3— Multiplicative Systems in Several Variables I, II, III: With C.J. Everett (LA-683, June 7, 1948) (LA-690, June 11, 1948) (LA-707, October 28, 1948)

### I— A Remark On Measure Theory

1. It is convenient for our purposes to have a simple set of axioms on which measure theory may be shown to rest. To this end, let r = {z} be a set of points z, and I = {i} a class of special subsets i of F called intervals. We demand that the entire set F and the empty set X be intervals. Denote by J the class of all subsets S = Ei of F which are set sums of a finite or countable number of intervals. (All set sums hereafter are understood to operate on finitely or countably many summands. The notation La is used to indicate that the summands are mutually disjoint.)

We suppose that intervals satisfy the following axioms: I 1. Every set S = Zi of J can be represented as a sum Ej of disjoint intervals. I 2. The complement i' = -i of an interval i in J. I 3. The set product ij of two intervals i and j is an interval k.

Moreover, we assume m 1. To every interval i is assigned a non-negative real number m(i)>0 called its measure. m 2. M(F) = 1, m(0) = 0. m 3. If i = Ej, where the i and j are intervals, then m(i) = mr(j).

An additive class C of subsets of F is one such that C 1. All intervals belong to C. C 2. A is in C whenever all sets A are in C. C 3. If A is in C, so is A'.

The Borel sets are those common to all additive classes, hence, themselves form an additive class.

We shall define a property of subsets U of F called measurability, and prove that the class of measurable sets is additive. Simultaneously we define a measure m(M) for every measurable set M and prove

M 1. 0 <m(M)< 1 for M measurable. M 2. m ( M) =Em(M) for M measurable. M 3. If M is an interval, the measure assigned to M as a measurable set coincides with its intitially given measure.

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2. In stating the above axioms, we have attempted to focus attention on the fundamental assumptions. It remains to show that the essential theorems follow from our axioms. This we do in the remainder of this chapter, for the sake of completeness, following the classical arguments for the real line1 . We note first the following properties of intervals: I 4. E S is in J whenever all the summands S are in J. For E S is an at most countable sum of at most countable sums of intervals. N

I 5. I1 i, then S' is in J whenever all N factors are in J. It suffices to see this for N = 2. Hence let S1 = Zim, S2 = j. Then S1S2= i,mjn = E (imjn). By I 3, imjn is an interval, so S1S2 is in J, the range (m, n) being at most countable. 6. If S = i, then S' is in J. N For S' = i', each i' is in J by I 2, and thus S' is in J by I 5. 1

3. We now assign a measure to every set S of J. Note first that if i D Si in, then i = Zin±+i(in) . By I 6 and I 5, the latter set is in J, so by I 1, we may write i= in i-n+ Ejm, and by m 3, and m 1, m(i = E (in) + E T(j) > E m (in).

Hence, if in is any sum of disjoint intervals, we have for every N, F D i in and hence 1 > Z m (in). Thus E m (in) exists, less than or equal to 1, for every such disjoint sum.

Now suppose -i,n = Zjn. Then im = im(Zjn)= E (imjn) ,nkmn, kmn being the intervalim, jn. (See 1 3.) Hence by m 3, m (im) = En m (k,,) and m(im) = m En m(kmn). Similarly Em(j,) = En Em m(kmrl). Now the double sum mn, m (kmn) converges to a limit < 1 by the preceding paragraph since the kmn are disjoint. Hence the iterated sums are both equal to this limit and hence to each other. Thus E m(im)= - m (jn).

It is therefore clear that, if S is any set of J we can write S = Zi and define m(S) = rm(i) unambiguously. Clearly 0 <m(S), and if S is an interval i, m(S) coincides with the initially given measure m(i).

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4. We establish in this section some properties of the measure m(S) for sets S of J. First, m 4. m(S) =m(S). For, write Sm =i,,mn. Then m(ZS) m( n imn)En,m (im,) = Zm n m (in,) = m (S,) . Next m 5. If S and T are in J and S c T, then m(S)<m(T).

For, let S = E im and define Sn =m. Then T = S +TS', the summands being disjoint and in J. Hence, by m 4, m(T) = m (S) + m(TS')> rn(Sn) =m(i,,). Hence m(S) = m(i) < m(T). Thus follows m 6. For all S in J, 0 <m(S)< 1.

Next we prove m 7. For arbitrary sets S of J, disjoint or not, m(Z£S)<Zm(S), where the latter sum may of course be infinite.

Write S = E imn. Then ZSm = m,i,,, = ip = il4i1i 2+ili2" 3i . , the latter summands being disjoint and in J. Hence, by m 4, m 5, m(Sm,) = m(il) +(i'i2) + ... <(il)+m(i 2) + ...= r m(ip) = m,, 7n(imr,) = nErE m (imn) = Emm(S), the latter possibly being infinite.

Finally we have m 8. If S and T are in J, then m(S + T) + m(ST) = m(S) + m(T). First suppose S and T are each finite sums of intervals. Write S + T = S- (T- ST) ST- (T - ST) = T

where S and ST are in J, and, ST being a finite sum of intervals, also T - ST = T(ST)' is in J. Thus by m 4, m(S + T) = m(S) + m(T - ST) m(ST) + m(T - ST) = m(T), and addition yields the equation of the theorem.

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Now let S = Zim, T = ij,, where we make either sum terminate in 0 in case it happens to be finite. Define Sn and Tn as the corresponding partial sums through the first n terms. Then by the first part of the proof we have for every n, n (Sn + Tn)+m(SnTn) = m(Sn)+mr(Tn). But limm(Sn) = limTli, = mnS), and limm(T,) = m(T). (Recall that m(0) = O).) Also, S + T = im + j, = E (im +jm) S, = S1-+SS2i   Thus m(S + T) = m (S1) + m (S2) +...= lim, (m (SI)m(S9S2)+...+ m(S   S S)) =limnm(Si+SS2 +-... + S ... n) = limn (1 + ...+Sn) =limn m (Sn+ T) . Finally ST = n im , = ,jimj n =mmnn = kj = mn = k +k k2 +.. where the intervals kmn = kp are listed in sequential order by upper squares: (1,1), (1,2), (2,1), (2,2); .... Then m(ST)= rn(kl) + m (klk2) + .. . limn2 (m (kl)+... + mn (ki ... k2_lkn2)) = lim,n mk+... ... (kl + +k k _)lk 2) = lim m (k + m (SnTn).

Hence we obtain m 8 in general by taking limits of both sides of the finite relation.

5. Let U be an arbitrary subset of F and define the outer measure O(U) = glb(m(S); U c S c J), and the inner measure I(U) = I - O(U'). From m 6 we have for all subsets U, m 9. O <O(U)< 1, and O <I(U)< 1.

Moreover, if U1 c U2 , every S of J which contains U2 also contains U1, so the numbers m(S) defining O(U 2) form a subset of those defining O(U1), and hence m 10. If Ui c U2 , then O(U1) O0(U2 ) and I(UI) < I (U2).

For every Si D U, S 2 D U', SI, S 2, in J, we have S1+S 2 D U+U' = F and M (F) = 1 = m (S + S2) <m (S)+m(S2 ). Fix S1 U. Then 1 - rn (S) < m (S2 ) for all S2 D U', so 1 - m(S)<O()1 U') U')< m(SI) for all S, DU, and 1 - O(U')<O(U). Thus follows m 11. For every subset U, I(U)<O(U).

Now let {Un } be a sequence of arbitrary subsets Un of F. Fix e > 0. For every 7n there exists an Sn in J such that Sn D Un and n (S,) < O(U) +e/2n. Hence Sn D EU,, Sn is in J, and m(r Sn)<Em(Sn) <ZO(Un)+e. Thus O(EUn )<m(Sn,) < EO(Un) +e for every e > 0 and we have

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m 12. For arbitrary subsets U, O(0 U)<0(U). Suppose U 1, U 2 are disjoint. For every S1DUl, S2 D U2, S1,S2 in J, we have Sl + S2 D U{+ U2 = (U1U2)' = 0 = so m (S + S2 )= 1. Also, (Ul +U2)' = UlU C S1 S2 is in J, hence O (U +U 2)'< gib (m(S1S2)). Then I (Ul + U2) - I (Ul)- I (U2) = 1 - O (U1 + U2)'- I - (U)- 1O(U2) = O(U{) + O(U2) -0(Ul + U2)' - 1 > glb m (S1) + glb m (S2)gib m (S1S2) -1 gib (m(S1) + m(S2) -m (SiS2 )) - = gib m (S1 + S2) - 1= 1 - 1 = 0. We therefore obtain m 13. UIU2 =0 implies I (Ul) + I(U2) < I (Ul + U2 ).

Generalizing this, we have I(ZU)> I(U1)+ I(E2 U) >I(U1)+ I (U) + (I Z 3U)> ... >I() +...+(UN) +I( EN+) > 1 I(U) for all N, and m 14. (I(U)<I(L U), for disjoint summands.

6. We say a set M is measurable in case I(M) = O(M) and denote by K the class of all measurable sets M. For a measurable set M we define a measure m(M) = I(M) = O(M). From m 9 follows m 15. For a measurable set M, 0 <m(M)< 1.

Let i be an interval. Since i is in J and contains itself, we have O(i)<m(i) since O(i) is a lower bound. Now suppose i c S E J. Then m(i)<m(S) and m(i)<O(i) since O(i) is a greatest lower bound. Hence rn(i) = O(i). Since i' is in J, we may write F = i+ jn, so m(F) = 1 = m(i) + m(i'). Now 1(i) = 1-O(i') = 1 - O(Ejn) > 1-E (O(j,)) 1 - Em(jn) 1 - m(i') = m(i) = O(i)>I(i), so we have established m 16. Every interval is measurable with O(i) = I(i) = m(i), its initially given measure.

Also, we see m 17. If Mn are disjoint measurable sets, then E M, is measurable and m(ZMn) = Emr(M,). For, Em(M.) = I (M,)<I( Mn) <O(M,) < 0O(M,)= Em (M,), and I(EMn,) = O(EM) = E (Mn).

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As a corollary we get m 18. Every set S = i of J is measurable and its measure as such coincides with that previously defined. If M is measurable, I(M') = 1-O(M) = 1-I(M) = 1-(1-O(M')) O(M'), and so m 19. The complement of a measurable set is measurable and m(M) + m(M') = 1.

Finally , we have to prove m 20. A sum of measurable sets, disjoint or not, is measurable. Let M1, M 2 be measurable, and hence also M,, M2. Fix e > 0. Then there exist sets S1,S2, T1 ,T2 in J such that S1 DMl,S2 D M2 , T1D MI, T 2 D M2 and m(Si) < O(MI)+e; m(Ti) < 0(Ml)+e m (S2) < O(M2) + e; m (T2) < 0 (M2) + e.

Now S1 + TI D M1+ M' and S2+T 2 D M 2+ M2 so S1 +T = F = S2 + T2 . But 1 + m(SlTl) = m(Si + T1)+ mT m(iTi) = i(S 1) + m(Ti) < O(M1 ) + e + 0(Ml) + e =m(Mi) + m(M{) + 2e = 1 + 2e. Thus m(SiTl) < 2e and similarly m(S2T2) < 2e.

But M1 + M 2C S1 + S2 and (M1 + M2)' = Ml,M c TiT2, so F (M1 + M2) + (Mi + M2 )' c S1+ S2 + TiT2 . Moreover, from the first two inclusions, 0(M1+M 2) < m(S1 +S2 ) and I(M1+M 2) = 1-O(M +M2)' > 1-m(TiT2). So, 0(M1+ M2)-I(Mi + M 2)<m(Sl + S2)+m(TiT2)-1 = m(S1+S2+TlT2)+m( (S1 + S2)TiT2)-1 = m (F)+m( (S + S2) TT2)-1 = m(S1TiT2+ S2TIT2)<mi(STl + S2T2)<m(SiTi) + m(S2T2) <4e. Thus O(M1+M2) <I(M1+M2) and M1+M2 is measurable. Accordingly, so is every sum of a finite number of measurable sets.

But we see now that a sum of a countable number of measurable sets E M, = M1 +MM2-+ (Ml+ M2)'M3- ... is measurable by m 17 and the fact that each of the latter summands (M1+ . + Mn_l)'Mn (Ml+ ... +M,-i + M,)' is measurable by m 19 and the preliminary results for finite sums.

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3— Multiplicative Systems in Several Variables I, II, III: With C.J. Everett (LA-683, June 7, 1948) (LA-690, June 11, 1948) (LA-707, October 28, 1948)