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Let $g_n: [0,\infty) \to [0,\infty)$ be a right-continuous and monotone non-decreasing function for every $n \in \mathbb N$. Define the pointwise limit $g(x) = \sum_n g_n(x)$ and assume that it is finite for all $x \geq 0$. Then it is true that:

  1. $g$ is also right-continuous and monotone non-decreasing.

  2. For every $f: [0,\infty) \to \mathbb R$ with existing Lebesgue-Stieltjes integral $\int f \mathrm d g$ we have that $$ \int f \mathrm d g = \sum_{n \in \mathbb N} \int f \mathrm d g_n $$

It's been quite a while since I've taken measure and integration theory and the Lebesgue-Stieltjes integral is new to me.

For 1.: The monotony is clear by induction for finite sums. I'm a little embarrassed to say that I have difficulty coming up with a rigorous argument for the infinite sum although it seems obvious that $\sum_n a_n \leq \sum_n b_n$ where $a_n \leq b_n$ for all $n$. Concerning right-continuity, it is known that the infinite sum of continuous functions need not be continuous in general. But my Analysis is not strong enough to figure this out for this case.

So I tried to resort to measure theory: For every $f_n$ there exists a unique measure $\mu_n$ by setting $\mu_n([0,x])=f_n(x)$ for all $x \geq 0$ and applying the uniqueness theorem for measures, since the intervals $[0,x]$ generate the Borel-$\sigma$-algebra and are closed under intersection of sets. Now, the infinite sum $\sum_n \mu_n$ of measures should again be a measure $\mu$. By definition we have $g(x)=\mu([0,x])$ for all $x\geq 0$. If $\mu$ is in fact a measure, then it should follow directly that $g$ has the required properties, right?

However, this would technically require to show the lemmata that the countable sum of measures is a measure and that the correspondence between a non-negative, right-continuous and monotone non-decreasing function and a measure is one-to-one. I would like to be able to verify the statement by more analytic means.

Another question that pops into my head is whether the finiteness of $g$ is necessary for 1. to hold. (I see that it is necessary for 2.)

For 2.: Again, this corresponds to the statement that $$ \int f \mathrm d \sum_n \mu_n = \sum_n \int f \mathrm d \mu_n. $$

I faintly remember having seen a statement like this. My first guess would be that this can be shown by measure theoretic induction: First for indicator functions, then for simple functions and so on. Is there any cuter way to do this?

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