Exercise 0.0.1.
\[ \sum_{\alpha \in A} x_\alpha < \infty. \]
Let \((x_\alpha)_{\alpha \in A}\) be a collection of non-negative numbers \(x_\alpha \in [0, +\infty]\), such thatShow that \(x_\alpha = 0\) for all but at most countably many \(\alpha \in A\), even if \(A\) itself is uncountable.
Proof & Intuition
In the section before this exercise, Tao introduced that the idea of taking a sum over a set - no matter countable or not - is taking the supremum over all sums of its finite subset.
Then the intuition comes forward - we are to construct finite subsets of an arbitrary set, that is guaranteed to sum up to infinity as it goes to even the slightest countable infinity.
Take \( S_n \subset A \), where
\[ S_n := \left\{ \alpha \in A : \chi_\alpha \geq \frac{1}{n} \right\} \]\[ \Rightarrow S_n \text{ must be finite, otherwise } \sum_{\alpha \in S_n} \chi_\alpha \geq \sum_{\alpha \in S_n} \frac{1}{n} = \infty \]Define:
\[ S := \bigcup_{n=1}^\infty S_n \]\[ \forall \chi_\alpha > 0,\ \exists n \in \mathbb{N} \text{ such that } \chi_\alpha \geq \frac{1}{n} \Rightarrow \forall \chi_\alpha > 0,\ \exists n \text{ such that } \alpha \in S_n \]\[ \Rightarrow \forall \chi_\alpha > 0,\ \alpha \in S \]Since each \( S_n \) is finite, \( S \) is countable.
Hence, \( \chi_\alpha = 0 \) for all but at most countably many \( \alpha \in S \), such that
\[ \sum_{\alpha \in A} \chi_\alpha < \infty \]Exercise 0.0.2 (Tonelli over arbitrary sets).
Let \(A,B\) be sets (possibly uncountable) and \((x_{n,m})_{n\in A,\,m\in B}\subset[0,+\infty]\). Show
\[ \sum_{(n,m)\in A\times B} x_{n,m} \;=\; \sum_{n\in A}\ \sum_{m\in B} x_{n,m} \;=\; \sum_{m\in B}\ \sum_{n\in A} x_{n,m}. \]
Proof & Intuition
Prior to this exercise, Tao proved the Fubini’s Theorem for doubly countably infinite series.
In exercise 0.0.1, we saw that if an infinite (non-negative) series converges, all terms would be zero but at most countably many are not.
In addition, Tao introduced that for countably infinite non-negative series, the index of sum does not matter (as summing over the new indices produced by a bijection of indicies does not change the series).
Now we focus if we can suffice the original problem over uncountable sets into a problem over countable sets, and we apply the original Tonelli’s Theorem.
Proof:
Define \( x_n := \sum_{m \in B} x_{n,m} \)
Since \( x_{n,m} \in [0, +\infty] \), we have \( x_n \in [0, +\infty] \).
Suppose \( \sum_{n \in A} x_n < \infty \)
\[ \Rightarrow \text{All } x_n = \sum_{m \in B} x_{n,m} = 0 \text{ but countably many } x_n \neq 0 \]\[ \Rightarrow \forall n \text{ with } \sum_{m \in B} x_{n,m} \neq 0, \text{ all } x_{n,m} = 0 \text{ but countably many } x_{n,m} \neq 0 \]\[ \Rightarrow \text{All } x_{n,m} = 0 \text{ but countably many } x_{n,m} \neq 0 \]\[ \Rightarrow \sum_{(n,m) \in A \times B} x_{n,m} = \sum_{\substack{(n,m) \in A \times B \\ x_{n,m} \neq 0}} x_{n,m} \text{ is a countable sum} \]By Fubini’s Theorem on \( \mathbb{N} \times \mathbb{N} \), for the countable set
there exists a bijection of non-zero terms.