Bartle's exercise to prove $\lambda(E)=\sum_{n=1}^\infty 2^{-n} \mu_n (E)$ is a measure

Question

Let $\lambda(E)=\sum_{n=1}^\infty 2^{-n} \mu_n (E)$ with $\mu_n:\mathcal{P}(X)\rightarrow \mathbb{R}$ a measure with $\mu_n(X)=1$. Prove $\lambda$ is a measure.

Well, we clearly have $\lambda(E)\geq 0$ and $\lambda(\emptyset)=0$.

It is interesting to note $\lambda$ is finite, because $\mu_n(E)\leq \mu_n(X\setminus E)+\mu_n(E)=\mu_n(X)=1$, so:

$$S_k=\sum_{n=1}^k2^{-n} \mu_n(E)\leq \sum_{n=1}^k 2^{-n}\leq 1$$

Because $S_k$ is an increasing bounded sequence it must converge to the supremum $\sup_k S_k$.

If the $E_i$ are disjoint, we have:

$$\lambda(\cup_{i\in \mathbb{N}} E_i)=\sum_{n=1}^\infty 2^{-n}\mu_n(\cup_{i\in \mathbb{N}} E_i)=\sum_{n=1}^\infty 2^{-n}\sum_{i=1}^\infty \mu_n (E_i)=\sum_{n=1}^\infty \sum_{i=1}^\infty2^{-n}\mu_n(E_i)$$

We only need to change the order of summation to conclude the additive property of the measure.

I am aware convergent series of positive terms (with one summing index) are commmutatively convergent because if $b_m=a_{\phi(m)}$ is a reordering ($\phi: \mathbb{N}\rightarrow \mathbb{N}$ where $\phi$ is a bijection) of the $a_n$, for every fixed $n$, there is $m$ large enough, say $\max_{i=1,...,n}\{ \phi^{-1}(i)\}$ such that $\sum_{k=1}^n a_k \leq \sum_{k=1}^m b_k \leq \sup \sum b_n $. Therefore, $ \sup \sum a_n\leq \sup \sum b_n$. A similar inequality follows from taking $n$ large enough and therefore we may sum any way we want and $\sup \sum a_n=\sup \sum b_n$.

I suspect we may formally exchange the order of the double infinite summation using a similar argument, but I haven't been able to do so, because there are two supremum's involved and it gets rather convoluted...

Answer 1

We only need to change the order of summation to conclude the additive property of the measure.

For the purposes of this exercise [1, Chap 3, Ex. 3C] you don't need to give a proof of this change of order in your answer, but just quote the theorem below on double series. However if you are interested a proof of this theorem is given as application (iii) of Theorem 2 in this answer to question Rearrangements of absolutely convergent series.

Theorem

The order of summation of a non-negative double series in the extended real number system $\overline{\mathbb{R}}$ can be changed and the sum within $\overline{\mathbb{R}}$ remains the same, ie. if $a_{k,n}$ is a double sequence in $[0, \infty]$ then the following two sums (\ref{eq:double-series-1}) & (\ref{eq:double-series-2}) are equal (possibly $\infty$) : \begin{eqnarray} \sum_{k=1}^{\infty}\;\sum_{n=1}^{\infty} a_{k, n} & = & \sum_{k=1}^{\infty} r^{(k)}, \hspace{2em} \mbox{where $r^{(k)} =$ sum in $\overline{\mathbb{R}}$ of $k^{\mathrm{th}}$ row} \label{eq:double-series-1} \tag{1} \\ \sum_{n=1}^{\infty}\;\sum_{k=1}^{\infty} a_{k, n} & = & \sum_{n=1}^{\infty} c^{(n)}, \hspace{2em} \mbox{where $c^{(n)} =$ sum in $\overline{\mathbb{R}}$ of $n^{\mathrm{th}}$ column} \label{eq:double-series-2} \tag{2} \end{eqnarray} $\blacksquare$

Because we are dealing with non-negative term series each of these series always converges in $\overline{\mathbb{R}}$, to either a real number or to $\infty$. Note the terms $a_{k,n}$ themselves can be $\infty$, so they can be the values of measures.

A proof of Theorem 2 is given in the above answer as well as in [2, p143, §90], where it is termed 'Cauchy's Double Series Theorem' (CDST). This proof in turn depends on Theorem 1 of the above answer, concerning 'extended rearrangements' of infinite series, which is also proved in [2, p142, Theorem 4]. Extended rearrangements of infinite series allow more complicated ways of rearranging an infinite series than simply permutating terms, and they are often very useful - some applications are shown in the above answer.

However in textbooks the Theorems 1 & 2 and their corollaries are often just tacitly assumed. The book [2] by Knopp is an older textbook that includes the proofs. In the above answer the proofs are written out with some further details and additions.

The CDST is straightforward to prove from Theorem 1, and application (iii) of CDST follows readily from CDST, but Theorem 1 itself requires more effort.

Theorem 3, and its counterpart Theorem 4, in the above answer involve diagonal sums of double series in $\mathbb{R}$, and are also useful in measure theory, allowing derivation of some basic properties of the Outer Measure on $\mathbb{R}$ [3, Chap 2], eg. countable subadditivity on $P(\mathbb{R})$, and countable additivity on the 'system of intervals' $\mathcal{I}=\{$all countable unions of intervals in $\mathbb{R}\}$.

If you attempt to prove the above theorem without the help of Theorems 1 & 2 I think it can get quite complicated. These theorems give a clear path to the result. Note also in your answer the functions $\mu_n$ and $\lambda$ have domain the $\sigma$-algebra $\mathcal{X}$ on $X$ rather that the power set of $X$.

References

[1] Robert G. Bartle (1966), Elements of Integration, John Wiley.

[2] Konrad Knopp (1954), Theory and Application of Infinite Series, 2nd English Edition translated from 4th German Edition, Blackie & Sons.

[3] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.

Answer 2

I think I was able to solve this without any sophisticated measure theory in the case we have convergence (so a little weaker then Tonelli's Theorem which was stated in the comments). If anyone has a more elegant solution, I would really appreciate to see it. Let $a_{nm}=2^{-n}\mu_n(E_m)$. In this case:

$$\sum_{m=1}^\infty \sum_{n=1}^\infty a_{n m}=\lim_{M}\sum_{m=1}^M \lim_N \sum_{n=1}^N a_{nm}=\lim_M\lim_N \sum_{m=1}^M\sum_{n=1}^N a_{nm}=$$ $$=\lim_M\sup_N \sum_{m=1}^M \sum_{n=1}^N a_{nm}=\sup_M\sup_N \sum_{m=1}^M \sum_{n=1}^N a_{nm}$$

Let $f(M,N)=\sum_{m=1}^M\sum_{n=1}^N a_{nm}$. I state that $\sup_m\sup_n f(m,n)= \sup_{m,n}f(m,n) $. Indeed, let us prove this by way of contradiction. If:

$$\sup_{m}\sup_n f(m,n)< \sup_{m,n} f(m,n)\Rightarrow \exists n_o,m_o|\sup_m \sup_n f(m,n)<f(m_o,n_o)\leq\sup_{m,n} f(m,n)$$

$$f(m_o,n_o)\leq \sup_n f(m_o,n) \leq \sup_m \sup_n f(m,n) \Rightarrow$$ $$ \sup_m \sup_n f(m,n) < \sup_m \sup_n f(m,n)$$

This is clearly absurd and so $\sup_{m,n}f(m,n)\leq \sup_m \sup_n f(m,n)$. The other inequality is more straightforward:

$$f(m,n)\leq \sup_{m,n} f(m,n) \Rightarrow \sup_n f(m,n)\leq \sup_{m,n} f(m,n)\Rightarrow \sup_m \sup_n f(m,n)\leq \sup_{m,n} f(m,n) $$

This proves our assertion and therefore:

$$\sum_{m=1}^\infty \sum_{n=1}^\infty a_{n m}=\sup_{N,M}\sum_{m=1}^M \sum_{n=1}^N a_{nm}$$

But by symetry this is enough to conclude:

$$\sum_{m=1}^\infty\sum_{n=1}^\infty a_{nm}=\sup_{N,M} f(N,M)=\sum_{n=1}^\infty\sum_{m=1}^\infty a_{nm}$$

EDIT: Actually we could have writen $\sup_M\sup_N f(M,N)=\sup_N \sup_M f(M,N)$ quite easily because of this other MSE post. My original argument is okay, but more lengthsome than needed. Also we may drop the assumption of convergence using this same MSE post I mentioned, so this is precisely Tonelli's Theorem.