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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!

  • 1
    Use the [Riesz representation theorem](http://en.wikipedia.org/wiki/Riesz_representation_theorem) to find $k$.2011-10-24
  • 0
    By the way, you're already assuming that $\nu$ can be decomposed, in the last line. You shouldn't do that...2011-10-24
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    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.)2011-10-24
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    @MichaelHardy: Thanks!2011-10-25
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    @t.b.: I posted my tentative answer below... thanks for your help!2011-10-25
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    Maybe you want to go to the library and consult Rudin's real and complex analysis, where this (von Neumann's) proof of the Radon-Nikodym theorem is given.2011-10-25
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    @t.b.: You said "of informative value $0$" -- is this not a correct answer? I think using Riesz there is nothing left to show. Is this wrong?2011-10-25
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    @t.b.: OK. I can't go to the library right now but I'll try to find it online...2011-10-25
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    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](http://www.math.ethz.ch/education/bachelor/lectures/hs2011/math/fa/FAI_HS11_Serie5.pdf) 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...2011-10-25
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    @t.b. Hey. Just wanted to say thanks for the time you spent teaching me here on SE. I really enjoyed that. And I really would like to pass my two exams and do a lot more maths (with or without you) but it's just not very realistic. And after failing I'll probably be too crushed so I am likely to quit it all. So I guess, that was it. Nice to have met you. (don't worry about losing rep, I won't delete my account, I'll just let it rot)2012-07-26
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    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...).2012-07-26
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    @t.b. Thanks. I am working, of course.2012-07-26

1 Answers 1

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