If $\mu_{n+1}(E) \ge \mu_n(E)$ and Define $\mu (E):=\lim_{n \rightarrow \infty }\mu_n (E)$ prove that $\mu$ is a measure on $\beta$ Let $X$ be a set and $\beta$ be a sigma algebra of subsets of $X$. Let $\{\mu_n\}$ be a sequence of positive measure on $(X, \beta)$ such that $\mu_{n+1}(E) \ge \mu_n(E)$.  Define $\mu (E):=\lim_{n \rightarrow \infty }\mu_n (E)$ prove that $\mu$ is a measure on $\beta$. Is the conclusion still true if  $\mu_{n+1}(E) \le \mu_n(E)$?
I figured the bigger problem is to show that $\mu$ is countably additive.
This is what I did.
Let $E_i$ be a sequence of disjoint measurable sets and $E=\cup_{i=1}^{\infty}E_i$.
Want to show that 
$$\mu\left(\cup_{i=1}^{\infty}E_i\right)=\sum_{i=1}^{\infty}\mu\left(E_i\right).$$
Without loss of generality, assume $\mu_n(E_i)<\infty ,  \forall n$.
then,$$\mu(\cup_{i=1}^{\infty}E_i)=\lim_{n \rightarrow \infty}\mu_n(\cup_{i=1}^{\infty}E_i)$$
since $\mu_n$ is measurable 
$$\mu_n(\cup_{i=1}^{\infty}E_i)=\sum_{i=1}^{\infty}\mu_n(E_i)$$.
$\Rightarrow \mu(\cup_{i=1}^{\infty}E_i)=\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\mu_n(E_i)$.
Now this is where my difficulty lie, I know I cannot just write 
$$\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\mu_n(E_i)=\sum_{i=1}^{\infty}\lim_{n \rightarrow \infty}\mu_n(E_i)$$ without justification. Can Monotone convergence Theorem be used here and how? if not, can someone show me how? 
 A: In the case where $(\mu_n)_{n\in \mathbb{N}}$ is an increasing sequence of measures, $\mu = \lim\limits_{n\to \infty} \mu_n$ is a measure. If the definition of measure requires that at least one set has finite measure, then we have
$$\mu(\varnothing) = \lim_{n\to\infty} \mu_n(\varnothing) = \lim_{n\to\infty} 0 = 0,$$
so $\mu$ satisfies that condition. What remains is to show the countable additivity of $\mu$. For that, we can invoke the monotone convergence theorem for Lebesgue integrals by viewing an infinite sum as an integral with respect to the counting measure - given a sequence $(E_k)_{k\in\mathbb{N}}$ of disjoint sets in $\beta$, let $f_n(k) = \mu_n(E_k)$, then if $\zeta$ is the counting measure on $\mathbb{N}$ we have
$$\sum_{k = 0}^\infty \mu_n(E_k) = \int_{\mathbb{N}} f_n\,d\zeta,$$
and the monotone convergence theorem gives
$$\mu\biggl(\bigcup_{k = 0}^\infty E_k\biggr) = \lim_{n\to\infty} \sum_{k = 0}^\infty \mu_n(E_k) = \lim_{n\to\infty} \int_{\mathbb{N}} f_n\,d\zeta = \int_{\mathbb{N}} \lim_{n\to\infty} f_n\, d\zeta = \sum_{k = 0}^\infty \mu(E_k).$$
We can also appeal to the monotone convergence theorem for series (which is a special case of the monotone convergence theorem for Lebesgue integrals, but the series version is often proved separately before integration is treated). By monotonicity, $\lim\limits_{n\to\infty} \sum\limits_{k = 0}^\infty \mu_n(E_k)$ exists (in $[0,+\infty]$), and since $\mu_n(E_k) \leqslant \mu(E_k)$ for every $k$ and $n$, we have
$$\lim_{n\to\infty} \sum_{k = 0}^\infty \mu_n(E_k) \leqslant \sum_{k = 0}^\infty \mu(E_k).\tag{$\ast$}$$
But for every fixed $K$, we have
$$\sum_{k = 0}^K \mu(E_k) = \lim_{n\to\infty} \sum_{k = 0}^K \mu_n(E_k) \leqslant \lim_{n\to\infty} \mu_n(E_k),$$
and so
$$\sum_{k = 0}^\infty \mu(E_k) = \lim_{K \to \infty} \sum_{k = 0}^K \mu(E_k) \leqslant \lim_{n\to\infty} \sum_{k = 0}^\infty \mu_n(E_k),$$
which together with $(\ast)$ shows the desired equality.
Things are different if $(\mu_n)_{n\in \mathbb{N}}$ is a decreasing sequence of measures. Then $\mu = \lim \mu_n$ need not be a measure. Consider the measures on $\mathbb{N}$ given by
$$\mu_n(A) = \sum_{k \in A} a_k^{(n)},$$
where
$$a_k^{(n)} = \begin{cases} \frac{1}{(k+1)^2} &, k \leqslant n \\ \frac{1}{k+1} &, n < k.\end{cases}$$
Then $\mu_n(\mathbb{N}) = +\infty$ for all $n$, but
$$\sum_{k = 0}^\infty \mu(\{k\}) = \sum_{k = 0}^\infty \frac{1}{(k+1)^2} = \frac{\pi^2}{6} < +\infty = \mu(\mathbb{N}).$$
The failure is analogous to the possible failure of continuity from above of a measure while continuity from below always holds. If $A_n \supset A_{n+1}$ for all $n$, then we need not have
$$\lambda\biggl(\bigcap A_n\biggr) = \lim_{n\to\infty} \lambda(A_n),$$
but we have this if there is an $n$ with $\lambda(A_n) < +\infty$. In the same way, for a decreasing sequence $(\mu_n)$ of measures on $(X,\beta)$, the limit $\mu$ is a measure if there is an $n$ such that $\mu_n(X) < +\infty$. (This is a sufficient, but not necessary condition.)
