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Let $\mu_1,\mu_2,\dots$ measures on the measurable space $(E,\mathcal{E})$. Then $\mu = \sum_n \mu_n$ is a measure on $(E,\mathcal{E})$.

I want to show, that $$\mu f = \sum_n \mu_n f$$ for every positiv measurable function $f \in \mathcal{E}_+$ where $\mu f = \int_E f \, \mathrm{d}\mu$.

My strategy is to show for the function $L \colon \mathcal{E}_+ \to [0,\infty]$, $L(f)=\sum_n \mu_n f$ that the following properties hold:

  • $f=0 \Rightarrow L(f) = 0$
  • $f,g \in \mathcal{E}_+$, $a,b \in \mathbb{R}_+$ $\Rightarrow$ $L(af+bg)=aL(f)+bL(f)$
  • $(f_k) \subset \mathcal{E}_+$, $f_k \nearrow f$ $\Rightarrow$ $L(f_k) \nearrow L(f)$

Then there exists a unique measure $\nu$ on $(E,\mathcal{E})$ such that $L(f)=\nu f$ for every $f \in \mathcal{E}_+$ and because $ \nu(B) = L(1_B) = \mu(B), \ B \in \mathcal{E} $ the statement $\nu = \mu$ is valid.

My problem with this is the third point:

For every $N \in \mathbb{N}$

$$ \sup_k \sum_{n=1}^{N} \mu_n(f_k) = \sum_{n=1}^{N} \sup_k \mu_n(f_k) = \sum_{n=1}^{N} \mu_n(f) $$ from the monotone convergence theorem.

Now I wan't to interchange $\lim$ and $\sup$ like this:

$$ \sum_n \mu_n (f) = \lim_{N \to \infty}\sup_k \sum_{n=1}^{N} \mu_n(f_k)= \sup_k \lim_{N \to \infty} \sum_{n=1}^{N} \mu_n(f_k) = \sup_k \sum_{n} \mu_n(f_k) $$

Is this possible and why?

Heraklit
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    What prevents you from defining $\nu$ directly via the formula $\nu(A) = \sum_{n = 1}^{\infty} \mu_{n}(A)$? –  Feb 03 '21 at 18:14
  • How would you then show $\int f, \mathrm{d} \mu = \sum_n \int f ,\mathrm{d} \mu_n$? – Heraklit Feb 03 '21 at 18:26
  • When one wants to show the equality of two integrals, one often shows it first for simple functions and then infers it for positive measurable functions via monotone convergence. The way over the function L is actually the same but a little more elegant. That's why I tried it. – Heraklit Feb 03 '21 at 18:32

1 Answers1

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Concerning the issue with the sup, don't forget that the supremum in $k$ there is also the limit in $k$. Hence the monotone convergence theorem applies. To see this, define functions $a_{k} : \mathbb{N} \to [0,\infty]$ by \begin{equation*} a_{k}(n) = \int_{E} f_{k} \, d \mu_{n} \end{equation*} Notice that $a_{k}(n) \leq a_{k+ 1}(n) \leq \int_{E} f \, d \mu_{n}$ and $\lim_{k \to \infty} a_{k}(n) = \int_{E} f \, d \mu_{n}$. Therefore, by the monotone convergence theorem (applied to $\mathbb{N}$ with the counting measure), \begin{equation*} \lim_{k \to \infty} \sum_{n = 1}^{\infty} a_{k}(n) = \sum_{n = 1}^{\infty} \int_{E} f \, d \mu_{n}. \end{equation*}

On the other hand, one can check directly that the formula $\mu(A) = \sum_{n = 1}^{\infty} \mu_{n}(A)$ defines a measure on $\mathcal{E}$. Further, if $f$ is a simple function, then $\int_{E} f \, d \mu = \sum_{n = 1}^{\infty} f \, d \mu_{n}$ follows immediately. Arguing using the monotone convergence theorem (as above) and simple approximation, one then finds that the same identity holds when $f$ is bounded and non-negative. From there, one can generalize to arbitrary non-negative or integrable $f$.

  • Thanks for the answer! Using monotone convergence for the counting measure is a very cool argument! – Heraklit Feb 03 '21 at 19:47
  • But when I want to show that integration over the counting measure is a summation I would use the statement above with $\mu=\sum_n \delta_n$. – Heraklit Feb 03 '21 at 19:58
  • Can I argue like this instead: Let $f_k \nearrow f$ ($f_k$ are simple functions and $f$ is positive and measurable). Then let $a_{k,N}:= \sum_{n=1}^{N} \int f_k d \mu_n$. Can I argue that $\sup_k \sup_N a_{k,N} = \sup_N \sup_k a_{k,N}$ without using this argument? – Heraklit Feb 03 '21 at 20:01
  • https://math.stackexchange.com/questions/810341/interchange-of-suprema Ah, I didn't know it was that easy. – Heraklit Feb 03 '21 at 21:50