Define a Borel probability measure $\mu_n $ by $\mu_n ({x}) = \frac{1}{n} $ for $x = 0, \frac{1}{n}, \frac{2}{n}, ..., 1-\frac{1}{n} $. Let $\eta$ be a Lebesgue measure on $[0,1]$. i) I'm to compute $\zeta_n(t) = \int e^{itx}\mu _n (dx),$ the characteristics function of $\mu_n$. ii) Compute $\zeta(t) = \int e^{itx}\eta (dx),$ the characteristic function of $\eta$.
This was how i did the first one and was wondering whether it was correct or not.
i) $\zeta_n(t) = \int e^{itx}\mu _n (dx) = \frac{1}{n}\int_{0}^{1-\frac{1}{n}} e^{itx}\mu _n (dx) = \frac{i}{nt} (1-e^\frac{it(n-1)}{n}). $
ii) $\zeta(t) = \int_{0}^{1} e^{itx}\eta (dx) = \frac{1}{n}\int_{0}^{1} e^{itx}\eta (dx) = \frac{i}{t} (1-e^{it}). $
i) By definition of $\mu_n$, if $f\colon [0,1]\to\mathbf C$ is a measurable function, then $$\int f(x)\mathrm d\mu_n=\frac 1n\sum_{j=0}^{n-1}f\left(\frac jn\right),$$ therefore, with the choice $f(x)=e^{itx}$,
$$\zeta_n(t)=\frac 1n\sum_{j=0}^{n-1}e^{ijt /n},$$ which can be computed using the formula for a geometric series if $e^{it /n}\neq 1$.
ii) You have to compute (for $t\neq 0$) the integral $$\zeta(t) =\int_0^1 e^{itx}\mathrm dx= \left[\frac{e^{itx}} {it} \right]_0^1. $$