# Discrete uniform converges weakly to continuous uniform

I tried to work on the problem

Let $$(X_n)$$ be a sequence of random variables follows the discrete uniform distribution on $$\{0,\frac{1}{n},\frac{2}{n},\cdots,\frac{n-1}{n}\}$$. Let $$X$$ be a continuous uniform random variable on $$(0,1)$$.

I want to show that $$X_n$$ converges weakly to $$X$$. That is

$$E[f(X_n)] \rightarrow E[f(X)]$$ for all bounded and continuous function $$f$$.

I know that $$X_n$$ converges to $$X$$ in distribution, $$X_n \overset{D}{\rightarrow} X$$ and this implies the weak convergence. But how can I just use the definition from above to show this type of convergence?

\begin{align*} & E[f(X_n)] = \sum_{i = 0}^{n - 1}f(i/n)n^{-1} = \sum_{i = 0}^{n - 1}\int_{i/n}^{(i + 1)/n}f(i/n)dx, \\ & E[f(X)] = \int_0^1 f(x)dx = \sum_{i = 0}^{n - 1}\int_{i/n}^{(i + 1)/n}f(x)dx. \end{align*}
Since $$f$$ is continuous on $$\mathbb{R}$$, it is uniformly continuous on $$[0, 1]$$. That means, given $$\varepsilon > 0$$, there exists $$\delta > 0$$, such that for all $$x, y \in [0, 1]$$ and $$|x - y| < \delta$$, we have $$|f(x) - f(y)| < \varepsilon$$. Let $$N \in \mathbb{N}$$ be sufficiently large so that $$\frac{1}{N} < \delta$$, then for all $$n > N$$, $$x \in [i/n, (i + 1)/n)$$ implies that $$|x - i/n| < \frac{1}{n} < \frac{1}{N} < \delta$$, whence \begin{align} |E[f(X_n)] - E[f(X)]| \leq \sum_{i = 0}^{n - 1}\int_{i/n}^{(i + 1)/n}|f(i/n) - f(x)|dx < \sum_{i = 0}^{n - 1}\varepsilon \times \frac{1}{n} = \varepsilon. \end{align} This shows $$E[f(X_n)] \to E[f(X)]$$ as $$n \to \infty$$.