Given a filtered probability space $(\Omega,\mathcal{F}_{t},P)$ and a continuous nondecreasing process $U_{t}$ with $U_{0}=0$ and $U_{t}\rightarrow \infty$ $P-a.s.$ as $t$ goes to $\infty$.
Given a process $X_{t}$ (cadlag) with $E(|X_{1}|)<\infty$ for which the SLLN holds, so \begin{align} \lim_{t \uparrow \infty} \frac{X_{t}}{t}\rightarrow E(X_{1})\quad a.s. \end{align} I guess it holds, that \begin{align} \lim_{t \uparrow \infty} \frac{X_{U_{t}}}{U_{t}}\rightarrow E(X_{1})\quad a.s. \end{align} But i am struggling.
For each $n \in \mathbb{N}: \exists t_{n}: U_{t}=n.$ Fix this increasing sequence $t_{1}<t_{2}<\ldots$. Then i dont know if this is already sufficient to apply the SLLN of $X_{t}$ \begin{align} \lim_{n\uparrow \infty} \frac{X_{U_{t_{n}}}}{U_{t_{n}}}=\lim_{n\uparrow \infty}\frac{X_{n}}{n}\rightarrow E(X_{1})\quad a.s. \end{align}
I hope you can help me out. Best regards
Fix $\epsilon>0$. Since $(X_t)_{t \geq 0}$ satisfies the SLLN, there exists for almost all $\omega \in \Omega$ a constant $S=S(\omega)>0$ such that
$$\left| \frac{X_s(\omega)}{s} - \mathbb{E}(X_1) \right| \leq \epsilon \quad \text{for all $s \geq S$}.$$
As $U_t \uparrow \infty$, we have $U_t(\omega) \geq S$ for $t \geq T=T(\omega)$ sufficiently large. Using the previous equation for $s = U_t(\omega)$, we get
$$\left| \frac{X_{U_t(\omega)}(\omega)}{U_t(\omega)} -\mathbb{E}(X_1) \right| \leq \epsilon \qquad \text{for all $t \geq T$}.$$
Since this holds for almost all $\omega$, this finishes the proof.