Pointwise cdf convergence at a jump is not required

On $\mathrm{\Omega }=\{\ast \}$ take $\mathcal{F}=\{\mathrm{\varnothing },\mathrm{\Omega }\}$ and $\mathbb{P}(\mathrm{\Omega })=1$, $\mathbb{P}(\mathrm{\varnothing })=0$. This is a probability space: in any disjoint family at most one event is nonempty. For every integer $n\ge 1$ set $X_n(\ast )=1\mathrm{/}n$ and $X(\ast )=0$. Each map is measurable because every Borel preimage is either $\mathrm{\varnothing }$ or $\mathrm{\Omega }$. Its law is respectively ${\mu }_n={\delta }_{1\mathrm{/}n}$ or $\mu ={\delta }_0$: a Borel set has probability one exactly when it contains the specified value. Define $F_n=F_{X_n}$ and $F=F_X$ using F2.

For a point mass, $\int fd{\delta }_a=f(a)$ for bounded measurable $f$: this holds for simple functions by the definition of their integral, and then for nonnegative bounded functions by increasing simple approximation, and for real bounded functions by their positive and negative parts. For bounded continuous $f$, continuity at zero therefore gives $\int fd{\mu }_n=f(1\mathrm{/}n)\to f(0)=\int fd\mu$. By F1 the laws converge weakly.

By F2 and step 1.1, $F_n(t)={\mathbf{1}}_{\{t\ge 1\mathrm{/}n\}}$ and $F(t)={\mathbf{1}}_{\{t\ge 0\}}$. Thus $F_n(0)=0$ for every $n\ge 1$ while $F(0)=1$. Moreover $F(t)=0$ for every $t<0$, so $F$ has a jump from its left limit zero to its value one at zero. Hence weak convergence does not force CDF convergence at this jump.