Suppose $\mu_j$ is a sequence of measures on $\mathbb{R}$. By the definition of weak convergence of measures, $\mu_j$ weak converges to $\mu$ means that for any bounded continuous function $f$, there holds that $$\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu.$$ My question is whether we can just check the equality $\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu$ for all smooth functions $f$ with compact support.
I come across this problem while I try to prove the following problem:
If $\phi_j$ are a sequence of smooth convex functions defined on $\mathbb{R}$ with uniformly bounded second order derivative and $\phi_j$ converges to the convex function $\phi$ in $L^\infty(\mathbb{R})$, then $\phi_j''dx$ weakly converges to $\phi''dx$ as measures.
Can this be obtained just from the fact that the equality $$\int_{\mathbb{R}}f \phi_j''dx\to\int_{\mathbb{R}}f \phi''dx$$ holds for all smooth functions $f$ with compact support?
Best Answer
If the measures are probability measures, then yes you can; it's kind of a standard exercise.
The argument I've seen goes something like this: fix $\epsilon$ and choose a smooth compactly supported cutoff function $g$ with $0 \le g \le 1$ and $\int g\,d\mu \ge 1-\epsilon$ (possible by monotone convergence); let $K$ be the support of $g$. Then by assumption $\int g\,d\mu_j \to \int g\,d\mu \ge 1-\epsilon$ so we see that $\limsup_{j \to \infty} \mu_j(K) \ge 1-\epsilon$. In other words, $\{\mu_j\}$ is tight. Hence after passing to a subsequence, $\mu_j$ converges weakly to some measure $\nu$. Now we notice that $\int f\,d\mu = \int f\,d\nu$ for all smooth compactly supported $f$, and it follows from a monotone class type argument that $\mu=\nu$. Finally use the "double subsequence" trick to conclude that the original sequence $\mu_j$ also converges weakly to $\mu$.
If you don't assume they are probability measures then this can be false; let $\mu_j$ be a point mass at $j$ and let $\mu$ be the zero measure.