Lecture 2: Outer measures, construction of the Lebesgue measure.

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Marcus Carlssons mathematics courses

Marcus Carlssons mathematics courses

Күн бұрын

Пікірлер: 8
@TheMorhaGroup
@TheMorhaGroup 28 күн бұрын
"I hope you're as excited as I am😐😑😐"
@nicholaskhawli4330
@nicholaskhawli4330 18 күн бұрын
Isnt finite union concluded from the 2 sets union proof ? I guess u could prove it but shouldnt it be trivial
@小小学生-r5o
@小小学生-r5o Ай бұрын
I am still confused about what happened in 28:40. Why can we just simply add something to the left and then change the inequality the other way around?
@thierryrioual701
@thierryrioual701 Ай бұрын
Because that's the definition of the infimum: For all a in A, and for all epsilon > 0, a - epsilon < inf A =0, and epsilon >0, epsilon/2^k > 0. See "Infima and suprema of real numbers": en.wikipedia.org/wiki/Infimum_and_supremum Normally the inequality is strict but by taking epsilon arbitrarily close to zero the inequality becomes loose.
@asbjrnandersen4765
@asbjrnandersen4765 23 күн бұрын
lambda star is the infimum of the RHS. thats why adding a number greater than zero flips the inequality
@darrenpeck156
@darrenpeck156 2 жыл бұрын
Could this criteria lead from countable subadditivity to countable additivity?
@thomasjefferson6225
@thomasjefferson6225 7 ай бұрын
disjoint unions for countable additivity
@nicholaskhawli4330
@nicholaskhawli4330 20 күн бұрын
Why use inequality and add the epsilon if we will just equal it to zero . Then why not just use equality .
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