# Relation between setwise and weak convergence of measures

Let $$E$$ be a metric space. We say that a sequence $$(\mu_n)_{n\in\mathbb N}$$ of finite measures on $$(E,\mathcal B(E))$$ converges weakly to a finite measure $$\mu$$ on $$(E,\mathcal B(E))$$ if $$\int f\:{\rm d}\mu_n\xrightarrow{n\to\infty}\int f\:{\rm d}\mu\tag1\;\;\;\text{for all }f\in C_b(E).$$ Now consider the condition of setwise convergence: $$\mu_n(B)\xrightarrow{n\to\infty}\mu(B)\;\;\;\text{for all }B\in\mathcal B(E)\tag2.$$

Which relation holds between $$(1)$$ and $$(2)$$? Does one imply the other?

For example, $$(2)$$ clearly implies $$(1)$$ for all $$\mathcal E$$-measurable $$f$$ with $$|f(E)|\in\mathbb N$$. But can we say more?

• eventuallyalmosteverywhere.wordpress.com/2013/01/10/…
– user140541
Commented Oct 27, 2020 at 9:31
• @d.k.o. So, since $(2)$ in my question does clearly imply (e) in the Portmanteau Lemma, we should be able to conclude $(1)$ from $(2)$, right? The restriction to probability measures shouldn't be a restriction, since we can simply normalize $\mu_n,\mu$ here. Do we need to assume that the measures are "Borel" (in whatever sense the term is used there)? Commented Oct 27, 2020 at 10:15
• A Borel measure is any measure on $\mathcal{B}(E)$. Also (2) implies (1) but not vice versa.
– user140541
Commented Oct 27, 2020 at 11:36
• (1) implies the convergence of (2) on all sets B such that their boundary $\partial B$ has $\mu$-measure 0. Which may or may not exhaust all the sets in the Borel sigma algebra. Commented Jul 6 at 11:09

It is clear that $$(2)\implies(1)$$, even without using the portmanteau theorem; let $$f$$ be any bounded measurable function; then for $$\varepsilon>0$$ we can approximate $$f$$ uniformly (in a tolerance of $$\varepsilon$$) by a simple function $$\phi=\sum_{j=1}^N\lambda_i\chi_{E_i}$$, where $$E_i$$ are pairwise disjoint Borel sets. Then by triangle inequality $$|\int fd\mu_n-\int fd\mu|\le2\varepsilon+|\int\phi d\mu_n-\int\phi d\mu|\le2\varepsilon+\sum_{j=1}^N|\lambda_j|\cdot|\mu_n(E_j)-\mu(E_j)|$$ so for large $$n$$ we have $$|\int fd\mu_n-\int fd\mu|<3\varepsilon$$, i.e. that $$\int fd\mu_n\to\int fd\mu$$; this is true for all bounded measurable functions, not just continuous bounded ones. So you can see that $$(2)$$ is much stronger than $$(1)$$.