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Let $(\Omega,\mathcal{F},\mathsf{P})$ and $(\Theta,\mathcal{A},\mathsf{m})$ probability spaces.

Suppose that $\Phi : \Omega \mapsto \Theta $ measurable function and $K\in \mathcal{A}$.

My question: Is it correct to say $$\mathsf{P}\left( \left\{\omega \in \Omega \:|\: \Phi(\omega) \in K \right\}\right) = \mathsf{m}\left( \left\{\theta \in \Theta \:|\: \theta \in K \right\}\right) \:?$$If the answer is yes then I would appreciate a good justification.

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    Nope, you can find simple counterexamples for this (take $\Phi\equiv 0$). The problem would be interesting if you take $\Phi$ as a bijection. – 2017-02-15
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    Even if you take $\Phi$ bijection, this statement is not valid (you can take the same space, the same $\sigma$-algebra, $\Phi =\text{identity}$ and take different measures). – 2017-02-15
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    Let me know if you want these counterexamples as an answer... – 2017-02-16
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    This is completely false in general. When this does hold, then we call $m$ the pushforward of $P$ under $\Phi$. – 2017-02-16

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