It - UC Santa Barbara
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Homework 3 Solutions Katy Craig 2020 Claim ek A EX either A or AC is countable is the collection of µ measurable sets Assume that A teller If E E X i un countable µ E I MY Ah E ty4A n E since fi uncountable On the other hand if EE X is countable AA E and A n E are countable so µ E O MY Ah E MMA n E This shows that if AE Mei A islet measurable Deep Youcould also have appealed to Thm 1 to conclude A iser measurabl Now suppose A Fetter then neither A nor A is countable Consequently we txt I 42 µ4Xn A tether A't Thus A is not re measurable
Transcript of It - UC Santa Barbara
Homework 3 Solutions
KatyCraig 2020
Claim ek A EX either A or AC is countable isthe collectionof µ measurable sets
Assume that A teller If E E X i un countable
µ E I MYAh E ty4A n Esince
fi uncountable
On the other hand if EE X is countableAA E and A n E are countable so
µ E O MYAh E MMA n EThis shows that if AE Mei A islet measurableDeepYoucould also haveappealed toThm 1 to conclude A iser measurablNow suppose A Fetter then neither A nor Ais countable Consequentlywetxt I 42 µ4XnA tether A'tThus A is not re measurable
aim No 2X is the collectionof f measurablesets
Fix AE X
If E is countable so are EMA and ENA hence
v E O V CEN A VIERA't
If E is uncountable then eitherEMA or EMA is uncountable ThusE to v EAA t V CEAAc
Define M sup µ B BEA But measSince 0 EA and µ40 O Thin 1 ensuresis ut news so MIO Similarly since
MEMMA Ctd
bydefinition of the supremumH n C IN F Bue AstMEµHBn In
Define B Bn Since Bn is re measurablefor all NEIN Theorem 2 ensures B is alsoµ measurable Likewise since Bn EA V NEIN
we also have BEA Consequently by definitionof the supremum
GCB E M
On the other hand since Bn E B V nElNM Eph Bn th Eµ4B th Hn EIN
Sending n Sta we conclude µ CB M
Consider the set 5 ALB Since Bis f measurable and A SU B S mustnot be µ measurable or else It would betoo Thus by Thm I µ s 0
Suppose TFS and T is ut measurableThen BUT E A and BUT is G measurableso bydefinition of MMZ µ BUT
On the other hand by the Gmeasurabilityof B
MzµHBUT µ KBUT AB M KBUT It.es yz
uPlB tu T
M tµ T
Thus µ4Tl o
he function is clearly nonnegative andsymmetric It is nondegenerate since
dkxi.gl xz.yzD 0ESdxlx xzl OC 7xixedylyyd O y yz
It satisfies the triangle inequality since forall X Xz Xs E X y ye yes 4dllx.iydlxc.yzD
dxlx.az t dyly ya
dxlxi.XDtdxlxzixdtdylyi.ydtdylys.geDAX y Ks yes
t dkxz.yzl.lxs.gs
It suffices to show that every setthat is open in the product topology climax isopen in d
Note that by definitiondmaxllxyy.l.kzyzDEdllx yd lxz yzD
Suppose U is open Wrt climax Fixlx.gl c U Then I r O sitdmaxkx.gl xayzDsr ensures Kiya EU
Furthermore by inequality 89 aboveif dllx.iyd.kzyet r then dmaxkxyy.li xayzDsrso ha ya EU Since Kay was arbitrarythis shows U is open wrt d
