I want to show if $\mu_n=\frac{1}{n}\delta_n+(1-\frac{1}{n})\delta_0$ converges weakly using characteristic functions. I know that the characteristic function of $\delta_0$ is the constant function $1$. I also believe that $\phi_{\nu + \lambda}=\phi_\nu \phi_\lambda$ where $\phi$ always denotes the characteristic function respectively.
Firstly, I have the question if $\phi_{a\lambda}(t)=\phi_\lambda(at) \ \forall a \ $ in the real numbers? If yes, my attempt is the following:
$\phi_{\frac{1}{n}\mu_n}(t)=\phi_{\delta_n}(\frac{1}{n}t)\ \phi_{\delta_0}(t) \ \phi_{\delta_0}(\frac{-1}{n}t)= \int e^{i(\frac{1}{n})tx}d\delta_n \cdot 1 \cdot 1 = e^{i \frac{1}{n}t n}=e^{it}=\int e^{itx} d\delta_1$
which would mean that $\mu_n \to \delta_1$ weakly. Is that correct? Thanks in advance!
(PS: This is my first question on stackexchange, so I would appreciate some feedback if I did anything wrong.)
By definition of the characteristic function, we have $$ \varphi_{\mu_n}(t) = \int_{\mathbb R}e^{itx}\mu_n(\mathsf dx) = 1-\frac1n +\frac1n e^{int}. $$ It follows that $$ |\varphi_{\mu_n(t)}-1| = \frac1n|1-e^{int}|\leqslant \frac2n\stackrel{n\to\infty}\longrightarrow 0, $$ so that $\varphi_{\mu_n}\stackrel{n\to\infty}\longrightarrow1$ and hence $\mu_n\stackrel d\rightarrow \delta_0$.