Let $X(\omega)=a\omega+b$, in the $\left([0,1],\mathbf{B}(\mathbb{R}),\mu_{[0,1]}\right)$ probability space, where $\mu_{[0,1]}$ is the Lebesgue measure restricted to the interval $[0,1]$.
How can I check that $\forall_{B \in \mathbf{B}(\mathbb{R})} \ \ P_X(B)=\frac{\mu(B\cap[b,a+b])}{a}$?
Well, here's my derivation: $P_X(B)=\mu_{[0,1]}(X^{-1}(B))=\mu(X^{-1}(B)\cap [0,1])$ $=\mu(X^{-1}(B\cap [b,a+b])=\mu(X^{-1}(B\cap [b,a+b]))=\mu(\frac{(B\cap [b,a+b])-b}{a})$ $=\frac{\mu((B\cap [b,a+b])-b)}{a} =\frac{\mu(B\cap[b,a+b])}{a}$, with the last equality due to invariance to translation of the Lebesgue measure. Is my derivation correct? Does the property I used in the equality, before the last, have a name?
Any help would be appreciated.