# Convergence of measures on $\mathbb{T}$

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.

• I think you want norm convergence, not uniform convergence. – zhw. Dec 5 '15 at 18:43
• Hmm, what is your definition of norm convergence? – Milen Ivanov Dec 5 '15 at 18:55
• @zhw. Arent they equivalent? – TheOscillator Dec 5 '15 at 21:31
• 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) – Milen Ivanov Dec 5 '15 at 21:55