Why Isn't A^c the Empty Set in This Measure Theory Example?

[woundedtiger4]

Hi all,
I am reading Probability and Measure by Patrick Billingsley, and I am stuck at one example, please help me understanding it.
http://desmond.imageshack.us/Himg201/scaled.php?server=201&filename=30935274.jpg&res=landing
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 let's 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=...q=probability and measure billingsley&f=false
Example no 2.2 (section: Probability Measure), page 21.

Thanks in advance.

 

[micromass]

A= {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]}

What does this even mean? A is a set with intervals as elements?? That's not what Billingsley means.

 

[micromass]

Please ignore this post, as it is wrong

If you mean

$$A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1]$$

then yes, A^c is the empty set, because A=(0,1].

 

[woundedtiger4]

If you mean

$$A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1]$$

then yes, A^c is the empty set, because A=(0,1].

Sorry for the typing mistake, yes I mean union. but then why text says A^c = (0,a_1]U(a'_1, a_2]U...U(a'_m-1, a_m]U(a'_m, 1] , why does the complement have 0 & 1 in text ?

[STRIKE]Perhaps I am also wrong of being making the A^c = empty set because Ω contains only 0 & 1 so shouldn't it be A^c = {(0.1, 0.2], (0.3, 0.4], (0.5, 0.6], (0.7, 0.8]} ?[/STRIKE]

 

[Citan Uzuki]

If $A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1]$, then the formula given states that $A^{c} = (0, 0] \cup (.1, .2] \cup (.3, .4] \cup (.5, .6] \cup (.7, .8] \cup (1, 1]$, which is correct.

 

[woundedtiger4]

If $A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1]$, then the formula given states that $A^{c} = (0, 0] \cup (.1, .2] \cup (.3, .4] \cup (.5, .6] \cup (.7, .8] \cup (1, 1]$, which is correct.

OK
So,
If a1=0, then
A^c=(0,0]∪(a1′,a2]∪...∪(am′,1]=(a1′,a2]∪...∪(am′,1]
since (0,0]=∅. Likewise, if am′=1 we have
A^c=(0,a1]∪(a1′,a2]∪...∪(1,1]=(0,a1]∪...∪(am−1′,am]
since (1,1]=∅

 

[xAxis]

If you mean

$$A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1]$$

then yes, A^c is the empty set, because A=(0,1].

How comes? Shouldn't
A^c = {0} U (0.1, 0.2] U (0.3, 0.4] U (0.5, 0.6] U (0.7, 0.8] ?

 

[xAxis]

...since (0,0]=∅...

I think (0,0]= {0}.

 

[D H]

How comes? Shouldn't
A^c = {0} U (0.1, 0.2] U (0.3, 0.4] U (0.5, 0.6] U (0.7, 0.8] ?

No. (0,0] is the empty set. Other than that, you are correct, as is Citan Uzuki in post #5.