I want to prove that, in $(\mathbb{R},B(\mathbb{R}))$, we have that $\frac{1}{n}\sum_{i=1}^{n}\delta_{\frac{i}{n}}$ converges to $U_{[0,1]}$.
We need to prove, by definition, that $\lim_{n \to +\infty}\int f \, d(\frac{1}{n}\sum_{i=1}^{n}\delta_{\frac{i}{n}})=\int fdU_{[0,1]}, \forall f: \mathbb{R} \to \mathbb{R}$ continuous and bounded.
Let $f: \mathbb{R} \to \mathbb{R}$ continuous and bounded.
Is it correct to say that $$ \lim_{n \to +\infty}\int f \, d\left(\frac{1}{n} \sum_{i=1}^n \delta_{\frac{i}{n}}\right)=\lim_{n \to +\infty}\int f \, d\left(\sum_{i=1}^{n}\frac{1}{n} \delta_{\frac{i}{n}}\right) =\lim_{n \to +\infty}\sum_{i=1}^{n}\int f \, d\left(\frac{1}{n} \delta_{i/n}\right) \\=\int f \, dU_{[0,1]} \text{ ?} $$
If yes, how to detail these calculations?
The crucial step (and the unique one which needs to be detailed) is to notice that $$\int fd\left(\frac 1n\delta_{i/n}\right)=\frac 1n\cdot f\left(\frac in\right),$$ and to notice that for each $f\colon [0,1]\to\mathbb R$ continuous, $$\lim_{n\to+\infty}\frac 1n\sum_{i=1}^nf\left(\frac in\right)=\int_0^1f(x)\mathrm dx.$$ This can be seen using uniform continuity of $f$ and approximating $f(i/n)$ by $\int_{i/n}^{(i+1)/n}f(x)\mathrm dx$.