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"
and this is the solution that the book provide
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.
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), $|T_r|=\binom{p}{r}$.