Denote by $M(\mathbb T)$ the set of complex-values measures on the circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$.Prove that $D(T)$, the set of discrete measures on $\mathbb{T}$ is:

  • closed in $M(\mathbb{T})$ under norm convergence.
  • dense in $M(\mathbb{T})$ under weak convergence.

I had problem formulating both of these questions:

About the norm convergence, suppose we take a sequence $\{\mu_n\}_n\subset D(\mathbb T)$, we can assume it convergences to infintite sum of delta measures (otherwise the sum is finite and the claim is trivial) $$\mu_n\to\mu=\sum_{k=1}^\infty c_k\delta_{x_k},\quad x_k\in\mathbb{T}.$$ How can I formulate that $$\Vert\mu -\mu_n\Vert =\Vert \sum_{k=1}^\infty c_k\delta {x_k} -\sum_{j=1}^n c_j\delta_{x_j}\Vert \to 0?$$

About the second point: Let $\nu\in M(\mathbb{T})$. I thought taking a sequence of discrete measures $$\mu_n=\sum_{j=1}^n d_j\delta_{e^{\frac{2\pi i jt}{n}}}$$ but I don't know how to choose the coefficients exactly such that $\mu_n\to\nu$ for this $\nu\in M(\mathbb T)$?

How do I prove the convergence in norm and find the coefficients ?


For the first part: Let $\mu_i = \sum_j a_j \delta_{x_{ij}}$ Then consider the measurable set $A = \mathbb{T} - \{x_{ij}\}$ By definition of uniform convergence, fo$$ $B \in A$ is measurable, then by the definition of uniform convergence there exists $N s.t \forall n > N $

$$|\mu_n(B) - \mu(B)| < \epsilon$$ But because $B \in A \mu_n(B) = 0$, hence $\mu$ is $0$ except on a countable subset of $\mathbb{T}$ and the result follows. Note that we didn't need to find the coefficients, as the proof just becomes messier.

For the second part: I couldn't if $mu = \nu_1 + \nu_2$ where $\nu_1$ is the absolutely continuous part of $\mu$ and $\nu_2$ is the singular part of $\mu$ with respect to the Lebesgue measure, then we can use the Radon-Nikodym theorem to conclude $\nu_1(E) = \int_E f(x)dx$ for some $f$ measurable and finite a.e. By a famous theorem in measure theory, $f$ can be uniforlmy (!) approximated by finite linear combinations of elementary functions $f_k$. Given $E$ : measurable $$\nu_1(E) = \int_E f(x)dx = lim_{k \to \infty} \int_E f_k(x)dx$$ For each $f_k$ we can easily find a corresponding discrete measure which coincides with $f_k$ so this part is done (note the choice of $f_k$ can be made independent of the set $E$ Also note that the interchanging of limit and integral is justified as the convergence is uniform: say by Dominated convergence theorem).

We are left with $\nu_2$ which is zero outside a set of Lebesgue measure 0. I'm not sure how to proceed if it's not a discrete set: I will update the post if I find a solution.

  • $\begingroup$ I think you want norm convergence, not uniform convergence. $\endgroup$ – zhw. Dec 5 '15 at 18:43
  • $\begingroup$ Hmm, what is your definition of norm convergence? $\endgroup$ – Milen Ivanov Dec 5 '15 at 18:55
  • $\begingroup$ @zhw. Arent they equivalent? $\endgroup$ – TheOscillator Dec 5 '15 at 21:31
  • $\begingroup$ He may have a definition of the type: consider the measures as a bounded linear functional on $C_0(\mathbb{T})$, the compactly supported continuous functions on the torus. They form a normed space with respect to ... norm (I have seen only the uniform norm on measures here, but who knows) $\endgroup$ – Milen Ivanov Dec 5 '15 at 21:55

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