Question

Measure theory question

Answer 1

Ω=(0,1]
My question is that how come the A^c = (0,a_1]U(a'_1, a_2]U.....U(a'_m-1, a_m]U(a'_m, 1] ?????? because lets say that A= {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]} then
A^c = Ω - A
A^c = (0, 1] - {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]}
A^c = ∅ .........an empty set??????????

You can see this example at http://books.google.co.uk/books?id=a...ngsley&f=false
Example no 2.2 (section: Probability Measure), page 21.

Answer 2

@KingGeorge @SmoothMath any clue?

Answer 3

please see the attached pic

Answer 4

When you go from
A^c = (0, 1] - {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]}
to
A^c = ∅ Notice that it should actually be
A^c = {(0.1, 0.2], (0.3, 0.4], (0.5, 0.6], (0.7, 0.8]}

Answer 5

correct but the text says that A^c = (0,a_1]U(a'_1, a_2]U.....U(a'_m-1, a_m]U(a'_m, 1] which means A^c have 0 & 1 whereas 0 & 1 should be removed

Answer 6

Could you possibly put another link up to the book? Preferably without any dot dot dots. The current link isn't working for me

Answer 7

http://books.google.co.uk/books?id=a3gavZbxyJcC&printsec=frontcover&dq=probability+and+measure+billingsley&hl=en&ei=hHy1T7eVBNT_8QPn7sHZDw&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CDYQ6wEwAA#v=onepage&q=probability%20and%20measure%20billingsley&f=false

Answer 8

Example no 2.2 (section: Probability Measure), page 21.

Answer 9

As far as I can tell, A^c doesn't actually contain 0 at all. Nor does it have to contain the neighborhood. For example, if I chose A=(0, .5] then
A^c=(0, 1] - (0, .5]
Which according to their notation, would be
A^c=(0, 0)U(.5, 1]=(.5, 1]

Answer 10

It's a similar thing if A contains 1. It then forces the last interval of A^c to be \((1, 1] =\emptyset\)

Answer 11

so the complement given in text is wrong?

Answer 12

@KingGeorge ?????

Answer 13

I don't think it's wrong, just a way of writing it that isn't totally natural.

Answer 14

sorry I didn't get you can you please explain that what the heck is in text :|

Answer 15

For some reason, the actual book isn't showing me the pages you're looking at, so I'm having to go by the little snippet you included. If I'm reading correctly, it's defining the complement of A in the interval (0, 1] = \(\Omega\).

So suppose \(A\in \Omega\). This means that \(A=(a_1, a_1']\cup...\cup(a_m, a_m']\in\Omega\) such that \(i<j \implies a_i<a_j\). Then, the complement \(A^c\) is defined as \[A^c = \Omega-A=(0, a_1]\cup(a_1', a_2]\cup...\cup(a_m', 1]\]

Answer 16

yes correct

Answer 17

I have attached the whole example

Answer 18

The issue you seem to be having with this appears to be "what if \(a_1=0\)" and "what if \(a_m'=1\)." If \(a_1=0\), then \[A^c =(0, 0]\cup(a_1', a_2]\cup...\cup(a_m', 1]=(a_1', a_2]\cup...\cup(a_m', 1]\]since \((0, 0]=\emptyset\).

Likewise, if \(a_m'=1\) we have \[A^c =(0, a_1]\cup(a_1', a_2]\cup...\cup(1, 1]=(0, a_1]\cup...\cup(a_{m-1}', a_m]\]since \((1, 1]=\emptyset\)

Answer 19

Does this make more sense to you?

Answer 20

got it thanks a tonne

Answer 21

You're welcome.