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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?

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    $\begingroup$ Yes, the monotone convergence justifies it. View the series as an integral with respect to the counting measure. $\endgroup$ – Daniel Fischer Dec 3 '16 at 20:37
  • $\begingroup$ Okay, could you write it out for me. And also does it hold if $\mu_{n+1}(E)\le \mu_n(E)$ $\endgroup$ – J. Kyei Dec 4 '16 at 1:26
  • $\begingroup$ @J.Kyei In which book did you find this exercise ? $\endgroup$ – 129492 Jun 11 at 15:43
  • $\begingroup$ @129492, I'm sorry I do not know exactly from what textbook it came from, it was a question from my Real analysis class. $\endgroup$ – J. Kyei Jun 13 at 15:19
  • $\begingroup$ Thank you very much. It also was an exercise in my final test $\endgroup$ – 129492 Jun 14 at 4:34
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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.)

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