Law of large numbers for a Subordinator. 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.
 A: This is a simple application of Law of large numbers.
Note that 
$$E(Y_1) = \int^\infty_0 x\nu(\text{d}x)$$
Assume that $E(Y_1)<\infty$. Let $n$ be the largest integer smaller than $t$, then
$$\frac{Y_t}{t}=\frac{Y_1+(Y_2-Y_1)+...+(Y_n-Y_{n-1})}{n}\frac{n}{t}+\frac{Y_t-Y_n}{t}$$
By the strong law of large numbers, the first part of the sum converges to $E(Y_1)$. Note that $\frac{Y_t-Y_n}{t}$ is a convering almost surely to $0$.
For $E(Y_1)=\infty$, then obverse that $Y$ admits the following decomposition
$$Y=Y_1+Y_2$$
where the Laplace exponent of $Y_1$ and $Y_2$ are given by
$$\int_{(0,1]}(1-e^{\lambda x})\nu(dx)$$
and 
$$\int_{(1,\infty)}(1-e^{\lambda x})\nu(dx)$$
repsectively. note that $Y_2$ is a compound Poisson process with intensity $\lambda = \nu((1,\infty))$. Notice that $EY_1$ is finite, so
$$\frac{Y_t}{t}= \frac{Y^1_t}{t}+\frac{Y_t^2}{t}$$.
Note that $Y_t^1/t$ is convering to $\int_{(0,1]}x\nu(dx)$ by the previous argument.
$$Y_t^2=\frac{N_t}{t}\frac{\sum_{i=1}^{N_t}\zeta_i}{N_t}$$
Well, $N_t/t\rightarrow\nu((1,\infty))$, but $\frac{\sum_{i=1}^{N_t}\zeta_i}{N_t}$ is going to infinity by SLLN because the expection of $\zeta_i$
$$\frac{1}{\nu (1,\infty)}\int_{(1,\infty)}x\nu(dx)=\infty$$ 
