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If the open interval (a, b) is equinumerous with the cardinality of the continuum, why is it that the probability of two related intervals can have different values?

Take $$x=x_0cos(wt)\qquad \alpha=wt$$ as an example, where the PDF of $\alpha$ is $$\rho_\alpha(\alpha)=\frac{1}{2 \pi}$$and so $$\rho_x(x)=\rho_\alpha(\alpha)\; \frac{d\alpha}{dx}$$

Clearly then, $$\mathbb{P}(a_\alpha<x<b_\alpha)\;\neq\;\mathbb{P}(a_x <\alpha<b_x)$$

Since $[a_\alpha,b_\alpha]$ and $[a_x,b_x]$ are equinumerous, why is it that the probability for them is different.

I am looking for an answer that sheds light on the number of elements in the sets represented by those two intervals from the perspective of probability theory, but not fundamental theorem of calculus, change of variabes etc.

Asaf Karagila
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kfs
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    http://math.stackexchange.com/questions/383440/can-the-number-of-numbers-in-two-intervals-over-mathbbr-be-compared/ http://math.stackexchange.com/questions/867750/incorrect-proof-of-the-infinities-between-0-and-1-and-0-and-2 http://math.stackexchange.com/questions/40309/cardinality-density and probably I'm missing a handful of other related threads. – Asaf Karagila Apr 09 '16 at 10:41
  • @Asaf Would you mind writing an answer to my question? I know those topics are related. But I asked the question to get an answer, not citations. – kfs Apr 09 '16 at 11:09

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