Can I change the order of summation here?

Question

Is it true that $\sum _{i = 1 } ^n \alpha _i \sum _{r=1 } ^{\infty } \mu(A _i \cap E _r ) = \sum _{r = 1 } ^{\infty } \sum _{i=1 } ^{n } \alpha _i\mu(A _i \cap E _r ) $

I know that I can change order of two infinite sumattions if the double sum is absolutely convergent, but I don't believe this can be assumed here.

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I want to prove that the Lebesgue integral is countably additive for a nonnegative simple function $s = \sum _{i=1 } ^n \alpha _i \mu ( A _i \cap E ) $, and the above comes from the countable additivity of $\mu $.

Also I dont't assume $\mu ( A _i \cap E ) < \infty $, but $\alpha _i < \infty $.

Maybe this is trivial, but I got stuck...

Thanks in advancd!

Answer 1

It's true because the values are nonnegative.

You can do whatever you want with the order of summing if that is the case. The sum is always the supremum of finite sums of some terms.