Theorem: Let $\mu$ and $\nu$ be two $\sigma$-finite measures on a measurable space $(X, B)$. Then $\nu$ can be decomposed as $ \nu = \nu_\mathrm{abs} + \nu_\mathrm{sing}$ into the sum of two $\sigma$-finite measures with $\nu_\mathrm{abs} \ll \mu$ being absolutely continuous with respect to $\mu$ and $\nu_\mathrm{sing} \bot \mu$ being singular to each other.
Remark: We only prove the theorem for finite measures.
a) Define a measure $m = \mu + \nu $ and define on the real Hilbert space $H = L_m^2(X)$ a linear functional $\Phi(g) := \int g \; d\nu $. First restrict it to simple functions and show that the operator is bounded on the space of simple functions in $L^2$. Extend it to $H$ and prove that $\exists k \in H : \Phi (g) = \int g k \; d m$.
I've done the first two parts of part a) and now I'm stuck with proving that $\exists k \in H : \Phi (g) = \int g k \; d m$.
I was thinking something like this: \begin{align} \Phi g = \int_X g \; d \nu = \int_X g \; d \nu_\mathrm{abs} + \int_X g \; d \nu_\mathrm{sing} = \int_X fg \; d \mu + \int_{X_1} g \; d \nu + \int_{X_2} g \; d \nu = \int_X fg \; d \mu + \int_{X_2} g \; d \nu \end{align}
But then I don't know how to proceed. Am I on the right track? Many thanks for your help.
b) Prove that $k$ takes values in $[0,1]$ $m$-almost surely.
Can you tell me if the following is correct:
\begin{align} P(\{ x | k(x) \in [0,1]\}) = m(k^{-1}([0,1])) = \int_{k^{-1}([0,1])} 1 dm = \\ \int_{k^{-1}([0,1])} (1 \circ k) dm = \int_{k^{-1}([0,1])} 1 \cdot (1 \circ k) dm = \int_{k^{-1}([0,1])} (1 \circ k) d\nu = \int_{[0,1]} 1 d k(\nu) \end{align}
And then I want this to be $1$ but I don't know $\nu$ and I don't know $d k(\nu)$ so I think I'm stuck here.
Edit
a) OK, using t.b.'s comment the answer to ta) is:
Using the Riesz representation theorem for Hilbert spaces the existence of $k $ follows immediately.
Thanks for your help!