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Let $(B)_{t\in \mathbb R_+}$ be a real-valued Levy Process (i.e. independent and stationary increments) with almost surely continuous paths on the probability space $(\Omega, \mathcal A, P)$.

Let us fix some $t>0$ and let us define for $j,n \in \mathbb N$:

\begin{align} X_{n,j}:=\triangle\Big[B(tj/n)-B(t(j-1)/n)\Big] \: \qquad \text{with} \quad \triangle(x):= x \: \cdot \mathbf{1}_{(-1,1)}(x). \end{align}

How can I deduce now, that $\lim\limits_{n \rightarrow \infty}{\sum_{j=1}^{n}X_{n,j}=B(t)-B(0)}$ almost surely.

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    $\begingroup$ Well, if $(B_t)_{t \geq 0}$ has almost surely continuous sample paths, then $(B_t)_{t \geq 0}$ is just a Brownian motion, right? $\endgroup$ – saz May 30 '17 at 17:36
  • $\begingroup$ That is the aim of this exercise. To show that this $(B_t)_t$ has increments, which are normal distributed. This is only the First of six steps to show the existence of the brownian motion. So, unfortunately, I can not simply say it is a brownian motion here. $\endgroup$ – Frodo361 May 30 '17 at 19:47
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Recall that a continuous function on a compact interval is uniformly continuous. Since the process $(B_t)_{t \geq 0}$ has almost surely continuous sample paths, this means that

$$[0,t] \ni s \mapsto B_s(\omega)$$

is uniformly continuous with probability $1$. In particular, we can choose $N=N(\omega) \in \mathbb{N}$ such that

$$|B_u(\omega)-B_v(\omega)| < 1 \qquad \text{for all} \, \, u,v \in [0,t], |u-v| \leq \frac{1}{N}.$$

By the very definition of the mapping $\Delta(x) := x 1_{(-1,1)}(x)$, we get

$$X_{n,j}(\omega) := \Delta \left[ B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right) \right] = B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right)$$

for all $j=1,\ldots,n$ and $n \geq N$. This implies that

$$\sum_{j=1}^n X_{n,j}(\omega) = \sum_{j=1}^n \left[ B \left( \frac{tj}{n},\omega \right)- B \left( t \frac{j-1}{n},\omega \right) \right] = B(t,\omega)-B(0,\omega)$$

for all $n \geq N$. This proves the assertion.

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