Let $(\mu_n)$ be the sequence of probability measures on $[0,1]$, defined by $$\mu_n(\{(i − 1)/n\}) = 1/n, i = 1, . . . , n$$ that is, $(\mu_n)$ is the uniform distribution on $[0,1]$ supported on n points.
How to compute the weak limit, if it exists, of $(\mu_n)$?
On
Weak convergence is equivalent to convergence in distribution. Therefore you can simply construct the CDF of r.v. $X_n$ distributed as $\mathbb P(X_n=(i-1)/n)=1/n$, $i=1,\ldots,n$ and look at pointwise limiting distribution. $$ F_{X_n}(x)=\mathbb P(X_n\leqslant x) = \begin{cases} 0, & x<0\cr \frac1n, & 0\leqslant x < \frac1n \cr \frac2n, & \frac1n\leqslant x < \frac2n \cr \ldots & \cr \frac{i}{n} & \frac{i-1}n\leqslant x < \frac{i}n \cr \ldots & \cr \frac{n-1}{n} & \frac{n-2}n\leqslant x < \frac{n-1}n \cr 1 & x\geqslant \frac{n-1}{n} \end{cases} $$ Draw a graph of this function and see that it's a ladder along the diagonal $F(x)=x$. The greatest distance $\sup_x|F_n(x)-F(x)|$ between this ladder and the diagonal is $1/n$ and is reached at points $(i-1)/n$, $i=1,\ldots,n$. So, $F_n(x)\to F(x)$ not only pointwise but uniformly in $x$. The limiting distribution is uniform, the limiting measure in Lebesgue measure on $[0,1]$.
You can think of the set $\{0,\frac{1}{n},\ldots,\frac{n-1}{n}\}$ as a subdivision of the interval $[0,1]$. Now, if $f$ is continuous on $[0,1]$ you will get $\int_{[0,1]}f d \mu_n=\frac{1}{n}\sum_{j=1}^{n}f(\frac{j-1}{n})$ which you may recognize as the Riemann sum for $f$. So, in the limit you will get the Riemann integral, and that is the limit measure.