inaccuracy in Schilling Proposition 6.5

Question

I found an inaccuracy in the Proposition 6.5 of the book "Measures, Integrals and Martingales" From René Schilling [EDIT: I'm talking about the first edition of the book] and I would like to correct this error.

To be sure that I'm clear in asking the question I will list the definitions I use in this post so that there are no notational misunderstandings.

I will try to make sure that even those unfamiliar with the book can answer the question. But obviously those who have studied from that book are more likely to be able to answer the question.

I organized this post in the following way:

In the first part there's the question.

In the second part there is the list of definitions and notations that I use, so if you come across a symbol or a definition that you do not know go to read the second part.

In the third part there is the reason that led me to formulate the question, how the question is related to proposition 6.5 and what was my attempt to answer the question.

FIRST PART: THE QUESTION

[Edit: as suggested I changed the name of the "Theorem 0.2" calling it "Statement 0.2"]

Statement 0.2: Let $X$ be a set, $\mathcal F$ a semi-ring on $X$ and $\mu : \mathcal F \to [0,+\infty]$ a function then

$\mu$ is a pre-measure on $X$ if and only if

[\begin{split} &(i) \hspace{0.5cm} \mu(\emptyset)=0 \\ &(ii) \hspace{0.5cm} A \in \mathcal F, B \in \mathcal F , A \cap B = \emptyset , A \cup B \in \mathcal F \implies \mu(A\cup B)=\mu (A) + \mu (B) \\ &(iii) \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_n \subseteq A_{n+1} \forall n \in \mathbb N , \bigcup_{n \in \mathbb N} A_n \in \mathcal F \implies \mu(\bigcup_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \end{split}]

Moreover, with the additional hypothesis that $\mu (A) < + \infty \forall A \in \mathcal F$, (iii) can be replaced by either of the following equivalent conditions:

[\begin{split} &(iii') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N , \bigcap_{n \in \mathbb N} A_n \in \mathcal F \implies \mu(\bigcap_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \\ &(iii'') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N , \bigcap_{n \in \mathbb N} A_n = \emptyset \implies \lim _{n \to +\infty} \mu (A_n) = 0 \end{split}]

End of the Statement 0.2

MY QUESTION IS: is Statement 0.2 True? And what is the proof of the implication "from the right to the left"? If this implication of Statement 0.2 is not true what is an example of a set $X$ with a semi-ring $\mathcal F$ and a function $\mu$ on $\mathcal F$ which satisfies properties (i), (ii), (iii'') of Statement 0.2 but such that $\mu$ is not a pre-measure on $X$ ?

I could only prove the implication "from the left to the right" of theorem 0.2 but i couldn't prove the viceversa. If you want to see a sketch of the proof of the implication i proved, you can find it in the third part of the post.

SECOND PART: Notations and definitions I use

if $X$ is a set and $\mathcal F \subseteq \mathcal P (X)$ then we define:

$\mathcal F$ is a ring on $X$ if and only if (by definition) [\begin{split} &1) \hspace{0.5cm} \emptyset \in \mathcal{F} \\ &2) \hspace{0.5cm} A,B \in \mathcal{F} \implies B \setminus A \in \mathcal{F} \\ &3) \hspace{0.5cm} A,B \in \mathcal{F} \implies A \cup B \in \mathcal{F} \end{split}]

$\mathcal F$ is a semiring on $X$ if and only if (by definition) [\begin{split} &1) \hspace{0.5cm} \emptyset \in \mathcal{F} \\ &2) \hspace{0.5cm} A,B \in \mathcal{F} \implies \exists A_1,A_2,\dots,A_n \in \mathcal{F} \text{ pairwise disjoint } : \,\; B \setminus A = \bigcup_{k=1}^{n}{A_k}\\ &3) \hspace{0.5cm} A,B \in \mathcal{F} \implies A \cap B \in \mathcal{F} \end{split}]

it is easy to prove that a ring on $X$ is also a semiring on $X$

If $X$ is a set, $ \mathcal F \subseteq \mathcal P (X) : \emptyset \in \mathcal F$ and $\mu : \mathcal F \to [0,+\infty] $ then $ \mu $ is a pre-measure on $(X, \mathcal F )$ [or simply on $X$] if and only if (by definition) [\begin{split} &1) \hspace{0.5cm}\mu(\emptyset) = 0 \\ &2) \hspace{0.5cm} A_1,A_2,\dots \in \mathcal{F} \text{ pairwise disjoint , } \bigcup_{n\in\mathbb{N}}{A_n} \in \mathcal{F} \implies \mu\bigg( \bigcup_{n \in \mathbb{N}}{A_n}\bigg) = \sum_{n=1}^{\infty}{\mu(A_n)}\\ \end{split}]

so basically a pre-measure is the same as a measure with the only differences that a pre-measure isn't necessarily defined on a $\sigma$-agebra and the $\sigma$-additivity holds only when the union of the sets is in the domain of the function.

Moreover $\mathcal J ^n$ is the set whose elements are the subset of $\mathbb R ^n$ in the form $[a_1,b_1) \times ... \times [a_n,b_n)$ where $a_i,b_i \in \mathbb R \forall i = 1,...,n$

THIRD PART: Correlation of the question with proposition 6.5 and my attempt to answer the question

Now i will list some theorems of the Schilling required to understand the problem with proposition 6.5 and how it is related to Statement 0.2

Theorem 4.4: Let $X$ be a set, $\mathcal F$ a $\sigma$-algebra on $X$ and $\mu : \mathcal F \to [0,+\infty]$ a function then

$\mu$ is a measure on $X$ if and only if

[\begin{split} &(i) \hspace{0.5cm} \mu(\emptyset)=0 \\ &(ii) \hspace{0.5cm} A \in \mathcal F, B \in \mathcal F , A \cap B = \emptyset \implies \mu(A\cup B)=\mu (A) + \mu (B) \\ &(iii) \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N : A_n \subseteq A_{n+1} \forall n \in \mathbb N \implies \mu(\bigcup_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \end{split}]

Moreover, with the additional hypothesis that $\mu (A) < + \infty \forall A \in \mathcal F$, (iii) can be replaced by either of the following equivalent conditions:

[\begin{split} &(iii') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N \implies \mu(\bigcap_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \\ &(iii'') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N , \bigcap_{n \in \mathbb N} A_n = \emptyset \implies \lim _{n \to +\infty} \mu (A_n) = 0 \end{split}]

end of the Theorem 4.4

Then the book states that Theorem4.4 is still valid in a more general case, regarding pre-measures instead of measures. But, due to the fact that the proof of the Theorem 4.4 requires that the family $\mathcal F$ is closed under finite intersection, union and difference of sets, then this more general theorem is true in the hypotesis that $\mathcal F$ is a ring on $X$.

I could easily prove this more general theorem, I'll copy the statement below under the name of Theorem 0.1

Theorem 0.1 : Let $X$ be a set, $\mathcal F$ a Ring on $X$ and $\mu : \mathcal F \to [0,+\infty]$ a function then

$\mu$ is a pre-measure on $X$ if and only if

[\begin{split} &(i) \hspace{0.5cm} \mu(\emptyset)=0 \\ &(ii) \hspace{0.5cm} A \in \mathcal F, B \in \mathcal F , A \cap B = \emptyset \implies \mu(A\cup B)=\mu (A) + \mu (B) \\ &(iii) \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_n \subseteq A_{n+1} \forall n \in \mathbb N , \bigcup_{n \in \mathbb N} A_n \in \mathcal F \implies \mu(\bigcup_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \end{split}]

Moreover, with the additional hypothesis that $\mu (A) < + \infty \forall A \in \mathcal F$, (iii) can be replaced by either of the following equivalent conditions:

[\begin{split} &(iii') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N , \bigcap_{n \in \mathbb N} A_n \in \mathcal F \implies \mu(\bigcap_{n \in \mathbb N} A_n)=\lim _{n \to +\infty} \mu (A_n) \\ &(iii'') \hspace{0.5cm} A_n \in \mathcal F \forall n \in \mathbb N , A_{n+1} \subseteq A_n \forall n \in \mathbb N , \bigcap_{n \in \mathbb N} A_n = \emptyset \implies \lim _{n \to +\infty} \mu (A_n) = 0 \end{split}]

End of theorem 0.1

Now the problems begin.

The book defined the function $\lambda ^n : \mathcal J ^n \to [0,+\infty)$ that associates each element of $\mathcal J ^n $ to its volume.

The book also proved that $\mathcal J ^n$ is a semi-ring on $\mathbb R^n$ (Proposition 6.4)

My problem with the Proposition 6.5 is that the book wants to prove that $\lambda ^n$ is a pre-measure on $\mathcal J ^n$ using some sort of Theorem 0.1 but the problem is that $\mathcal J ^n$ is not a Ring on $\mathbb R^n$ but only a semi-ring.

Let me be clearer, this is the proposition6.5 of the book:

Proposition 6.5: $\lambda ^n$ is a pre-measure on $\mathcal J ^n$

Proof:

the proof of the book consists in showing that the properties (i) (ii) (iii'') of the theorem 0.1 holds for $\lambda ^n$ and i'm ok with this proof (obviously you need to modify the property (ii) requiring that $A \cup B \in \mathcal J ^n$ ) but the problem is that then he uses a not well specified version of the theorem 0.1 to conclude that $\lambda ^n $ is a pre-measure on $\mathbb R^n$. But, again, you can't use theorem0.1 because $\mathcal J ^n$ is not a Ring on $\mathbb R^n$ but only a semi-ring.

What was my attempt to solve this issue?

Firstly I have formulated a version of the theorem suitable for semi-rings which goes under the name of Statement 0.2 (You've read it in the first part of the post)

I will shortly show how i proved the first implication of the Statement 0.2 (The implication "from the left to the right"):

First of all I proved the following theorems:

Theorem 0.3 : Let $X$ be a set and $\mathcal F \subseteq \mathcal P (X)$ then it does exist the smallest Ring on $X$ containing $\mathcal F$ and it's denoted with the simbol $R(\mathcal F)$

Theorem 0.4 : Let $X$ be a set and $\mathcal F$ a semi-ring on $X$ then $R(\mathcal F)$ is the set of all the finite disjoint unions of elements of $\mathcal F$

Theorem 0.5 : Let $X$ be a set and $\mathcal F$ a semi-ring on $X$ and $\mu : \mathcal F \to [0,+\infty] $ a pre-measure on $X$ then there exists and is unique an extension of $\mu$ to $R(\mathcal F)$ such that this extension is a pre-measure.

Proof of the implication from the left to the right of Statement 0.2:

if $\mu$ is a pre-measure on the semi-ring $\mathcal F$ then denoted the unique extension of $\mu$ to $R(\mathcal F)$ still with $\mu$ we get the assertion thanks to Theorem0.1

Answer 1

In R. L.Schilling. Measure, Integrals and Martingales, Cambridge, First Edition 2005, there is no explicit version of Theorem 0.2. as stated by the OP. This is also the case for the second edition. The main result that resembles the OP's Theorem 0.2 is Theorem 4.4 on page 24 (first edition) which is stated under the assumption that $(X,\mathcal{F})$ is a measurable space (i.e. $\mathcal{F}$ is a $\sigma$-algebra of subsets of $X$):

Theorem 4.4: Suppose $(X,\mathcal{F})$ is a measurable space. A map $\mu:\mathcal{F}\rightarrow[0,\infty)$ is a measure iff

• (i) $\mu(\emptyset)=0$,
• (ii) $\mu(A\cup B)=\mu(A)+\mu(B)$ for all $A,B\in\mathcal{F}$ with $A\cap B=\emptyset$,
• (iii) $\mu\big(\bigcup_nA_n\big)=\lim_n\mu(A_n)$ for all monotone nondecreasing sequence of set $(A_n:n\in\mathbb{N})\subset\mathcal{F}$.

In addition, if $\mu(A)<\infty$ for all $A\in\mathcal{F}$, then (iii) is equivalent to either of the follwing conditions:

• (iii)' $\mu\big(\bigcap_nA_n\big)=\lim_n\mu(A_n)$ for any monotone nonincreasing sequence of of sets $(A_n:n\in\mathbb{N})\subset \mathcal{F}$.
• (iii)'' $\lim_n\mu(A_n)=0$ for any monotone nonincreasing sequence of of sets $(A_n:n\in\mathbb{N})\subset \mathcal{F}$ such that $\bigcap_nA_n=\emptyset$.

The confusion of the OP, I believe, stems from a remark (op. cit. Remark 4.5, page 24 of first edition) in the book that states that Theorem 4.4 and its precursor Proposition 4.3 (which I don't state but only mention that it is related to necessary conditions concerning monotonicity and subaditivity properties of any measure $\mu$) are valid when $\mu$ is a premeasure on $\mathcal{F}$ where $\mathcal{F}$ is not a $\sigma$-algebra, with the obvious re-wordings. Here is the precise statement in the book:

Remark 4.5: With some obvious rewordings, P4.3 and T4.4 are still valid for pre-measures, i.e. for families $\mathcal{F}$ which are not $\sigma$-algebras. Of course, one has to make sure that $\emptyset\in\mathcal{F}$, and that $\mathcal{F}$ is stable under finite unions, intersections and differences of sets$^1$ (for P4.3) and, for T4.4, that increasing and decreasing sequences of the sets under consideration have their limits in $\mathcal{F}$. The proofs are literally the same.

The superscript $^1$ in the textbook makes it clear that author refers to the case where $\mathcal{F}$ is a ring. In such case, Theorem 0.2. as stated by the OP holds and is not complicated to prove.

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Now, If we consider a semiring $\mathcal{R}$ on $X$ and $\mu$ is a premeasure (as defined in the OP) on $\mathcal{R}$, conditions (i)-(iii) still hold. This I state as a Theorem R:

Theorem R: Suppose $\mathcal{R}$ is a semi-ring on $X$. If $\mu$ is a premeasure on $\mathcal{R}$, then

• (i) $\mu(\emptyset)=0$,
• (ii) $\mu(A\cup B)=\mu(A)+\mu(B)$ for all $A,B\in\mathcal{R}$ with $A\cap B=\emptyset$ and such that $A\cup B\in\mathcal{R}$,
• (iii) $\mu\big(\bigcup_nA_n\big)=\lim_n\mu(A_n)$ for all monotone nondecreasing sequence of sets $(A_n:n\in\mathbb{N})\subset\mathcal{R}$ such that $\bigcup_nA_n\in\mathcal{R}$.
• (iv) $\mu$ is countably sub-additive on $\mathcal{R}$, that is $\mu\big(\bigcup_nB_n\big)\leq\sum_n\mu(B_n)$ for any sequence $(B_n:n\in\mathbb{N})\subset\mathcal{R}$ such that $\bigcup_nB_n\in\mathcal{R}$.

Comments:

• Conditions (i), (ii) and (iv) in Theorem R are the minimal ingredients to extend $\mu$ to a $\sigma$-algebra that contains the semiring $\mathcal{R}$ -This is the Carathéodory extension theorem (op. cit. Theorem 6.1 page 37 on first edition). Once the extension is achieved, then the equivalence between (iii) and either (iii)' or (iii)'' under finiteness follows by virtue of Theorem 4.4.
• The proof of Carathéodory's extension theorem relies on the construction of an outer measure $\mu^*$ by exploiting the countably additivity of the premeasure $\mu$ (i.e., $\sum_n\mu(A_n)=\mu\big(\bigcup_nA_n\big)$ for any sequence $(A_n:n\in\mathbb{N})\subset\mathcal{R}$ which is pairwise disjoint, and for which $\bigcup_nA_n\in\mathcal{R}$) and the countably subadditivity of $\mu$(i.e., condition (iv) of Theorem R).
• The labors of Carathéodory allows one to construct the Lebesgue (and more generally Lebesgue-Stieltjes) measure on the real line starting from the ring of intervals $\mathcal{R}=\{(a,b]: -\infty<a<b<\infty\}$ and $\mu((a,b]):=F(b)-F(a)$ where $F$ is monotone-nondecreasing and right continuous ($F(x)=x$ yields Lebesgue's measure is the case discussed in Theorem 6.5 of thaformentioned textbook).

Proof of Theorem R Assume $\mu$ a premeasure on $\mathcal{R}$ (in the sense outlined by the OP). Parts (i) and (ii) hold trivially: (i) by definition of premeasure and (ii) by taking $A_1=A$, $A_2=B$, and $A_n=\emptyset$ for all $n\geq 3$.

For any sets $A, B$, $A=(A\setminus B)\cup (A\cap B)$. Thus, if $A,B\in\mathcal{R}$,$A\setminus B$ is the finite union of pairwise disjoint sets, say $D_1,\ldots, D_k$, in $\mathcal{R}$. If $\mu(A)<\infty$ or $\mu(A\cap B)<\infty$, it follows from part(ii), that $$\sum^k_{j=1}\mu(D_j)=\mu(A)-\mu(A\cap B)$$ which shows that the sums of the premeasures of the finite pairwise disjoint components of $A\setminus B$ does not depends of how the $A\setminus B$ is decomposed as a finite pairwise disjoint union of ell ents in $\mathcal{R}$. Thus, the quantity $\bar{\mu}(A\setminus B):=\sum^k_{j=1}\mu(D_j)=\mu(A)-\mu(A\cap B)$ is well defined whenever $\mu(A)<\infty$ or $\mu(A\cap B)<\infty$.

Another consequence of parts (i) and (ii) is that if $A\in\mathbb{R}$ and $B\subset A$ is the finite union of pairwise disjoint sets, say $B_1,\ldots, B_k\in\mathcal{R}$, then $$\sum^k_{j=1}\mu(B_k)\leq \mu(A)$$ Indeed, it can easily proved by induction in the $k$ that $A\setminus \bigcup^k_{j=1}B_j$ is the finite union of pairwise disjoint sets, say $C_1,\ldots,C_m\in\mathcal{R}$. Thus $\{B_j,C_\ell:1\leq j\leq k,\,1\leq \ell\leq m\}$ is a collection of pairwise disjoint sets in $\mathcal{R}$ and so, $$A=\Big(\bigcup^k_{j=1}B_j\Big)\cup\Big(\bigcup^m_{\ell=1}C_\ell\Big)$$ Hence $$\mu(A)=\sum^k_{j=1}\mu(B_j)+\sum^m_{\ell=1}\mu(C_\ell)\geq \sum^k_{j=1}\mu(B_j)$$

Part (iii) is the only one that requires a little work. Suppose $B_n$ is an increasing sequence in $\mathcal{R}$, i.e. $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$, such that $B:=\bigcup_nB_n\in\mathcal{R}$. Since $\mu(B_{n-1})\leq \mu(B_n)$, it is enough to assume that all $\mu(B_n)<\infty$.

Let $A_1:=B_1$, and $A_n=B_n\setminus B_{n-1}$. Clearly the collection $\{A_n\}$ is pairwise disjoint, and $B=\bigcup_nA_n$. Each $A_n$ is the finite union of disjoint elements, say $C_{n1},\ldots,C_{nk_n} $, of $\mathcal{R}$. Then the collection $\{C_{n,j}:n\in\mathbb{N}, \,1\leq k_j\leq n\}$ is a countable pairwise disjoint collection of sets in $\mathcal{R}$. Since $$ A_n=\bigcup^{k_n}_{j=1}C_{n,j},\qquad B=\bigcup^\infty_{n=1}\big(C_{n1}\cup\ldots\cup C_{nk_n}\big)$$ we have that by part (ii) that $\bar{\mu}(A_n)=\sum^{k_n}_{j=1}\mu(C_{nj})=\mu(B_n)-\mu(B_{n-1})$, where $B_0=\emptyset$. Since $\mu$ is a premesure $$\begin{align} \mu(B)&=\sum^\infty_{n=1}\sum^{k_n}_{j=1}\mu(C_{nj})=\sum^\infty_{n=1}\bar{\mu}(A_n)\\ &=\mu(B_1)+\sum^\infty_{n=2}\mu(B_n)-\mu(B_{n-1})=\lim_n\mu(B_n) \end{align}$$

(iv) Follows similarly. It suffices to consider the case where $\mu(B_n)<\infty$ for all $n$. Let $A_1=B_1$ and $A_n=B_n\setminus A_{n-1}$. By induction each $A_n$ is be written as the finite union of sets, say $D_{n1},\ldots, D_{nk_n}\in \mathcal{R}$. Then $$A_n=B_n\setminus A_{n-1}=\bigcup^{k_n}_{j=1}D_{nj}\subset B_n,\qquad B=\bigcup^\infty_{n=1}\big(D_{n1}\cup\ldots\cup D_{nk_n}\big)$$ As $\mu$ is a premeasure and $\{D_{nj}:n\in\mathbb{N},\,1\leq j\leq k_n\}$ is a countable collection of pairwise disjoint sets in $\mathcal{R}$, we have that $$ \mu(B)=\sum^\infty_{n=1}\Big(\sum^{k_n}_{j=1}\mu(D_{nj})\Big)\leq\sum^\infty_{n=1}\mu(B_n)$$

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I leave to the OP the task to convince himself/herself- most like this is already the case- that when $\mathcal{R}$ is a ring, then Theorem 4.4 holds when $\mu$ is a premeasure on $\mathcal{R}$. Only sufficiency requires a proof, but this becomes easy as $A\setminus B$, and $A\cup B$, and $A\cap B$ belong to $\mathcal{R}$ whenever $A,B\in\mathcal{R}$.

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Edit:

• In the first edition, the author does actually refer to (iii)' in his construction of Lebesgue measure on $\mathbb{R}^n$. He argues that (a) $\lambda:\mathbb{R}\rightarrow\mathbb{R}$, where $\mathcal{R}=\{\prod^n_{j=1}[a_j,b_j): a_j<b_j\}$, given as $\lambda(\prod^n_{j=1}[a_j,b_j))=\prod^n_{j=1}(b_j-a_j)$ satisfies condition $(iii)'$; (b) this in turn implies (iii); (c) and so $\lambda$ is a pre-measure by the converse of Theorem 4.4. That logic is not correct for as the OP pointed out, $\mathcal{R}$ is not a ring.

• In the second edition of the book (pages 46-50), the construction if the Lebesgue measure on $\mathbb{R}^n$ is rewritten and no reference to (iii)' is used. Conditions (i), (ii) along with countably additivity are derived in his presentation by topological arguments.