Approximation of Dirac measure by absolutely continuous measures (w.r.t. Lebesgue measure) Let $\mathcal{P}$ be the set of probability measures on the Borel $\sigma$ Algebra on the unit circle $S^1$.
Let $a\in S^1$ and $\delta_a$ the Diracmeasure in $a$.
I want to find a sequence $(\mu_n)_{n\in\mathbb{N}}\subset\mathcal{P}$ with $\mu_n\ll\lambda$ for all $n$, such that
$\mu_n\rightarrow\delta_a$, in the sense that $$\int_{S^1}fd\mu_n\rightarrow\int_{S^1}fd\delta_a$$ for every continuous $f$.
My thoughts so far: Since $\int_{S^1}fd\delta_a= f(a)$, I want to find a sequence of functions $(g_n)_{n\in\mathbb{N}}$, such that
$$\int_{S^1}fg_nd\lambda\rightarrow f(a)$$ for every continuous $f$.
Then I can define $\mu_n$ as $$\mu_n(E)=\int_E g_nd\lambda,$$ which satisfies $\mu_n\ll\lambda$ and
$$\int_{S^1}fd\mu_n=\int_{S^1}fg_nd\lambda\rightarrow f(a)=\int_{S^1}fd\delta_a.$$
I just have trouble finding such a sequence $(g_n)$. I would appreciate any help on that. Or maybe my approach isn't even the right way to go about this problem? Thanks in advance!
 A: Let $a \in S^1$ be arbitrary point, and let $\delta_a$ be a probability measure on $(S^1,\mathcal B(S^1))$ (dirac measure).
You want to find sequence $(\mu_n)$ of probability measures on $(S^1,\mathcal B(S^1))$ that are absolutelly continuous (w.r.t Leb measure on $S^1$) and converge weakly (by probabilistic definition) to $\delta_a$ measure (because every continuous function on $S^1$ is bounded, by compactness). In other words, you want to find sequence $(X_n)$ of random variables, such that $\mu_n$ is distribution of $X_n$ and $X_n \Rightarrow X$ in distribution, where $X \sim \delta_a$.
We'll use two facts:

*

*If $Y,Z$ are independent and $Y$ is absolutelly continuous w.r.t Leb measure, then $Y+Z$ is absolutelly continuous.


*If $Y_n \to Y$ almost surely, then $Y_n \Rightarrow Y$ in distribution.
Let $Z$ be Uniformly distributed on $S^1$ and independent of $X \sim \delta_a$. Take $X_n = X + \frac{1}{n}Z$. Then by fact (1), we know that $X_n$ is absolutelly continuous. Moreover, since $||X_n - X|| = \frac{1}{n}||Z|| = \frac{1}{n} \to 0$, then $X_n \to X$ almost surely, so by fact (2), $X_n \to X$ in distribution. Hence $\mu_n \sim X + \frac{1}{n}Z$ is such a sequence of probability measures that are absolutelly continuous w.r.t Leb measure, that converges in the weak sense to $\delta_a$.
In other words, $\mu_n = \delta_a * \nu_n$, where $\nu_n$ is a distribution of $\frac{1}{n}Z$ which can be viewed as uniform distribution on $S^1(0,\frac{1}{n}) := \{ x \in \mathbb R^2 : ||x|| = \frac{1}{n} \}$
