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This is from the book Exercises in Integration by Claude George, page 37

This is an exercise that wants us to prove the so called "Poincare's formula"

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and this is the solution that the book provide enter image description here

What i want to ask is the part:

$\displaystyle\sum_{A} (-1)^{Card A} \displaystyle\sum_{B \supset A} meas(E_B^{'})=\displaystyle\sum_{B} meas(E_B^{'}) \displaystyle\sum_{A \subset B} (-1)^{Card A}$

If p=Card B,then $\displaystyle\sum_{A \subset B} (-1)^{Card A} = \displaystyle\sum_{r=1}^{p} (-1)^r $ ${p}\choose{r}$

I don't know how to derive Both of the R.H.S of the two equations, any help would be nice , thanks in advance.

RobPratt
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G.t.g.h
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1 Answers1

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The first equation just switched the order of the sum. If $S,T$ are finite sets, $f:S\times T\to \mathbb{R}$ is a function, $M\subset S\times T$, we can define the slices $$M_s=\{t\in T:(s,t)\in M\},$$ $$M^t=\{s\in S:(s,t)\in M\},$$ then $$\sum_{(s,t)\in M}f(s,t)=\sum_{s\in S}\sum_{t\in M_s}f(s,t)=\sum_{t\in T}\sum_{s\in M^t}f(s,t).$$

Apply this with $M=\{(A,B):A\subset B\}$. Then $M_A=\{B:B\supset A\}$ and $M^B=\{A:A\subset B\}$. Let $f(A,B)=(-1)^{\text{Card}A}meas(E'_B)$.

For the second piece, we can split up the sum over $A$. We split up the sets $A$ according to their cardinalities. We are (presumably) omitting $A=\varnothing$, which would be the $r=1$ term. Let $T_r$ be the set of $A\subset B$ with $\text{Card}A=r$. Then $$\sum_{A\subset B}(-1)^{\text{Card}A}=\sum_{r=1}^p\sum_{A\in T_r}(-1)^{\text{Card}A}=\sum_{r=1}^p\sum_{A\in T_r}(-1)^r.$$ The last equality is because $\text{Card}A=r$ for all $A\in T_r$. Then $\sum_{A\in T_r}(-1)^r=(-1)^r|T_r|$, and we need to calculate $|T_r|$. But since $T_r$ consists of the $r$-element subsets of $B$ (which has $p$ elements), $|T_r|=\binom{p}{r}$.

  • I think you repeat the "$\displaystyle\sum_{r=1}^{p} \displaystyle\sum_{A \in T_r} (-1)^{cardA}$" twice – G.t.g.h Dec 30 '23 at 15:05
  • Also , can you elaborate more on the part "$\sum_{A\subset B}(-1)^{\text{Card}A}=\sum_{r=1}^p\sum_{A\in T_r}(-1)^{\text{Card}A}$", I'm not sure i get what you are saying, thank you very much – G.t.g.h Dec 30 '23 at 15:08
  • We are summing over certain $A$. What we do is first split up the $A$ according to their cardinality, so that $T_1$ consists of the singleton $A$s, $T_2$ consists of the $A$s with $\text{Card}A=2$, etc. First sum $(-1)^{\text{Card}A}$ over each group, them sum the results. This is equal to the ungrouped sums. –  Dec 30 '23 at 15:10
  • understood , thank you very much – G.t.g.h Dec 30 '23 at 15:19