the Portmanteau theorem
/ port-MAN-toh /
Once you define weak convergence by integral f dmu_n -> integral f dmu for all bounded continuous f, you immediately want other ways to check it, in terms of sets, in terms of liminf and limsup, in terms of smaller test-function classes. The Portmanteau theorem (the name is a French word for a suitcase that holds many things, fittingly, since it bundles many equivalent statements into one) collects these equivalent characterizations. It is the workhorse you reach for whenever you must verify or exploit weak convergence in practice.
For probability measures mu_n, mu on a metric space S, the following are equivalent: (1) mu_n => mu, i.e. integral f dmu_n -> integral f dmu for all f in C_b(S); (2) integral f dmu_n -> integral f dmu for all bounded uniformly continuous f; (3) limsup_n mu_n(F) <= mu(F) for every closed set F; (4) liminf_n mu_n(G) >= mu(G) for every open set G; (5) mu_n(A) -> mu(A) for every Borel set A that is a continuity set of mu, meaning mu(boundary of A) = 0. The closed-set and open-set statements are dual via complements, and the continuity-set statement is the bridge to the CDF definition: on the line, (-infinity, x] is a continuity set exactly when x is a continuity point of the limiting CDF.
The asymmetry in (3) and (4), inequalities rather than equalities, is the whole subtlety. Mass can leak onto the boundary of a closed set in the limit (limsup can exceed nothing, so we only get <=), and mass can escape from an open set (we only get >=). Equality is recovered precisely on continuity sets, where there is no boundary mass to worry about. This is why the line definition demands continuity points of F: it is the one-dimensional shadow of (5). The theorem is the conceptual core of the whole field, and almost every later result is proved by checking one of its clauses.
Let mu_n = delta_{1/n}, the point mass at 1/n, and mu = delta_0. Take the open set G = (0, 1). Then mu_n(G) = 1 for all n but mu(G) = 0, so liminf mu_n(G) = 1 >= 0 = mu(G), consistent with clause (4) but with strict inequality, because 0 is on the boundary of G. For the continuity set A = [0, 1/2] with mu(boundary) = mu({0, 1/2}) = 0, we do get mu_n(A) -> 1 = mu(A).
Strict inequality on an open set whose boundary carries limit mass, and exact convergence on a continuity set.
A common slip is to expect mu_n(A) -> mu(A) for ALL Borel A. It holds only for continuity sets. The point mass example shows mu_n((0,1)) = 1 never converges to mu((0,1)) = 0.