Let a $\in \mathbb{R}$, $\phi_n : \mathbb{R} \rightarrow \mathbb{R}_+$, $\phi_n (x) : = \frac{n}{\sqrt{2 \pi}} e^{\frac{-n^2 x^2}{2}}$, $n \geq 1$ and let $\mu_n (d x) : = \phi_n (x - a) \lambda (d x)$, where $\lambda$ is the Lebesgue measure. What does it mean $\mu_n (dx)$ and $\lambda (dx)$?
How can I show that for every bounded borelian function $\phi : \mathbb{R} \rightarrow \mathbb{R}$, $\underset{n}{lim} \underset{\mathbb{R}}\int \phi (x) \mu_n(dx) = \underset{\mathbb{R}}\int \phi(x) \delta_a(dx)$, where $\delta_a (A) = 0$ if $a \notin A$ and $\delta_a(A) = 1$ if $a \in A$?
Thank you!
It means that the measure is define in the following way: $$ \mu_n(A) = \int_A \phi_n(x-a) dx $$ in short the notation for that is $d\mu_n(x) =\phi_n(x-a) \lambda(dx) $. This is because an alternative notation for $\mu_n(A)$ is $$ \int_A \mu_n(dx) $$
And for the Lebesgue measure, it means that $$ \lambda (A) = \int_A dx $$