If two independent random variables are summed in the variable $Z = X+Y$ then to find the P.D.F. of $Z$ which would be $P(Z=z)$ we need to integrate the joint distribution of $X$ and $Y$ over all values where $x+y=z$ ie two random measurements of our variables $X$ and $Y$ are a particular value from $Z$. In particular, we can find the value of $P(Z=z)dz$ a.k.a the probability that $Z$ takes a value between $z$ and $z+dz$. This is done in the following:
$$P(Z=z)dz = \int_{-\infty}^{\infty}\int_y^{y+dz}f(x,y)dxdy$$ where $f(x,y)$ is the joint distribution of the variables $X$ and $Y$. Since these two are independent we can write $f(x,y)=g(x)h(y)$. Also, in the integration limits we can replace $y$ by $z-x$ by the definition of $z$:
$$P(Z=z)dz = \int_{-\infty}^{\infty}g(x)dx\int_{z-x}^{z-x+dz}h(y)dy$$
From this point onwards I am not sure how to show that $$P(Z=z)dz = \left[\int_{-\infty}^{\infty}g(x)h(z-x)dx\right]\cdot dz$$
which is the more familiar convolution formula. I tried replacing $dy$ with $dz-dx$ but the function $h$ still depends on both of these so I didn't see how to get rid of the second integral. The final formula does make intuitive sense but the step between the last two equations is what I don't understand. Many thanks.