Uniform Logical New Proofs for the Daniell–Stone Theorem and the Riesz Representation Theorem

Abstract

Integration logic is a logical (model theoretic) framework for studying measure and probability structures by logical means. The Daniell–Stone theorem for Daniell integrals and Riesz representation theorem are two important classical results in analysis concerning existence of measures with certain properties. Many proofs of the Riesz representation theorem can be found in the literature from elementary ones which use ordinary techniques from measure theory and analysis, to more sophisticated ones that highly employ techniques from other fields of mathematics such as nonstandard analysis. There are also a few proofs for the Daniell–Stone theorem. This paper pursues two main goals. One is to give novel and uniform proofs for both these theorems using some ideas from logic. The second and maybe even more important one is to try to reveal more the power of the logical methods in mathematics in particular measure theory, and make stronger connections between analysis and logic. Our proofs are uniform in the sense that they are based on the same general idea and rely on the application of the same technical tool from logic to measure theory, namely the logical compactness theorem. We use the setting of “integration logic” mentioned above, elaborate it and use its expressive power and a version of the compactness theorem holding in it to prove our results. The paper is mostly written for general mathematicians, in particular the people active in logic or analysis as the main audience. It is self-contained and the reader does not need to have any advanced prerequisite knowledge from logic or measure theory.

Appendix A: Proof of Preliminary Lemmas

1.1 Proof of Lemma 2.9

Proof

We first start to prove the lemma for just one f and one U. Fix $f \in {\mathcal {A}}$ and $\alpha \in {\mathbb {R}}$. For every $n \in {\mathbb {N}}$ set

$$\begin{aligned} g^1_n(x):= & {} n\left( \min \left( f(x),\alpha +\frac{1}{n}\right) -\min (f(x),\alpha )\right) \\ g^2_n(x):= & {} n\left( \max (f(x),\alpha )-\max \left( f(x),\alpha -\frac{1}{n}\right) \right) \\ h^1_n(x):= & {} n\left( \min (f(x),\alpha )-\min \left( f(x),\alpha -\frac{1}{n}\right) \right) \\ h^2_n(x):= & {} n\left( \max (f(x),\alpha +\frac{1}{n}\right) -\max (f(x),\alpha )). \end{aligned}$$

Then, it is not very difficult to see that the sequences of functions $(g^1_n)_{n<\omega }$ and $(g^2_n)_{n<\omega }$ increase pointwise to $\chi (\{x:\alpha < f(x)\})$ and $\chi (\{x: f(x) < \alpha \})$, respectively. Also the support of each $g^1_n$ and each $g^2_n$ is a subset of $\{x:\alpha < f(x)\}$ and $\{x: f(x) < \alpha \}$, respectively. Similarly, $(h^1_n)_{n<\omega }$ and $(h^2_n)_{n<\omega }$ decrease pointwise to $\chi (\{x:\alpha \leqslant f(x)\})$ and $\chi (\{x: f(x) \leqslant \alpha \})$, respectively while the support of each $h^1_n$ and each $h^2_n$ is a superset of $\{x:\alpha \leqslant f(x)\}$ and $\{x: f(x) \leqslant \alpha \}$, respectively. Also $g^1_n$s, $g^2_n$s, $h^1_n$s and $h^2_n$s are functions in ${\mathcal {A}}$ taking values in [0, 1]. So, the statement is proved for one f and one U of the form $(-\infty ,\alpha ), (\alpha ,\infty ), (-\infty ,\alpha ],$ and $[\alpha ,\infty )$.

For $\alpha <\beta$, if $(g_n)_{n<\omega }$ and $(h_n)_{n<\omega }$ are sequences of functions obtained above increasing to $\chi (\{x: f(x)<\beta \})$ and $\chi (\{x: \alpha < f(x)\})$, respectively, then $(g_n\wedge h_n)_{n<\omega }$ increases to $\chi (\{x: \alpha< f(x)<\beta \})$ and the support of each $g_n\wedge h_n$ is a subset of $\{x: \alpha< f(x)<\beta \}$. Similarly, if $(g_n)_{n<\omega }$ and $(h_n)_{n<\omega }$ are sequences of functions obtained above decreasing to $\chi (\{x: f(x)\leqslant \beta \})$ and $\chi (\{x: \alpha \leqslant f(x)\})$, respectively, then $(g_n\wedge h_n)_{n<\omega }$ decreases to $\chi (\{x: \alpha \leqslant f(x)\leqslant \beta \})$ and the support of each $g_n\wedge h_n$ is a superset of $\{x: \alpha \leqslant f(x)\leqslant \beta \}$. These prove the statement of lemma for one f and one U of the form $(\alpha ,\beta )$ and $[\alpha ,\beta ]$. It is worth mentioning that if $(g_n)_{n<\omega }$ is a sequence which increases to $\chi (\{x: f(x)<\beta \})$, then $(1-g_n)_{n<\omega }$ decreases to $\chi (\{x:\beta \leqslant f(x)\})$ and vice versa.

Let $U_1,\ldots ,U_m$ be open intervals of the above forms and $f_1,\ldots ,f_m \in {\mathcal {A}}$. Assume that for each $i=1,\ldots ,m$, the sequence $(g^i_n)_{n<\omega }$ is the sequence of functions increasing to $\chi (\{x:f(x) \in U_i\})$ obtained in the way that is explained above. Then the sequences of [0, 1]-valued functions $(g^1_n \wedge \cdots \wedge g^m_n)_{n<\omega }$ and $(g^1_n \vee \cdots \vee g^m_n)_{n<\omega }$ increase to $\chi (\bigcap _{i=1}^m f_i^{-1}(U_i))$ and $\chi (\bigcup _{i=1}^m f_i^{-1}(U_i))$, respectively. Moreover, since the support of each $g^i_n$ is a subset of $\{x:f(x) \in U_i\}$, then the supports of $(g^1_n \wedge \cdots \wedge g^m_n)_{n<\omega }$ and $(g^1_n \vee \cdots \vee g^m_n)_{n<\omega }$ are subsets of $\bigcap _{i=1}^m f_i^{-1}(U_i)$ and $\bigcup _{i=1}^m f_i^{-1}(U_i)$, respectively. Similarly, the same statements hold when we replace being open by being closed for $U_i$s, increase to decrease and subset by superset. In particular, in this case, one sees that the proof works when some of $U_i$s are single real numbers since every real number can be seen as a closed interval. $\square$

1.2 Proof of Lemma 2.13

Proof

We prove the claim for $t=2$, namely, for two functions $f_1$ and $f_2$. The general case is similar. Also the case $t=1$ would be a special case of $t=2$ for when $f_1=f_2$. Recall that since I is a positive operator, we have $I(1_X) \not =0$. Define

$$\begin{aligned} Z_1:=\{x \in (r,s): x \text { is not an inessential value of } f_1 \text { with respect to } I\}, \end{aligned}$$

and

$$\begin{aligned} Z_2:=\{x \in (r,s): x \text { is not an inessential value of } f_2 \text { with respect to } I\}, \end{aligned}$$

where we remind the reader that the notion of inessential value of a function with respect to I was defined in Definition 2.12. It is enough to show that $P \not = \emptyset$ where $P:=(r,s) {\setminus } (Z_1 \cup Z_2)$. We prove an even stronger statement and show that $Z_1 \cup Z_2$ is countable. Assume for contradiction that $Z_1 \cup Z_2$ is uncountable. Hence, at least one of $Z_1$ or $Z_2$, is uncountable. Without loss, we may assume that $Z_1$ is uncountable. For each $u\in (r,s)$, let $(h^u_n)_{n < \omega }$ be the decreasing sequence corresponding to u for $f_1$ (see Definition 2.10) and let $S(u):=\lim _{n \rightarrow \infty } I(h^u_n)$. So, for every $u\in Z_1$, since u is not an inessential value of $f_1$ with respect to I, we have $S(u)>0$. Thus, since $Z_1$ is uncountable, for some $m\in {\mathbb {N}}$, there exists an infinite subset V of $Z_1$ such that $S(u)>\frac{I(1_X)}{m}$ for each $u\in V$. Let $v_1,\ldots , v_{2m}\in V$ be distinct. So in particular $S(v_i)>\frac{I(1_X)}{m}$ for each $i=1,\ldots ,2m$. Using Remark 2.11, for each $i=1,\ldots ,2m$ and n, the support of each $h^{v_i}_n$ is a subset of $f_1^{-1}(v_i-\frac{1}{n},v_i+\frac{1}{n})$. Thus, for each $i=1,\ldots ,2m$, we can choose a function $g_i$ from the sequence of functions $(h^{v_i}_n)_{n < \omega }$ in such a way that at the end of the selections, the supports of the selected $g_i$s are mutually disjoint. Since $h^{v_i}_n$s take values in [0, 1] (as was mentioned after Remark 2.11), for every i we have $0 \leqslant g_i(x) \leqslant 1$ (for every x). Now the function $g:=\sum _{i=1}^{2m}g_i$ belongs to ${\mathcal {A}}$ and $0 \leqslant g(x) \leqslant 1$ for every x. We remind that I is positive linear. Therefore, $0 \leqslant I(g) \leqslant I(1_X)$. On the other hand, for each i, since $(h^{v_i}_n)_{n < \omega }$ is a decreasing sequence, by positive linearity of I, the sequence $(I(h^{v_i}_n))_{n < \omega }$ is decreasing which follows that $I(g_i) \geqslant \lim _{n \rightarrow \infty } I(h^{v_i}_n)=S(v_i)$. So we have $I(g)=\sum _{i=1}^{2\,m}I(g_i) \geqslant \sum _{i=1}^{2\,m} S(v_i) \geqslant 2\,m.\frac{I(1_X)}{m}=2I(1_X)$. Combining above facts, we get $I(1_X)=0$ which is a contradiction. So, $Z_1 \cup Z_2$ is countable. Therefore, $P \not =\emptyset$ and we are done. $\square$

1.3 Proof of Lemma 2.14

Proof

For convenience, we call a measurable subset D of Y a type I subset if there is some $f \in {\mathcal {K}}$ such that $D=f^{-1}(0,\infty )$. We call D a nice type I subset if there exists such f with the additional property that $\mu (f^{-1}(\{0\}))=0$. Similarly, we call a measurable subset D a type II subset if there is some $f \in {\mathcal {K}}$ such that $D=f^{-1}[0,\infty )$. Since $(f^{-1}(0,\infty ))^c=(-f)^{-1}[0,\infty )$, using disjunctive normal form representation of members of Boolean algebras, every element of ${\mathcal {C}}$, such as $U_n$s, can be represented as a finite union of finite intersections of type I or type II sets. We call such representation of any $U_n$ a good representation of it.

Let $\epsilon >0$ be fixed. Also for each n, fix a good representation of $U_n$. For each n, let $U'_n$ be the modification of $U_n$ by replacing the clauses of the form $f^{-1}[0, \infty )$ in the mentioned good representation of $U_n$ by some bigger sets $f^{-1}(-\delta _n,\infty )$ for some small enough positive real numbers $\delta _n$s, in such a way that the difference between the sum of the measures of $U'_n$s, namely $\sum _{n<\omega }\mu (U'_n)$, and that of $U_n$s, namely $\sum _{n<\omega }\mu (U_n)$, is not more than $\epsilon$. It is easily seen that the family of $U'_n$s is a covering of X. We note that for every $f \in {\mathcal {K}}$ and real number $\delta$, we have $f^{-1}(-\delta ,\infty )=f'^{-1}(0,\infty )$, where $f'=f+\delta$. Since ${\mathcal {K}}$ is a vector lattice, obviously $f' \in {\mathcal {K}}$. Similarly, for every $f_1,f_2 \in {\mathcal {K}}$, we have $f_1^{-1}(0,\infty ) \cap f_2^{-1}(0,\infty )=g^{-1}(0,\infty )$ and $f_1^{-1}(0,\infty ) \cup f_2^{-1}(0,\infty )=h^{-1}(0,\infty )$ where $g= f_1 \wedge f_2$ and $h=f_1 \vee f_2$, respectively. Again, since ${\mathcal {K}}$ is a vector lattice, clearly $g,h \in {\mathcal {K}}$. Now using these facts, it is not hard to see that for each n, we have $U'_n=f_n^{-1}(0,\infty )$ for some $f_n \in {\mathcal {K}}$. So $U'_n$s are type I subsets of Y. Let ${\mathcal {U}}$ denote the family of $U'_n$s. Recall that ${\mathcal {U}}$ is a covering of X. If all $U'_n$s are nice type I subsets, then we are done. Otherwise, $I \not = \emptyset$ where $I \subseteq {\mathbb {N}}$ is the set of all $n \in {\mathbb {N}}$ such that $U'_n$ is not nice (so, for each $n \in I$, we have $\mu (f_n^{-1}(\{0\}))>0$). In that case, in the following procedure, we will replace some members of ${\mathcal {U}}$ with some families of subsets of Y in such a way that after these replacements, our new ${\mathcal {U}}$ still remains a covering of X, the sum of the measures does not differ with more than $\epsilon$ and moreover, our new ${\mathcal {U}}$ only contains nice type I subsets of Y. The procedure is as follows.

Corresponding to each $n \in I$, we can find a sequence $a^n_1,a^n_2,\ldots$ of distinct positive real numbers decreasing to 0 such that the followings hold.

• (i) $\mu (f_n^{-1}(\{a^n_j\}))=0$ for each $j<\omega$.

• (ii) $\mu (U''_n)$ does not differ from $\mu (U'_n)$ with more than $\frac{\epsilon }{4^n}$, where $U''_n:=f_n^{-1}(a^n_2,\infty )$,

• (iii) $\sum _{j=1}^{\infty } \mu (U''_{n,j}) \leqslant \frac{\epsilon }{4^n}$, where for every $j<\omega$, $U''_{n,j}:=f_n^{-1}(a^n_{j+2},a^n_j)$.

It is easy to see that for each $n \in I$, $U''_n=p_n^{-1}(0,\infty )$ where $p_n:=f_n-a^n_2$. Also for $j<\omega$, we have $U''_{n,j}=f_n^{-1}(a^n_{j+2},a^n_j)=(h_{n,j})^{-1}(0,\infty )$, where $h_{n,j}:=(f_n-a^n_{j+2}) \wedge (a^n_j-f_n) \in {\mathcal {K}}$. So, $U''_{n,j}$s and $U''_n$s are all type I subsets of Y. Also (i) guarantees that they are all nice type I subsets. Now for each $n \in I$, we remove the member $U'_n$ from the family ${\mathcal {U}}$ and instead, add $U''_n$ and all $U''_{n,j}$ (for each $j<\omega$) to ${\mathcal {U}}$. Now, this new ${\mathcal {U}}$ is a family of nice type I subsets of Y. Furthermore, for each $n \in I$, we have

$$\begin{aligned} \left( \bigcup _{j<\omega } U''_{n,j}\right) \bigcup U''_n=\left( \bigcup _{j<\omega } f_n^{-1}(a^n_{j+2},a^n_j)\right) \bigcup f_n^{-1}(a^n_2,\infty )=f_n^{-1}(0,\infty )=U'_n. \end{aligned}$$

So, having this, it is not hard to see that ${\mathcal {U}}$ is still a countable covering of X. Also (ii) and (iii) guarantee that for each $n \in I$, sum of the measures of $U''_n$ and all $U''_{n,j}$s does not differ from $\mu (U'_n)$ with more than $\frac{\epsilon }{2^n}$. So, the sum of measures of members of our new ${\mathcal {U}}$ does not differ from sum of measures of $U_n$s with more than $\epsilon$. $\square$