# Thread: P(A) = P(B) = 0 prove: PA(UB) = 0

1. ## P(A) = P(B) = 0 prove: PA(UB) = 0

Hey people. Im sitting here with an assignment.

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It is given that P(A) = P(B) = 0
Then i have to prove that P(A U B) = 0.
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We are supposed to use the theorem 3.1:
Let P be a probability measure. Then we have that

P(A\B) = P(A) - P(B) if A contains B
and
P(Ac) = 1 - P(A) where Ac is A complement.

furthermore, P is increasing in the sense that

if A contains B it implies that P(A) is greater than or equal to P(B)

and P is continuous in the sense that An (arrow up) A => P(An) (arrow up) P(A) and the same the other way around.
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I hope it makes sense and i hope someone can help me.
Its a fairly simple thing, but somehow i cant wrap my head around it.

2. P(aub)=p(a)+p(b)-p(ab)=0+0-0=0 done.

3. How can you say that P(aub) = P(a)+P(b)-P(ab)?.. in my book the theorem says that the correct equation is
P(AUB) = P(A) + P(B) - P(A n B) Where n is the upside down U. As far as i know P(AB) isnt equal to P(AB)?
Can you elaborate?

Thanks for the help.

4. what does this mean.............
'As far as i know P(AB) isnt equal to P(AB)?'
so $\displaystyle x\ne x$, very interesting, but..............

AB means intersection and AB is contained in A , also B
So it has probability zero
anyhow you still would have
P(AUB)=-P(AB)which is less than or eqal to zero...... giving you 0=0.

It's 3am and I don't have time for this.

5. heh sorry a typing error from my side. I meant that P(AB) wasnt equal to P(AnB).

so your saying that P(AB) = P(AnB)? apparently i missed that.

6. yes, it's just notation

if you prefer

$\displaystyle P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)=0+0-0=zero$

SEE (1) on http://www.weibull.com/SystemRelWeb/...ity_theory.htm

7. Alright, thanks a lot for the help .