weak convergence of discrete uniform to lebesgue measure using characteristic functions

Question

Here is the question.

Suppose we have a sequence of probability measures $\{\mu_n\}$ defined \begin{align*} \mu_n(\{x\}) = \frac{1}{n} \end{align*} for $x = 0,\frac{1}{n}, \frac{2}{n},...,\frac{n-1}{n}$. Show that $\mu_n \Rightarrow \lambda$ where $\lambda$ is the Lebesgue measure on $[0,1]$.

By the continuity theorem we have the equivalence between weak convergence and pointwise convergence of characteristic functions. So it suffices to show that the characteristic functions $\{\phi_n\}$ of the sequence of measures $\{\mu_n\}$ converges pointwise to the characteristic function of $\lambda$. That is, I want to show that for any $t \in [0,1]$, \begin{align*} \phi_n(t) = \int e^{itx} \mu_n(dx) \to \int e^{itx} \lambda(dx) = \phi(t) \end{align*} First I computed the characteristic function of $\mu_n$, \begin{align*} \phi_n(t) = \int e^{itx} \mu_n (dx) = \sum_{k=0}^{n-1} \frac{1}{n} e^{it\frac{k}{n}} \end{align*} And the characteristic function of $\lambda$, \begin{align*} \phi(t) = \int e^{itx} \lambda(dx) = \int_0^1 e^{itx} dx = \frac{e^{it} - 1}{it} \end{align*} I am stuck here. I can't show that \begin{align*} \lim_n \sum_{k=0}^{n-1} \frac{1}{n} e^{it\frac{k}{n}} = \frac{e^{it} - 1}{it} \end{align*} but I computed some values in Matlab and I'm fairly certain its true. How can I evaluate this limit / show it is equal to the expression on the right?

Answer 1

Partial sum of the geometric series \begin{align}\sum_{k=0}^{n-1}\frac1n e^{it\frac{k}n}&=\frac1n\sum_{k=0}^{n-1}\left(e^{it\frac1n}\right)^k=\frac{1}{n}\cdot\frac{1-(e^{it}\frac1n)^n}{1-e^{it\frac1n}}=\frac1n\cdot\frac{1-e^{it}}{1-e^{it\frac1n}}=(1-e^{it})\cdot\frac{\frac1n}{1-e^{it\frac1n}}\end{align} and L'Hopital's rule \begin{align}\lim_{n\to\infty}\frac{\frac1n}{1-e^{it\frac1n}}=\lim_{x\to 0}\frac{x}{1-e^{itx}}\overset{\text{L'Hopital}}=\lim_{x\to 0}\frac{1}{-ite^{itx}}=\frac{1}{-it}\end{align} give $$\lim_{n\to \infty}\sum_{k=0}^{n-1}\frac1n e^{it\frac{k}n}=(1-e^{it})\cdot\frac{1}{-it}=\frac{e^{it}-1}{it}$$