The integral $\int f \mu(dt)$ versus $\int f \mu(t) dt$

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In the theory of continuous-time branching processes there is a quantity called the Malthusian parameter $\alpha \in \mathbb{R}$ defined by:

$$ \int_0^{\infty} e^{-\alpha t} \mu(dt) = 1 $$

The measure $\mu$ having to do with the branching process is defined as follows. The number of children of a given individual in the branching process is a point process $\xi$ on $\mathbb{R}^+$. $\mu(t)$ is simply the reproduction function for the branching process, which is the expected number of children of an individual by the time that individual is of age $t$. In other words, $\mu$ is the expected number of points in $[0,t]$ of $\xi$: $$ \mu(t) = \mathbb{E} \xi[0,t] $$ In the course of some calculations involving $\alpha$, I was being sloppy and wrote: $$ \int_0^{\infty} e^{-\alpha t} \mu(dt) = \int_0^{\infty} e^{-\alpha t} \mu(t) dt $$ I was immediately told that I am, indeed, making a boneheaded integration mistake. Instead it was suggested that to make this equality true, I need to put $$ \int_0^{\infty} e^{-\alpha t} \mu(dt) = \int_0^{\infty} \alpha e^{-\alpha t} \mu(t) dt $$ Note the extra $\alpha$ on the right-hand side. But I just can't see where this came from.