Let $\left( X_{t}\right) _{t\geq0}$ be a subordinator with the Laplace exponent given by
$$ \Phi\left( \lambda\right) =d\lambda+\int_{0}^{\infty}\left( 1-e^{-\lambda x}\right) \nu\left( dx\right) $$
Show that almost surely
$$ \lim_{t\rightarrow\infty}\frac{X_{t}}{t}=d+\int_{0}^{\infty}x\nu\left( dx\right) $$
I first use the Levy-Khintchine formula, where we have
$$ X_{t}=dt+Y_{t}% $$
and $\left\{ Y_{t}\right\} _{t\geq0}$ is a subordinator whose the Laplace exponent is given by
$$ \Phi^{\prime}\left( \lambda\right) =\int_{0}^{\infty}\left( 1-e^{-\lambda x}\right) \nu\left( dx\right) $$
Thus,
$$ \frac{X_{t}}{t}=d+\frac{Y_{t}}{t}% $$
Now, I think it suffices to show that almost surely $\lim_{t\rightarrow\infty }\frac{Y_{t}}{t}=\int_{0}^{\infty}x\nu\left( dx\right) .$
However, I don't know if I can do this and what should I do next.