Exercise $\boldsymbol{10.44}$. Let $X$ and $Y$ be independent continuous random variables with probability density functions $f_X(x)$ and $f_Y(y)$. Compute $P(X+Y \leq z)$ by first conditioning on $Y=y$ and then using the averaging principle of equation ($10.15$).
This is my work↓
Hint given: $\displaystyle P(X<x) = \int_{-\infty}^\infty P(X<x|Y=y) f_Y(y) dy.$
\begin{align} P(X+Y<z) &= P(X<z-y,Y=y) \\ &= P(X<z-y) P(Y=y) \tag{by indep.} \\ &= \int_{-\infty}^\infty P(X<z-y|Y=y) f_Y(y) P(Y=y) dy \\ &= \int_{-\infty}^\infty \frac{P(X<z-y,Y=y)}{P(Y=y)} f_Y(y) P(Y=y) dy \\ &= \int_{-\infty}^\infty P(X<z-y, Y=y) f_Y(y) dy \\ &= \int_{-\infty}^\infty \\ \\ \\ &= \int_{-\infty}^z f_{X+Y}(s) ds. \end{align}
I'm required to use the hint but have no idea how this goes down to the solution.
About Averaging identities:
If $Z=X+Y$, then \begin{align} F_Z(z) &= P(X+Y)\le z) \\ &=\int_{-\infty}^{\infty} P(X\le z-Y|Y=y) f_Y(y)dy, \end{align} So that \begin{align} f_Z(z) &= F_Z'(z) \\ &= \frac{d}{dz}\int_{-\infty}^{\infty} F_X(z-y|Y=y)f_Y(y)dy \\ &= \int_{-\infty}^{\infty} f_X(z-y)f_Y(y)dy. \end{align}