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!
Best Answer
For b) use a): You showed that $$\int_X g \,d\nu = \int_X gk \,dm, \quad \text{for }g\in L^2(X,m). \tag{1}$$ We can show that $L^{\infty}(X, m) \subset L^2(X,m)$ because $$\| g \|_{L^2(X,m)}=\sqrt{\int_X |g|^2\, dm} \leqslant \|g\|_{\infty}\sqrt{\int_X dm}=\underbrace{\sqrt{m(X)}}_{<+\infty}\underbrace{\|g\|_{\infty}}_{<+\infty} < +\infty$$
We can use $(1)$ for $g=\chi_A$ because $L^{\infty}(X, m) \subset L^2(X,m)$ and we find $$\nu(A)=\int_A k\, dm.$$
Now $$\underbrace{\frac{\nu(A)}{\nu (A)+\mu(A)}}_{\in [0,1]}=\frac{\nu(A)}{m(A)}=\frac{1}{m(A)}\int_A k\, dm \in [0,1] \quad \text{for } m(A) \neq 0.$$
And now, use $k \in L^1(X,m)$ from Cauchy-Schwarz inequality $$\int_X |k|\, dm \leqslant \sqrt{m(X)}\underbrace{\sqrt{\int_X |k|^2\, dm}}_{<+\infty \text{because }k\in L^2(X,m)} < +\infty,$$ $m$ is finite measure and $\frac{1}{m(A)}\int_A k\, dm \in [0,1],$ so we can apply mean value theorem (I don't know English version of this theorem) and find $$k \in [0,1] \,\, [m]\text{-almost surely}$$