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Let $X$ be a set and $\mathcal{A}$, a sigma algebra of subsets of $X$. Let $\{\mu_n\}$ be a sequence of measures of $\mathcal{A}$ such that $\mu_{n+1}(E)\geqslant \mu_n(E)$ for every $E\in \mathcal{A}$. Let $\lim_{n\rightarrow \infty} \mu_n(E) = \mu(E)$. I want to show that $\mu$ is a measure on $\mathcal{A}$.
Would the conclusion still be true if $\mu_{n+1}(E)\leqslant \mu_n(E).$

My Attempt:

Let $E_n$ be a sequence of pairwise disjoint sets in $\mathcal{A}$. Then, I have to show $\mu\left(\cup_n E_n\right)=\sum_n \mu(E_n),$ where the $E_n$ are pairwise disjoint.
Let $E=\cup_n E_n$. Then $\mu(E)=\sum_{n=1}^{\infty}\mu(E_n)$. For one inclusion, we have that $\begin{align*} \mu(E) & \geqslant \lim_{k\rightarrow \infty}\sum_{n=1}^N \mu_k(E_n)\\ & =\sum_{n=1}^N\mu(E_n). \\ \end{align*} $

But this holds true for any $N$. Thus $\mu(E)\geqslant \sum_{n=1}^\infty \mu(E_n)$.

Is the above inclusion okay? I can't can up with a way to tackle the other inclusion. Perhaps someone will be kind enough to give a hint?

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    @DylanMoreland: Sorry for the confusion. I've fixed it. Hope it's better now.2011-11-25

2 Answers 2

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By definition, and using that each $\mu_k$ is countably additive, we have

$\mu(E) = \lim_{k \to \infty}\mu_k(E) = \lim_{k \to \infty}\sum_{n \geq 1}\mu_k(E_n).$

By the monotone convergence theorem, since each sequence $\{\mu_k(E_n)\}_{k}$ is increasing, this equals

$\lim_{k \to \infty}\sum_{n \geq 1}\mu_k(E_n) = \sum_{n \geq 1}\lim_{k \to \infty}\mu_k(E_n) = \sum_{n \geq 1}\mu(E_n).$

It would not hold true in general if the sequence of measures is decreasing.

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    I'll give you a hint. Denote the Lebesgue measure on the real line and put $\mu_k(S)=m(S-[-k,k])$. Now what if you cover the real line with a countable number of disjoint, bounded measurable sets?...2011-11-26
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Define $ a_{j,k}=\mu_{j+1}(E_k)-\mu_j(E_k)\tag{1} $ From the conditions above, all $a_{j,k}$ have the same sign.

Summing $(1)$ yields $ \mu_n(E_k)=\mu_1(E_k)+\sum_{j=1}^{n-1}a_{j,k}\tag{2} $ So we have $ \begin{align} \mu(E_k) &=\lim_{n\to\infty}\mu_n(E_k)\\ &=\mu_1(E_k)+\sum_{j=1}^\infty a_{j,k}\tag{3} \end{align} $ Since each $\mu_j$ is countably-additive, we can add $(2)$ yieldiing $ \begin{align} \mu_n(E) &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{k=1}^\infty\sum_{j=1}^{n-1}a_{j,k}\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{j=1}^{n-1}\sum_{k=1}^\infty a_{j,k}\tag{4} \end{align} $ Taking the limit of $(4)$, changing the order of summation (all terms have the same sign), and then applying $(3)$, we get $ \begin{align} \mu(E) &=\lim_{n\to\infty}\mu_n(E)\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{j=1}^\infty\sum_{k=1}^\infty a_{j,k}\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}\\ &=\sum_{k=1}^\infty\left(\mu_1(E_k)+\sum_{j=1}^\infty a_{j,k}\right)\\ &=\sum_{k=1}^\infty\;\mu(E_k)\tag{5} \end{align} $ Counterexample:

I commented that the argument above assumes that $\mu_1(E)<\infty$ when the $\mu_n$ are decreasing. This condition cannot be lifted.

Let $E_k=\{k\}$ and $\mu_n(\{k\})=1$ when $k>n$ and $\mu_n(\{k\})=0$ when $k\le n$. Then $ \mu(\{k\})=\lim_{n\to\infty}\mu_n(\{k\})=0 $ yet $ \mu(\mathbb{N})=\lim_{n\to\infty}\mu_n(\mathbb{N})=\infty $

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    @BlueMarker: If $\mu_{j+1}(E_k)=\mu_j(E_k)=\infty$, we can define $a_{j,k}=0$. The sum of the $a_{j,k}$ will still be infinite past the first $j$ for which $\mu_{j+1}(E_k)=\infty$, and the signs of the $a_{j,k}$ do not change.2015-02-03