Discrete uniform converges weakly to continuous uniform

Question

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?

Answer 1

By the definition of expectation:
\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 $n \to \infty$.

Note that the boundedness condition is actually not needed for this particular case because the support of the uniform distribution is a compact set (it is, of course, still essential in proving the equivalence between weak convergence and convergence in distribution for general random variables).