Elstrodt: Maß- und Integrationstheorie, exercise II.1.8

Question

I am trying to solve exercise II.1.8. from the book on Measure Theory by Elstrodt, but I don't really know how to approach the problem.

Definition

Let $\mathfrak{H}$ be a semi-ring over the set $X$. A map $\mu: \mathfrak{H} \to \overline{\mathbb{R}}$ is called a content if it has the following properties:

(i) $\mu(\emptyset) = 0$

(ii) $\mu \geq 0$

(iii) For every finite sequence of disjoint sets $A_1,...,A_n \in \mathfrak{H}$ with $\bigcup \limits_{j=1}^{n} A_j \in \mathfrak{H}$:

$\mu \left(\bigcup \limits_{j=1}^{n} A_j \right) = \sum \limits_{j=1}^{n} \mu(A_j).$

If $\mu$ is defined on a ring $\mathfrak{R}$, then (iii) can be replaced by

(iii*) $\mu(A \cup B = \mu(A) + \mu(B)$ for all disjoint sets $A,B \in \mathfrak{R}.$

Exercise 1.8

Let $\mu$ be a content on the ring $\mathfrak{R}$ and $A_1,...,A_n \in \mathfrak{R}, \mu(A_j) < \infty$ for $j=1,...,n$,

$$m_k := \sum \limits_{1 \leq i_1 \leq ... \leq I_k \leq n} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) \quad (k=1,...,n).$$

a) $\mu(A_1 \cup \cdots \cup A_n) = \sum \limits_{k=1}^{n} (-1)^{k-1} m_k$

b) For $p=1,...,n$ let $B_p$ be the set of $x \in X$ that are in exactly $p$ of the sets $A_1,...,A_n$. Then $B_p \in \mathfrak{R}$ and

$$\mu(B_p) = \sum \limits_{k=p}^{n} (-1)^{k-p} \binom{k}{p} m_k, \quad \sum \limits_{p=1}^{n} p \mu(B_p) = \sum \limits_{k=1}^{n} \mu(A_k)$$

c) For $p=1,...,n$ let $C_p$ be the set of $x \in X$ that are in at least $p$ of the sets $A_1,...,A_n$. Then $C_p \in \mathfrak{R}$ and

$$\mu(C_p) = \sum \limits_{k=p}^{n} (-1)^{k-p} \binom{k-1}{p-1} m_k, \quad \sum \limits_{p=1}^{n} \mu(C_p) = \sum \limits_{k=1}^{n} \mu(A_k).$$ (Hint: For $\emptyset \neq I \subset \{1,...,n\}$ let $E_I := \bigcap \limits_{j \in I} A_j, F_I := E_I \cap \bigcap \limits_{k \notin I} A_k^c$ and note that $E_I$ is the disjoint union of the sets $F_J$ with $I \subset J \subset \{1,...,n\}.$)

So far I have solved part a) using induction.

Proof:

a) Clearly, the claim is true for $n=1$. Now assume the claim holds for $n-1$ (i.e. it holds for any collection of sets of size $n-1$), then

\begin{align*} \mu(A_1 \cup \cdots \cup A_n) & = \mu(A_1 \cup \cdots \cup A_{n-1}) + \mu(A_n) - \mu((A_1 \cap A_n) \cup \cdots \cup (A_{n-1} \cap A_n)) \\ & = \sum \limits_{k=1}^{n-1} (-1)^{k-1} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots A_{i_k}) + \mu(A_n) \\ & + \sum \limits_{k=1}^{n-1} (-1)^k \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots A_{i_k} \cap A_n) \\ & = \sum \limits_{k=1}^{n} (-1)^{k-1} m_k. \end{align*}

Note that I have used the induction assumption for the sets $A_1,...,A_{n-1}$ as well as $A_1 \cap A_n,...,A_{n-1} \cap A_n$.

Now I think that b) can be shown using an induction argument as well, but I haven't been able to find a good way to approach the problem. My idea was to use the fact that being in $p$ of the sets $A_1,...,A_n$ could mean being in $p$ of the sets $A_1,...,A_{n-1}$ or in $p-1$ of the sets $A_1,...,A_{n-1}$ and in $A_n$, but this didn't get me anywhere so far.

Proof idea:

b) For $2 \leq p < n$ we have

\begin{align*} \mu(B_p) & = \sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) \\ & + \sum \limits_{k=p-1}^{n-1} (-1)^{k-(p-1)} \binom{k}{p-1} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_k} \cap A_n) \\ & = \sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) \\ & + \sum \limits_{k=p}^{n} (-1)^{k-p} \binom{k-1}{p-1} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_{k-1} \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_{k-1}} \cap A_n) \end{align*}

where I have used the induction assumption on the sets $A_1,...,A_{}n-1$ as well as the sets $A_1 \cap A_n,...,A_{n-1} \cap A_n$. Now I think I should be able to use $\binom{k-1}{p-1} = \binom{k}{p} - \binom{k-1}{p}$ for $k=p+1,...,n$ to prove the claim (and $\binom{p-1}{p-1} = \binom{p}{p}$), but this produces some additional terms.

\begin{align*} \mu(B_p) & = \sum \limits_{k=p}^{n} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) \\ & - (-1)^1 \binom{p}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_p \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_p} \cap A_n) \\ & - (-1)^2 \binom{p+1}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_{p+1} \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_{p+1}} \cap A_n) \\ & - \cdots - (-1)^{n-p} \binom{n-1}{p} \sum \limits_{1 \leq I_1 \leq \cdots \leq i_{n-1} \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_{n-1}} \cap A_n) \end{align*}

1. How can I show that the additional terms sum up to $0$? Am I on the right track here or is there a better way to prove this?

2. I can also see why the hint for c) is true, but I am not sure how to use it to solve the part c). My guess is that we need to combine b) and the hint.

Hope someone can help me out here. Thanks a lot!

Answer 1

Part B

In your first line in the calculation of $B_p$, you have a mistake. Here is the correct equation. $$ \begin{align*} \mu(B_p) & = \sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}\color{blue}{\cap A_n^c}) \\ & + \sum \limits_{k=p-1}^{n-1} (-1)^{k-(p-1)} \binom{k}{p-1} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_k} \cap A_n) \\ \end{align*} $$ In order to simplify this, use this equality to split the first summation into a difference of two summations. $$ \mu(A_{i_1} \cap \cdots \cap A_{i_k}\color{blue}{\cap A_n^c})=\mu(A_{i_1}\cap \dots \cap A_{i_k})-\mu(A_{i_1}\cap \dots \cap A_{i_k}\cap A_n) $$ Here is the result: \begin{align*} \mu(B_p) & =\sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) \\ & - \sum \limits_{k=\color{red}{p-1}}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}\cap A_n) \\ & + \sum \limits_{k=p-1}^{n-1} (-1)^{k-(p-1)} \binom{k}{p-1} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_k} \cap A_n) \\ \end{align*} Notice that in the second summation (the one being subtracted), I changed the lower limit from $p$ to $p-1$. This is allowed, because the additional $k=p-1$ terms is zero, as it has $\binom{p-1}{p}=0$ as a factor.

The second summation accounts for all of the additional terms that you found in your attempt.

The second and third sums now match up exactly, and can be combined. Since the first summation is going to be unchanged each line in the following computation, I colored it gray. \begin{align*} \mu(B_p) & = \color{gray}{\sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k})} \\ & + \sum \limits_{k=p-1}^{n-1} (-1)^{k-(p-1)} \left[\binom{k}{p-1} +\binom kp\right]\sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_k} \cap A_n) \\ & = \color{gray}{\sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k})} \\ & + \sum \limits_{k=p-1}^{n-1} (-1)^{k-(p-1)} \binom{k+1}{p}\sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_k} \cap A_n) \\ & = \color{gray}{\sum \limits_{k=p}^{n-1} (-1)^{k-p} \binom{k}{p} \sum \limits_{1 \leq i_1 \leq \cdots \leq i_k \leq n-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k})} \\ & + \sum \limits_{k=\color{blue}{p}}^{\color{blue}n} (-1)^{\color{blue}{(k-1)}-(p-1)} \binom{\color{blue}k}{p}\sum \limits_{1 \leq i_1 \leq \cdots \leq i_{\color{blue}{k-1}} \leq n-1} \mu(A_{i_1} \cap ... \cap A_{i_{\color{blue}{k-1}}} \cap A_n) \end{align*} Finally, both sums can be combined into one, and their combination is exactly equal to $\sum (-1)^{k-p}\binom kp m_k$.

Part C

I could not figure out how to apply the hint, but here is a different method. $$ \begin{align} \mu(C_p) &= \sum_{\ell=p}^n \mu(B_\ell) \\ &=\sum_{\ell=p}^n\sum_{k=\ell}^n (-1)^{k-\ell}\binom k\ell m_k \\ &=\sum_{k=p}^n(-1)^km_k\sum_{\ell=p}^k (-1)^\ell\binom k\ell\tag1 \end{align} $$ We simplify the inner summation as follows: $$ \begin{align} \sum_{\ell=p}^k (-1)^\ell\binom k\ell &=\left(\sum_{\ell=\color{blue}0}^k(-1)^\ell\binom k\ell\right) -\left(\sum_{\ell=0}^{p-1}(-1)^\ell\binom k\ell\right) \\ &=0 - (-1)^{p-1}\binom{k-1}{p-1} \\ &=(-1)^p\binom {k-1}{p-1} \end{align} $$ You can prove $\sum_{\ell=\color{blue}0}^k(-1)^\ell\binom k \ell =0$ by applying the binomial theorem to $(1-1)^k$. To prove the second simplification that $\sum_{\ell=0}^{p-1}(-1)^\ell\binom k\ell=(-1)^{p-1}\binom{k-1}{p-1}$, see Truncated alternating binomial sum.

Substituting $(-1)^p\binom {k-1}{p-1}$ for the inner summation in $(1)$ completes the proof.