Example of a sequence of measures that does not converge to a measure despite converging "pointwise" So, I am trying to solve the following problem: given a sequence of measures $(\mu_n)_{n \in \mathbb{N}}$ on $(X, A)$ such that for each $B \in A$ the limit $\lim_{n \to \infty}  \mu_n(A) \in [0, \infty]$ exists, we define $\mu(A) = \lim_{n \to \infty}  \mu_n(A)$. I want to find an example where $\mu$ is not a measure on $A$. Clearly $\mu(B) \geq 0 \; \forall B$ and $\mu(\emptyset) = 0$. So the only thing that should break is the countable additivity.
I have already proven that if the sequence of measures $(\mu_n)_{n \in \mathbb{N}}$ is monotonically increasing, i.e. $\mu_n(B) \leq \mu_{n+1}(B) \; \forall B \in A$ and $\forall n \in \mathbb{N}$, $\mu$ will be a measure. I also see that $\sum_j \mu(B_j) \leq \mu(\cup_j B_j)$ for any such $\mu$ - so, for $\mu$ to not be a measure I want to create an example where $\sum_j \mu(B_j) < \mu(\cup_j B_j)$ with the strict inequality for some $(B_j)$.
The following sequence of measures almost works on $\mathbb{N}$: $\mu_i(B) = n_i(B) / i$, where $n_i(B)$ is the count of natural numbers less than $i$ in $B$. For every finite set of natural numbers $C$ we have $\lim_i\mu_i(C) = 0$, but $\lim_i \mu_i (\mathbb{N}) = 1$, so countable additivity breaks. However, such sequence of measures does not converge "pointwise" for all subsets of natural numbers, and I fail to see what $\sigma$-algebra I need to consider for it to converge on all sets. Could someone please help me, either with my example (if it is even fixable), or with any other example? Thank you very much!
 A: The Saks-Vitali-Hahn theorem states that

(See Yosida's functional analysis) Let $(X,\mathscr{B},\nu)$ be a $\sigma$-finte measure space, and $(\mu_n)$ a sequence of complex measures (which have  finite total variation (i.e. $|\mu_n|(X)<\infty$)). If $\mu_n\ll \nu$ for all $n$ and $\lim_n\mu_n(B)=\mu(B)$ exists for any $B\in\mathscr{B}$, then $\mu$ is a $\sigma$-aditive and $\mu\ll \nu$.

This in particular implies that

(Doob's version) If $\mu_n$ are measures in a measurable space $(X,\mathscr{B})$ that converge as set-functions to an extended real valued function $\nu$, then $\nu$ is a measure if $\nu(X)<\infty$

These results imply that for an example of the type the OP, one should look for infinite measures $\mu_n$ and/or measures that are not $\sigma$-finite.

Here is an example. Consider $X=[0,\infty)$ equipped with power $\sigma$-algebra $\mathcal{P}$ and the counting measure $\mu$.
Let $\mu_n(\cdot)=\mu(\cdot \cap (n,\infty))$. If $A$ is bounded, then $\mu_n(A)\xrightarrow{n\rightarrow\infty}0$. If $A$ is unbounded, then $A\cap(n,\infty)$ is infinite for all $n$ and so, $\mu_n(A)=\infty\xrightarrow{n\rightarrow\infty}\infty$.
Thus $\mu_n$ converges to a set function $\mu$ such that $\mu(A)=0$ if $A$ is bounded and $\mu(A)=\infty$ other wise.
Notice that $\mu(\{n\})=0$ for each $n\in\mathbb{N}$ and $\mu(\mathbb{N})=\infty$.
