Kenneth Kuttler

14.3. THE LEBESGUE STIELTJES MEASURES AND BOREL SETS 327

= supn

supm

∑i=1

∫ n

−nXEi∩[−n,n]dF = sup

msup

∑i=1

∫ n

−nXEi∩[−n,n]dF

= supm

limn→∞

∑i=1

∫ n

−nXEi∩[−n,n]dF = sup

∑i=1

limn→∞

∫ n

−nXEi∩[−n,n]dF

= supm

∑i=1

µ (Ei)≡∞

∑i=1

µ (Ei)

In this computation, I have used the interchange of limits with supremums in case of anincreasing sequence. Also, I have used the monotone convergence theorem. Therefore,this has only shown the desired result in case µ (∪∞

i=1Ei) is finite because of the mono-tone convergence theorem we currently have. However, in case this is infinity, let l be areal number. Then by definition, there is n large enough that

∫ n−n X∪∞

i=1Ei∩[−n,n] (x)dF > l.If∫ n−n X∪m

i=1Ei∩[−n,n] (x)dF ≤ l for each m, then by the monotone convergence theorem,∫ n−n X∪∞

i=1Ei∩[−n,n] (x)dF ≤ l also, from the monotone convergence theorem for generalizedintegrals, which would be a contradiction. Hence for large enough m,

∞

∑i=1

µ (Ei)≥∫ n

−nX∪m

i=1Ei∩[−n,n] (x)dF =m

∑i=1

∫ n

−nXEi∩[−n,n]dF > l

Since l is arbitrary, it follows that in this case, both µ (∪∞i=1Ei) and ∑

∞i=1 µ (Ei) equal ∞. To

understand measure of intervals here is a lemma.For [c,d]⊆ (−n,n) it is not clear that

∫ n−n X[c,d]dF the Riemann Stieltjes integral even

exists. This is because X[c,d] is not continuous and we do not assume F is continuouseither. In particular, you could have F have a jump at c or at d. But with the generalizedintegral, one can get the appropriate result.

Theorem 14.3.6 Let F be an increasing integrator function. Then

1. µF ([c,d]) = F (d+)−F (c−)

2. µF ((c,d)) = F (d−)−F (c+)

3. µF ((c,d]) = F (d+)−F (c+)

4. µF ([c,d)) = F (d−)−F (c−)

Proof: For large n,µF ([c,d]) =∫ n−n X[c,d]dF, [c,d] ⊆ (−n,n). Define the gauge func-

tionδ ε (x)≡min(|x− c| , |x−d|) if x /∈ {c,d} and δ ε (c) = δ ε (d) = ε > 0

Let Pε be a δ ε fine division of [−n,n]. Then both c,d are tags because, due to the def-inition of δ ε , neither of these can be closer than δ ε (t) for any t /∈ {c,d} because ofthe definition of δ ε . For example, if c is not a tag, then there is some tag x such thatc∈ (x−δ ε (x) ,x+δ ε (x)) and so δ ε (x)> |c− x| which does not happen. Similarly d mustalso be a tag. If Ic is the interval containing c then c is on the interior of Ic. Otherwise, theadjacent interval having t for a tag and c an endpoint, would have c∈ (t−δ ε (t) , t +δ ε (t))which would say that |c− t| < δ ε (t) which does not happen due to the definition of δ ε .Similarly d is an interior point of Id the closed interval containing d. Thus the divisionpoints x j are

−n = x0 < · · ·< xk < c < xk+1 < xk+2 < · · ·< xm < d < xm+1 < xm+2 < · · ·< xl = n

14.3. THE LEBESGUE STIELTJES MEASURES AND BOREL SETS 327= supsup)) [” LE M| ny =supsup > [” RE A-n nlnom j=] m Nn j=]supsin Y [” 2 EAI mand =ap tim EOI —nn\dFm nee= sup) 1 (E )=Y ule)i i=1In this computation, I have used the interchange of limits with supremums in case of anincreasing sequence. Also, I have used the monotone convergence theorem. Therefore,this has only shown the desired result in case pt (U7, £;) is finite because of the mono-tone convergence theorem we currently have. However, in case this is infinity, let / be areal number. Then by definition, there is n large enough that |”, Bose Bana] (x)dF >I.If [", Rom Bal-na] (x)dF <1 for each m, then by the monotone convergence theorem,fon Xe E:0[-n.n| (x)dF <1 also, from the monotone convergence theorem for generalizedintegrals, which would be a contradiction. Hence for large enough m,Yu (Ei) 2 / Rom B:A(-n] (x) dF = y/ Lend P >!i=] —n j=] 7-2Since / is arbitrary, it follows that in this case, both p (U7, £;) and Y=, u (E;) equal oo. Tounderstand measure of intervals here is a lemma.For [c,d] C (—n,n) it is not clear that J”, 2j-,qdF the Riemann Stieltjes integral evenexists. This is because Ked| is not continuous and we do not assume F is continuouseither. In particular, you could have F have a jump at c or at d. But with the generalizedintegral, one can get the appropriate result.Theorem 14.3.6 Let F be an increasing integrator function. ThenI. Up ([c,d]) = F (d+) —F (c-)2. Up ((c,d)) =F (d—) —F (c+)3. Up ((c,d]) = F (d+) —F (c+)4. Up (|c,d)) =F (d—) —F (c—)Proof: For large n, Up ([c,d]) = J", Pic,a\4F, [c,d] C (—n,n). Define the gauge func-tionOe (x) = min(|x—c|,|x—d]) if x ¢ {c,d} and 6¢(c) = d¢(d) =e >0Let P; be a d¢ fine division of [—n,n]. Then both c,d are tags because, due to the def-inition of d¢, neither of these can be closer than 6, (t) for any t ¢ {c,d} because ofthe definition of 6. For example, if c is not a tag, then there is some tag x such thatc € (x— b¢ (x) ,x+ b¢ (x)) and so d¢ (x) > |c —x| which does not happen. Similarly d mustalso be a tag. If J. is the interval containing c then c is on the interior of J.. Otherwise, theadjacent interval having ¢ for a tag and c an endpoint, would have c € (t — b¢ (t) ,t + b¢ (t))which would say that |c—t| < 5¢(t) which does not happen due to the definition of d¢.Similarly d is an interior point of Jy the closed interval containing d. Thus the divisionpoints x; are—N=XQ <r LEO < Heyy <Xep2 <0 <n < d < Xp 4 <Xmg2 <i << Xp Hn