1],\ \mu(A)=\chi_A\ \forall A\subset\Omega$ the indicator function of $A$ (1 in $A$ and $0$ outside $A$) and $\nu$ the Lebesgue measure on $\Omega$.
If $\nu(A)=0$ then $\|\mu(A)\|_{L^2}=(\int_0^1|\chi_A|^2)^{\frac{1}{2}}=\nu(A)^{\frac{1}{2}}=0$ so $\mu<<\nu$. But I do not think there exists $\frac{d\mu}{d\nu}:\Omega\rightarrow X$, is there?
Why I think there is not: Assume there exists some $\frac{d\mu}{d\nu}:\Omega\rightarrow X$ such that $\mu(E)=\int_E\frac{d\mu}{d\nu}d\nu$ then for any $f\in X=X^*$ and $E\subset\Omega$:
$$\int_Ef(\omega)d\omega=\left\langle\mu(E),f\right\rangle_{L^2}=\int_E\left\langle\frac{d\mu}{d\nu}(\omega),f\right\rangle_{L^2}d\omega=\int_E\int_{\Omega}\frac{d\mu}{d\nu}(\omega)(t)f(t)dtd\omega$$
which implies $$\int_{\Omega}fd\delta_{\omega}=f(\omega)=\int_{\Omega}\frac{d\mu}{d\nu}(\omega)(t)f(t)dt$$ for all $f\in X$ and almost every $\omega\in\Omega$ which means $\delta_{\omega}<<\nu$ which is not possible.
You've left out a key phrase in your statement of the Radon-Nikodym property. What is true is that if $\mu$ is a $\sigma$-additive measure of bounded variation with range in $X$ and $\nu$ a $\sigma$-additive measure with range in $\mathbb{R}_+$ such that $\mu<<\nu$, then there exists some $\frac{d\mu}{d\nu}:\Omega\rightarrow X$ such that $\mu(E)=\int_E\frac{d\mu}{d\nu}d\nu$. (Check definition III.1.3 of Diestel and Uhl.)
Your measure $\mu$ does not have bounded variation. Indeed, fix $n$ and let $A_i = [(i-1)/n, i/n) \subset [0,1]$, $i=1,\dots, n$. Then $\|\mu(A_i)\|_{L^2} = n^{-1/2}$ for each $i$, so $\sum_{i=1}^n \|\mu(A_i)\| = n^{1/2}$, and the variation of $\mu$ is thus at least $n^{1/2}$; but $n$ was arbitrary.