pointwise limit of finite measures If there is a sequence of measures $\mu_n$ such that $\mu_n(A) \overset{n}{\rightarrow} \mu(A)$ for all $A$ in the $\sigma$-field and if $\mu_n(\Omega)\leq c$ $(c<\infty)$ for all $n$, then $\mu$ is a measure.
(problem 7, section 1.5 in Robert Ash)
The above assertion is proved by the Vitali-Hahn-Saks theorem.
What I would like to know is a counter example for the case when the condition,


*

*$\mu_n(\Omega) \leq c$ for all $n$


is not satisfied.

Take $\mu_n: \mathbb{N} \to \mathbb{R}$ to be
$$
  \mu_n(A) = \#(A \cap [n,\infty)).
$$
For any bounded set $B$, you have that $\mu_n(B) \rightarrow 0$. But, for any unbounded set $U$, $\mu_n(U) = \infty \rightarrow \infty$.
The limit $\mu$ is not a measure, because
$$
  \mu(\mathbb{N})
  = \infty
  \neq 0
  = \sum_{n \in \mathbb{N}} \mu(\{n\}).
$$
This example is quite easy to extend to $\mu_n(A) = \lambda(A \setminus K_n)$, where $K_n$ is a sequence of sets such that $K_n \uparrow K$, with $\infty \neq \lambda(K_n) \rightarrow \lambda(K) = \infty$.
In this case, $\mu(K) = \infty$, while $\mu(K_n) = 0$.
