Equivalence between two different representations of exponential Lévy Processes

Question

My questions are: Why do I know that $\frac{Z}{Z_-}$ looks like in the proof?

Why $\int \frac{d[Z^c]}{Z_-^2}=[Y^c]$?

Why does the part with the sum look like the one below? I only know that $f(x)*\mu^X=\sum f(\Delta X_s)$. Please help me!!

Answer 1

$\frac{Z}{Z_-}=\epsilon(Y)_t=\frac{\exp(Y_t-1/2[Y]^c_t)\prod_{s\le t}(1+\Delta Y_s)\exp(-\Delta Y_s)}{\exp(Y_{t-}-1/2[Y]^c_t)\prod_{s < t}(1+\Delta Y_s)\exp(-\Delta Y_s)}=(1+\Delta Y_t)=\Delta (\sum_{0<s\le t}1+\Delta Y_s)=\Delta((1+y)*\mu^Y)$

For the second one we know that $Z_t=1+\int_0^t Z_- dY$ which leads to $[Z_t]^c_t=[\int Z_-dY]^c_t=\int Z_{s-}^2 d[Y]^c_s$.

For the third question use $\frac{Z}{Z_-}$ and $f(x)*\mu^x=\sum_{s\le t}f(\Delta X_s)$.