Limit theorem for changed time This post seems long, but its almost everything proofed in this post. Only one step seems to be left, for the desired proof. I would be very gratefull for any help.
The setup
Given a Levy-Process $U_{t}$ with  with $E(U_t)=0$ (then $U_t$ is a martingale). Let $U_t$ have finite variance $Var(U_t)=tVar(U_1)$ and $Var(U_1)=\sigma^{2}$ and the limit theorem  holds:
\begin{align}
F_t:=\sqrt{t}\left(\frac{U_t}{t}-E(U_1) \right)=\frac{U_t}{\sqrt{t}}\xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty.\tag1
\end{align}
Let $K_t$ a non-decreasing positive ($K_{t}>0$ a.s.) process with cadlag-paths with the property that $K_{t}\rightarrow \infty$ almost sureley, as $t\rightarrow \infty$.
I want to show that 
\begin{align}
F_{K_t}:=\frac{U_{K_t}}{\sqrt{K_{t}}} \xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty. \tag2
\end{align}
For this one requires a positive non-random cadlag-function $a(t)$ with $a(t)\rightarrow \infty$ as $t\rightarrow \infty$ such that
\begin{align}
\frac{K_{t}}{a(t)}\rightarrow \theta\quad P\, a.s. \tag3
\end{align}
holds. Where $\theta$ is a positive random-variable.
Then the convergence in distribution of $F_{t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$ implies the convergence in distribution of $F_{K_t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$.
The original question from my old account is posted here. 
However with my reputation here, i am able to start a bounty for the question.
The suggestion of the proof are the following:
For simplicity it is said, that $\theta=1$ and $\sigma^{2}=1$ So that we have $K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))$ for large $t$. For $0<\theta<\infty$ we could do the same procedure and get the same result.
This is how we go on:
For small $m$ we have
$$
P(U_{K_t}<x\sqrt{K_t})\leq P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)+P\left(U_{a_t}<x\sqrt{(1+\epsilon)a(t)}+m\cdot \sqrt{\epsilon a(t))}\right)+ P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)
$$
The first term converges to 0 due to (3). The second term converges to $\Phi(x+m)$ (Why?) by the central limit theorem (1) applied to $U_{a(t)}$.
The third term is bounded by martingale inequalitys $L^{2}$ by a factor
$$\frac{1}{(m\cdot \sqrt{\epsilon a(t))})^{2}}$$
Otherwise we can state
$$
P(U_{K_t}<x\sqrt{K_t})\geq Z\xrightarrow{d} \Phi(x-m)
$$
So we have sandwiched it and the desired result (2) holds for arbitrary $0<\theta<\infty$.
HOWEVER (hopefolly the last step) We have with $(3)$ convegence to a strict positive finite randomvariable $\theta$. What is left to proof, considering, that $\theta$ is regarded as a random variable?
Some additional stuffto understand the inequalities
$$
P(U_{k_t}<x \sqrt{K_t})\\
\leq P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))]+P[U_{K_t}<x\sqrt{K_t},K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ \leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]+ P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))] \\
\leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]  \\
+P(U_{K_{t}}<x\sqrt{(1+ \epsilon)a(t)},|U_{K_t}-U_{a(t)}|\leq m\sqrt{\epsilon a(t)},|U_{K_t}-U_{a(t)}|> m\sqrt{\epsilon a(t)}]
\\ \leq P[U_{a(t)}<x\sqrt{(1+\epsilon)a(t)}+m \sqrt{\epsilon a(t)}] + P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)+P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)
$$
 A: *

*$(3)$ need not hold for general $K$. Say, set $K = t$ for half of $\omega$, and $K = t^2$ for another half. Then, depending on $a$, $\theta$ in $(3)$ will be either zero or infinite for half of $\omega$.

*Without the independence assumption, the statement is false in general, even if $(3)$ is true. Specifically, let $U=W$ be a Wiener process and define $K_t$ to be a càdlàg modification of $\inf\{s\ge 0: W_s\ge t\}$.  By the self-similarity property of $W$, $(3)$ holds with $a(t) = t^{2}$.  ($K$ is even a Lévy subordinator: it increases and has independent stationary increments thanks to the strong Markov property of $W$.) However, $(2)$ cannot hold, obviously. (Moreover, it follows from self-similatity that $F_{K_t}= t/\sqrt{K_t}\overset{d}{=} 1/\sqrt{K_1}$.)

*If $K$ is independent of $U$, then the statement is quite obvious. Indeed, denote $\varphi_s(u) = \mathsf{E}[e^{iu F_{s}}]$  the characteristic function of $F_s$. By $(1)$, $\varphi_s(u)\to e^{-u^2\sigma^2/2}$, $s\to\infty$, for any $u\in\mathbb{R}$. In particular, $\varphi_{K_t}(u)\to e^{-u^2\sigma^2/2}$, $t\to\infty$, a.s. On the other hand,
$$
\mathsf{E}[e^{iu F_{K_t}}] = \mathsf{E}[\mathsf{E}[e^{iu F_{s}}]|_{s=K_t}] = \mathsf{E}[\varphi_{K_t}(u)]
$$
thanks to independence. Then by the dominated convergence theorem, 
$$
\mathsf{E}[e^{iu F_{K_t}}] \to e^{-u^2\sigma^2/2}, t\to\infty,
$$
as claimed.
