Suppose $\mu_1, \mu_2, \dots$ are measures on a measurable space $(X, \mathcal A)$ and $\mu_n(A) \uparrow$ for each $A \in \mathcal A$. Define $$\mu(A) = \lim_{n\to \infty} \mu_n(A).$$ Is $\mu$ necessarily a measure? If not, give a counterexample. What if $\mu_n(A) \downarrow$ for each $A \in \mathcal A$, and $\mu_1(X) < \infty$?

I need some help justifying the switching of the order of limits in the $\mu_n(A) \uparrow$ case. After looking through some previous posts, I have found that we can switch them if the doubly-indexed sequence is positive and non-decreasing. Admittedly I don't quite understand this, and would like to see a formal proof. Also, this problem is introduced before the convergence theorems, so I'm trying to avoid applying this.

I imagine the $\mu_n(A) \downarrow$ case is not too much different, but I can't apply the results that I have found since the doubly-indexed sequence is not necessarily non-decreasing.

Here's my solution so far, leaving out the trivial part showing $\mu(\emptyset) = 0$:

Now suppose $A_i \in \mathcal A$ for all $i \in \mathbb N$ are pairwise disjoint. To see that $\mu$ is a measure, we must check that $\mu(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i)$. Define $a_{n,m} = \sum_{i=1}^m\mu_n(A_i)$ and observe the following: \begin{align*} \mu(\cup_{i=1}^\infty A_i) &= \lim_{n\to \infty}\sum_{i=1}^\infty \mu_n(A_i) & \sum_{i=1}^\infty \mu(A_i) &= \sum_{i=1}^\infty \lim_{n\to\infty} \mu_n(A_i)\\ &= \lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^m \mu_n(A_i) & &= \lim_{m\to\infty}\lim_{n\to\infty}\sum_{i=1}^m \mu_n(A_i)\\ &=\lim_{n\to \infty}\lim_{m\to\infty} a_{n,m} & &= \lim_{m\to\infty}\lim_{n\to\infty} a_{n,m} \end{align*} So if we can show the equality of the final expressions in each column, then $\mu$ will be a measure.


If $0 \le a_{mn} \nearrow b_n$ as $m \to \infty$, then $\sum_n a_{mn} \to \sum b_n$. This is actually the monotone convergence theorem applied to the counting measure!

For $a_{mn} \searrow b_n$, you want to use the dominated convergence result. As such, any counterexample would involve $\sum_n a_{mn} = \infty$ and $\sum b_n < \infty$.

But, here is a proof of the monotone convergence theorem. Suppose $\sum_n b_n = B < \infty$. Given $\epsilon>0$, there exists $N$ such that $\sum_{n=1}^N b_n > B - \epsilon$. For each $n$, there exists $M(n)$ such that for $m>M(n)$, we have $a_{mn} > b_n - 2^{-n} \epsilon$. So for $m > M = \max_{1 \le n \le N} M(n)$, we have $$\sum_{n=1}^\infty a_{mn} \ge \sum_{n=1}^N a_{mn} > B - \epsilon - \epsilon\sum_{n=1}^\infty 2^{-n} = B - 2\epsilon.$$ And clearly $\sum_{n=1}^\infty a_{mn} \le B$.

The case $B = \infty$ will be a similar proof.

The dominated convergence version follows as in the measure theory case: if $\sum_{n} a_{mn} = A_m < \infty$ for some $m = M$, apply the monotone convergence theorem to $a_{Mn} - a_{mn}$.

  • $\begingroup$ Thanks for responding Stephen. I see what you are saying, but as I mentioned, this problem was introduced before those theorems. I'd like to do this by a different approach. $\endgroup$ – dannum May 26 '14 at 2:38
  • $\begingroup$ I mimicked the proof of the monotone convergence theorem, removing unnecessary details. $\endgroup$ – Stephen Montgomery-Smith May 26 '14 at 3:00

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