Exercise 0.0.2 (Tonelli’s theorem for series over arbitrary sets)

Question

Let $A,B$ be sets (possibly infinite or uncountable), and $(x_{n,m})_{n\in A,m\in B}$ be a doubly infinite sequence of extended non-negative reals $x_{n,m}\in[0,+\infty]$ indexed by $A$ and $B$. Show that
$$\sum_{(n,m)\in A\times B}x_{n,m}=\sum_{n\in A}\sum_{m\in B}x_{n,m}=\sum_{m\in B}\sum_{n\in A}x_{n,m}.$$
(Hint: although not strictly necessary, you may find it convenient to first establish the fact that if $\sum_{n\in A}x_n$ is finite, then $x_n$ is non-zero for at most countably many $n$.)

Alden · Accepted Answer

We can consider two cases:

\begin{enumerate}
\item the uncountable sum is infinite: the equation holds obviously.
\item the sum is finite: recall last exercise 0.0.1, we just need to prove the conclusion holds when there are countable elements that are non-zero which has been proved in Exercise 0.0.1.
\end{enumerate}

龙珠收集者 · Answer

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{geometry}

\geometry{a4paper, margin=1in}

\begin{document}

\section*{Proof of Tonelli's Theorem for Series}

Let $x_{n,m} \geq 0$ for all $(n,m) \in A 	imes B$. We aim to prove:
\begin{equation}
\sum_{(n,m)\in A	imes B}x_{n,m} = \sum_{n \in A}\sum_{m\in B}x_{n,m}
\end{equation}

\begin{align}
 & 	ext{Let } F 	ext{ be a finite set such that } F \subset A 	imes B. \
 & 	ext{Define } A_{1} = \{x \mid \exists y \in B, (x,y) \in F\}, \quad A_{2} = \{y \mid \exists x \in A, (x,y) \in F\}. \
 & 	ext{Then } A_{1}, A_{2} 	ext{ are finite sets, } A_{1} \subset A, A_{2} \subset B, 	ext{ and } F \subset A_{1} 	imes A_{2}. \
 & \sum_{(n,m)\in F}x_{n,m} \leq \sum_{(n,m)\in A_{1}	imes A_{2}}x_{n,m} = \sum_{n \in A_{1}}\sum_{m\in A_{2}}x_{n,m} \leq \sum_{n \in A}\sum_{m\in B}x_{n,m}. \
 & 	ext{Therefore, } \sum_{(n,m)\in A	imes B}x_{n,m} = \sup_{\substack{F\subset A	imes B \ F 	ext{ finite}}} \sum_{(n,m)\in F}x_{n,m} \leq \sum_{n \in A}\sum_{m\in B}x_{n,m}. \
 & 	ext{If } \sum_{(n,m)\in A	imes B}x_{n,m}=+\infty, 	ext{ then } \sum_{n \in A}\sum_{m\in B}x_{n,m} \leq \sum_{(n,m)\in A	imes B}x_{n,m}. \
 & 	ext{If } \sum_{(n,m)\in A	imes B}x_{n,m}<+\infty: \
 & 	ext{Let } C = \{(n,m)\in A	imes B \mid x_{n,m}
eq 0\}. \
 & A_{3} = \{x \mid \exists y \in B, (x,y) \in C\}, \
 & B_{n} = \{m \in B \mid (n,m) \in C\}. \
 & 	ext{Then } A_{3} \subset A, B_{n} \subset B, 	ext{ and } C \subset A	imes B 	ext{ are all at most countable sets}. \
 & 	ext{Hence, } \sum_{n \in A}\sum_{m\in B}x_{n,m} = \sum_{n \in A_{3}}\sum_{m\in B_{n}}x_{n,m} = \sum_{(n,m)\in C}x_{n,m} \leq \sum_{(n,m)\in A	imes B}x_{n,m}.
\end{align}

\section*{Conclusion}
Combining the two cases, we have shown that:
\begin{equation}
\sum_{(n,m)\in A	imes B}x_{n,m} = \sum_{n \in A}\sum_{m\in B}x_{n,m}.
\end{equation}

\end{document}

Exercise 0.0.2 (Tonelli’s theorem for series over arbitrary sets)

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Proof of Tonelli’s Theorem for Series

Conclusion

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Exercise 0.0.2 (Tonelli’s theorem for series over arbitrary sets)

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Proof of Tonelli’s Theorem for Series

Conclusion

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