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I need a little help on a homework exercise, and I think you can help me:

Suppose $(\Omega,F,\mathbb{P})$ is a probability space, and $Z:\Omega\to\mathbb{R}$ is an RV s.th. $Z(\omega)> 0 \ \forall\ \omega$, and $E[Z]=1$. Define for any $A\in F: \\Q[A]=E[Z\cdot1_{A}]$.

Show that this is a probability measure, which I did, and then: Show that $Q$ and $\mathbb{P}$ are equivalent. But how can I show this if $\mathbb{P}$ has been nowhere defined?

If I write

"$Q[A]=E[Z\cdot 1_A] =0\Rightarrow P[A]=0$ by positivity of $Z$"

and

"$P[A]=0\Rightarrow 1_A\equiv0\Rightarrow E[Z\cdot1_A]=0=Q[A]$",

is that already enough?!

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    $\begingroup$ Looks OK to me. $\endgroup$ – GEdgar Mar 22 '12 at 12:32
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    $\begingroup$ I don't understand what you mean by "But how can I show this if $\mathbb{P}$ has been nowhere defined?" $\endgroup$ – Ashok Mar 22 '12 at 12:40
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    $\begingroup$ You might want to flesh out the "by positivity of $Z$" part. $\endgroup$ – Nate Eldredge Mar 22 '12 at 12:42
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    $\begingroup$ It is enough, you don't need the direct definition of $\mathbb P$ - it's just given. All you need to do is to use properties of integral and measure, as you have written. $\endgroup$ – Ilya Mar 22 '12 at 12:47
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$P(A)=0$ doesn't imply that $\Bbb 1_A\equiv 0$ (but it implies that this indicator function is in the class of $0$ for the equivalence $P$-almost everywhere, and it's probably what you meant). Since you take after the expectation with respect to $P$, it doesn't matter and the proof is good.

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