Portmanteau theorem

Remarks

Condition 1 implies 2 at once. For 2 implies 3, the Lipschitz functions $f_k(x) = \max(0, 1 - k \, d(x, F))$ decrease to the indicator of $F$ as $k \to \infty$. Conditions 3 and 4 are equivalent by taking complements. Conditions 3 and 4 applied to the closure and the interior of $A$ give 5. For 5 implies 1, write $\int f \, d\mu$ as an integral of $\mu(\set{f > t})$ over $t$; for all but countably many $t$ the set $\set{f > t}$ has a $\mu$-null boundary.

Example: $\delta_{1/n}$ converges weakly to $\delta_0$ on $\R$. For the open set $G = (0, 1)$, one has $\delta_{1/n}(G) = 1$ for $n \ge 2$ and $\delta_0(G) = 0$, so the inequality in 4 can be strict. For $A = \set{0}$, the boundary is $\set{0}$, which has $\delta_0$-measure $1$. So condition 5 makes no claim, and indeed $\delta_{1/n}(A) = 0$ does not tend to $\delta_0(A) = 1$.

Going deeper

For a student

One uses whichever form is easiest to check. Condition 2 is handy because Lipschitz functions are easy to control, while condition 5 gives convergence of probabilities of concrete events. For real random variables, condition 5 applied to half-lines gives convergence of distribution functions at continuity points. The theorem is a standard tool together with Prokhorov's theorem and the continuous mapping theorem.