As a quick revision, let me recollect the way we define the addition of countably many numbers:
Suppose a_0, a_1, \dots are non-negative real numbers. Then the series a_1 + a_2 + \dots is said to converge to L if the sequence s_n defined by s_n = a_1 + a_2 + \dots + a_n of partial sums of a_n converges to L.

In other words, the tail of the sequence must add up to atmost as small a number as one desires.

Now, in the quest of adding up uncountably many non-negative numbers, let us index the set of these numbers by I. Suppose f : I \to [0, \infty) is a function that associates every element of the index set I to a non-negative number. Our mission is to define \sum_{x \in I} f(x).

Turn I into a discrete measure space by defining the measure of each point in I to be 1. Clearly, now I, with the discrete measure \mu : \mathcal P (I) \to {1} is a measure on the measure space I.

Now define

\sum_{ x \in I} f(x)= \int_I f(t) d\mu (t)

In other words, the sum of all elements indexed by I, is the integral of f with respect to the discrete measure \mu taken over the entire index set I.

I then wondered if the sum of uncountably many positive numbers could be finite. (Note that a countable sum of positive numbers can be finite; the reciprocals of positive powers of two add up to unity.) The following proposition helps (cf. [Rud]) :

Proposition: Let (I, \mathcal P (I), \mu : I \to \{1\} ) be a measure space. If f : I \to [0,\infty) is such that \int_I f(t) d\mu(t)<\infty, then f\equiv 0 except perhaps on a countable subset of I.

Proof:
Decompose I as:
I = \{ x | f(x) = 0 \} \bigcup_{n \in N} \{ x | f(x) > 1/n \}
Call the right set as A_0 and the further right ones as A_n. Suppose that some of the A_n is infinite. Then
\int_I f(t) d\mu(t) > \int_{A_n} \frac{1}{n} d\mu(t) = \infty
contrary to our assumption that the integral is finite. Hence, every A_n is finite and so its union is countable and the proposition is proved.

Reference:
[Rud] Walter Rudin, “Real and Complex Analysis”, 3rd ed. TMH

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