1

In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :

For any two finite signed measures $\nu_1, \nu_2$ on $\left([-\pi, \pi], \mathcal{B}_{[-\pi, \pi]}\right)$ such that : $\int f\mathrm d\nu_1=\int f\mathrm d\nu_2$ holds for every $f$ continuous (and bounded), we have: $\nu_1=\nu_2$.

This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.
The answers in this thread are based on the fact that for $\mu, \mu'$ probability measures, $\mathcal{C}=\{B\in\mathcal{B}_{[-\pi, \pi]} : \mu(B)=\mu'(B)\}$ is a $\pi$-system. Does this even hold here ?

EDIT: It does when $\nu_1([-\pi, \pi])=\nu_2([-\pi, \pi])$, (required to show that $\{B∈\mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)\}$ is a Dynkin system), which is true in my case, since $\hat{ν_1}=\hat{ν_2}$.

deque
  • 551

1 Answers1

1

I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence $$g_n(x) := \begin{cases} 1 & \text{ if } x \in [a,b] \\ nx + (1-na) &\text{ if } x \in [a-1/n,a] \\ -nx +(1+nb) & \text{ if } x \in [b,b+1/n] \\ 0 & \text{otherwise} \end{cases}.$$ By the dominated convergence theorem (for signed measures) we get $$\nu_1([a,b]) =\nu_2([a,b]).$$ Thus, both measures are equal on a $\cap$-stable generator of the Borel-$\sigma$-algebra. Now, you need to extend the uniqueness theorem for $\sigma$-finite measures. In fact, the same proof applies. Note that $$\mathcal{C}=\{B\in\mathcal{B}_{[-\pi, \pi]} : \nu_1(B)=\nu_2(B)\}$$ is a $\pi$-system containing a $\cap$-stable generator with a sequence $E_n \in \mathcal{C}$ such that $E_n \uparrow \Omega$.

For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $\mu = \mu_1 - \mu_2$ with measures $\mu_1$ and $\mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)

p4sch
  • 7,655
  • 11
  • 26
  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $\mathcal{C}$ is stable under intersection.
    I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it     – deque Dec 05 '18 at 08:43
  • In the uniqueness theorem for $\sigma$-finite measures you need that there exists a sequence $(E_n)$ in $\mathcal{C}$ with $E_n \uparrow \Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $\nu_1(\Omega) = \nu_2(\Omega)$. – p4sch Dec 05 '18 at 08:50
  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)?
    Because that was the first approach I tried, but as I said, I couldn't prove that $\mathcal{C}$ is closed under intersections.     – deque Dec 05 '18 at 08:58
  • 1
    I mean this theorem. I called it 'uniqueness theorem for $\sigma$-finite measures'. – p4sch Dec 05 '18 at 09:12
  • This theorem still requires to show that $\mathcal{C}$ is closed under intersections though :/
    Can you help me with that please ?     – deque Dec 05 '18 at 11:47
  • No, the proof shows that $\mathcal{C}$ is a Dynkin system, containing a $\cap$-stable generator. You don't need to show that the whole set-system $\mathcal{C}$ is closed unter intersections. (We need to use that a Dynkin system with generator closed under intersection is a $\sigma$-Algebra. The proof can be found in the mentioned link.) – p4sch Dec 05 '18 at 11:52
  • Oh I get you ! So, $\mathcal{G}={ [a;b] : -\pi\leq a<b\leq \pi}\cup { \varnothing }$ is our $\cap$-stable generator, and $E_n = [a-1/n, b-1/n]$ for example ? – deque Dec 05 '18 at 12:10
  • Yes, you can also take $E_n = [-\pi,\pi]$, because we are in a finite setting. (Both measures are finite. Taking $f=1$, we also see that $\nu_1(\Omega) = \nu_2(\Omega)$ as already mentioned above.) – p4sch Dec 05 '18 at 12:15