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?


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.

  • $\begingroup$ Thanks for your answer. Can you please provide a counter-example why it will hold if the sequence of measures is decreasing? $\endgroup$ – Colin Nov 25 '11 at 21:39
  • $\begingroup$ 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?... $\endgroup$ – Bruno Joyal Nov 26 '11 at 0:27

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 $$ \mu(\mathbb{N})=\lim_{n\to\infty}\mu_n(\mathbb{N})=\infty $$

  • $\begingroup$ I think that this assumes $\mu_1(E)<\infty$ when $\mu_n$ is decreasing. I will see if I can lift that assumption. $\endgroup$ – robjohn Nov 25 '11 at 21:52
  • $\begingroup$ Aren't you secretly appealing to the monotone convergence theorem when switching the order of summation in (4)? $\endgroup$ – Bruno Joyal Nov 28 '11 at 17:22
  • $\begingroup$ @Bruno: all of the terms are the same sign, so we are allowed to rearrange the terms in any way, including changing the order of summation. I don't believe that monotone convergence is needed for that, but perhaps I am missing something. $\endgroup$ – robjohn Nov 28 '11 at 17:31
  • $\begingroup$ I`m not questioning whether it's true or not (it is), but I'm just saying it's a particular case of the monotone convergence theorem. :) $\endgroup$ – Bruno Joyal Nov 28 '11 at 17:36
  • $\begingroup$ @robjohn For the increasing case, isn't it possible that $a_{j,k}$ is not well-defined? More specifically, couldn't we have that $\mu_{j+1}(E_k)=\mu_{j}(E_k)=\infty$ for some $j$, in which case $a_{j,k}=\infty-\infty$ isn't defined? $\endgroup$ – Blue Marker Feb 3 '15 at 4:11

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