# Using dominated convergence theorem to show rearrangement of terms of an absolutely convergent sequence does not affect sum

I have read that you can use the Lebesgue Dominated Convergence Theorem to show that rearranging the terms in an absolutely convergent sequence does not affect the sum.

It is not apparent to me how to do this, I have a few thoughts but not much progress

Let $(a_n)$ be an absolutely convergent sequence. It seems natural to take the measure space $(\mathbb{N}, \mathcal{P}(N), \mu)$ where $\mu$ denotes the counting measure. Then we have that

$$\int |a|\textrm{ d}\mu=\sum_{n=1}^\infty|a_n|<\infty$$

I'm trying to find a good place to use the dominated convergence theorem. The only thing I could think of was to define $f_n=(a_1,a_2,...a_n,0,...)$. Then $\lim f_n=a$. We know that $|f_n|\leq |a|$ and $|a|$ is integrable by above. Then $f_n$ are clearly integrable as they correspond to finite sums. So by the dominated convergence theorem

$$\lim_n\int f_n\textrm{ d}\mu=\int a\textrm{ d}\mu=\sum_{n=1}^{\infty}a_n$$

but this doesn't seem to be going anywhere. Any help or hints would be appreciated.

Summing a rearrangement amounts to using a different sequence of sets $A_n \uparrow \mathbb N$with $|A_n|=n$ instead of $[1,…,n]$.
The functions(=sequences) $f_n:\mathbb N→ \mathbb C$ defined by $f_n := f\mathbb 1_{A_n}→ f$ pointwise and $|f|$ is a dominating function for $f_n$. Hence, $$∫ f_n\ d\mu → ∫ f \ d\mu = \sum_{i=1}^∞ f(i).$$