Sequences and Series/Multiple limits

Theorem (interchanging summation and integration):

Let ${\displaystyle (\Omega ,{\mathcal {F}},\mu )}$ be a measure space, and let ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ be a sequence of functions from ${\displaystyle \Omega }$ to ${\displaystyle \mathbb {K} ^{d}}$, where ${\displaystyle \mathbb {K} =\mathbb {R} }$ or ${\displaystyle \mathbb {C} }$. If either of the two expressions

${\displaystyle \int _{\Omega }\sum _{n=1}^{\infty }\|f_{n}(\omega )\|_{\infty }\mu (d\omega )}$ or ${\displaystyle \sum _{n=1}^{\infty }\int _{\Omega }\|f_{n}(\omega )\|_{\infty }\mu (d\omega )}$

converges, so does the other, and we have

${\displaystyle \int _{\Omega }\sum _{n=1}^{\infty }f_{n}(\omega )\mu (d\omega )=\sum _{n=1}^{\infty }\int _{\Omega }f_{n}(\omega )\mu (d\omega )}$.

Proof: Regarding the summation as integration over ${\displaystyle \mathbb {N} }$ with σ-algebra ${\displaystyle 2^{\mathbb {N} }}$ and counting measure, this theorem is an immediate consequence of Fubini's theorem, given that integration and summation are defined pointwise. ${\displaystyle \Box }$

Theorem (interchanging summation and real differentiation):

Let ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ be a sequence of continuously differentiable functions from an open subset ${\displaystyle U}$ of ${\displaystyle \mathbb {R} ^{d}}$ to ${\displaystyle \mathbb {R} ^{k}}$. Suppose that both

${\displaystyle \sum _{n=1}^{\infty }\|f_{n}(x)\|_{\infty }}$ and ${\displaystyle \sum _{n=1}^{\infty }\|Df_{n}(x)\|_{\infty }}$

converge for all ${\displaystyle x\in U}$, and that for all ${\displaystyle x\in U}$ there exists ${\displaystyle \delta >0}$ and a sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle \mathbb {R} }$ such that

${\displaystyle \sum _{n=1}^{\infty }a_{n}<\infty }$ and ${\displaystyle \forall n\in \mathbb {N} :\forall y\in {\overline {B_{\delta }(x)}}:a_{n}\geq |Df_{n}(y)|}$.

Then

${\displaystyle D\sum _{n=1}^{\infty }f_{n}(x)=\sum _{n=1}^{\infty }Df_{n}(x)}$

for all ${\displaystyle x\in U}$.

Proof: ${\displaystyle \Box }$