$\prod_{i=1}^{\infty}P(A_i)=x \iff \sum_{i=1}^{\infty}\log(P(A_i))=log(x)$

Question

Show for a $x>0$ (and $P$ be a probability measure)

$$\prod_{i=1}^{\infty}P(A_i)=x\iff \sum_{i=1}^{\infty}\log(P(A_i))=log(x)$$

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I have thought about this question for a while but I am not seeing how to show the bi-implication. What is the trick? I see that each $1 \geq P(A_i)\geq0$ since it is a probability measure.

Answer 1

The equivalence would be true for finite sums and products. The infinite case then follows from the continuity of the logarithm, i.e. you can interchange the limit and the $\log$.

(Note that $P(A_i)>0$ holds for all $i$ because $x>0$.)