Induced distribution measure and induced distribution function where original r.v. is Pareto

Solution 1:

You are making quite a mess of the part finding $\mu_Y$.

To keep things less complex I would rather write:

$$\begin{aligned}\mathsf P\left(Y\in A\right) & =\mathsf P\left(Y\in A\wedge X\leq d\right)+\mathsf P\left(Y\in A\wedge d<X\leq l\right)+\mathsf P\left(Y\in A\wedge X>l\right)\\ & =\mathsf P\left(0\in A\wedge X\leq d\right)+\mathsf P\left(X-d\in A\wedge d<X\leq l\right)+\mathsf P\left(l-d\in A\wedge X>l\right)\\ & =\mathsf P\left(0\in A\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X-d\in A\wedge d<X\leq l\right)+\mathsf P\left(l-d\in A\right)\mathsf P\left(X>l\right)\\ & =1_{A}\left(0\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in d+A\cap\left(0,l-d\right]\right)+1_{A}\left(l-d\right)\mathsf P\left(X>l\right) \end{aligned}\tag1 $$

Here $\mu_Y(A):=\mathsf P(Y\in A)$ and $\mu_X(A)=\mathsf P(X\in A)$.

Observe that a constant like $0$ can be looked at as a random variable and that $\mathsf P(0\in A)=1_A(0)$. These constant random variables are always independent wrt to any other random variable, so that: $$\mathsf P\left(0\in A\wedge X\leq d\right)=\mathsf P\left(0\in A\right)\mathsf P\left(X\leq d\right)=1_A(0)P\left(X\leq d\right)$$ To find CDF observe that: $$\mathsf F_{Y}\left(y\right)=\mathsf P\left(Y\in\left(-\infty,y\right]\right)$$ so to find it we must substitute $A=(-\infty,y]$ in $(1)$ leading to:

$$\begin{aligned}\mathsf F_{Y}\left(y\right) & =1_{\left(-\infty,y\right]}\left(0\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in d+\left(-\infty,y\right]\cap\left(0,l-d\right]\right)+1_{\left(-\infty,y\right]}\left(l-d\right)\mathsf P\left(X>l\right)\\ & =1_{\left[0,\infty\right)}\left(y\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in\left(d,\min\left(y+d,l\right)\right]\right)+1_{\left[l-d,\infty\right)}\left(y\right)\mathsf P\left(X>l\right)\\ & =1_{\left[0,\infty\right)}\left(y\right)\mathsf F_{X}\left(d\right)+\left[\mathsf F_{X}\left(\min\left(y+d,l\right)\right)-\mathsf F_{X}\left(d\right)\right]_{+}+1_{\left[l-d,\infty\right)}\left(y\right)\left(1-\mathsf F_{X}\left(l\right)\right) \end{aligned}\tag2 $$

Finding $\mathsf F_X$ and substituting in $(2)$ makes things complete now.