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.

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Use the Riesz representation theorem to find $k$. – t.b. Oct 24 '11 at 13:45
By the way, you're already assuming that $\nu$ can be decomposed, in the last line. You shouldn't do that... – t.b. Oct 24 '11 at 13:48
I did some TeX improvements. Notice that $\nu_{sing}$ looks different from $\nu_\mathrm{sing}$, and $<<$ looks different from $\ll$. (And a few other minor things.) – Michael Hardy Oct 24 '11 at 17:18
No it's not wrong, but it doesn't add any new information to this thread. I would've expected that you invest a bit more effort and thought into your homework. You've got 4 more parts of your exercise sheet to solve by tomorrow. I tried to 1. discourage you from posting all five parts as separate threads, 2. urge you that you do the thinking yourself, 3. tell you that it doesn't matter for you if you copy down the solution from answers here or from a book. But here you're asking people to invest time... – t.b. Oct 25 '11 at 12:35
Thanks a lot for this. I haven't ignored your mails and I'll write to you by tomorrow. I'm a bit in a hurry these days. Just don't be too pessimistic, keep your head up and keep working. Best wishes, t. (I just noticed that I could remove my downvote here...). – t.b. Jul 26 '12 at 12:26

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 $$k \in [0,1] \,\, [m]\text{-almost surely}$$