Prove increment of Brownian motion is Brownian motion I am trying to solve the following exercise in Oksendal's book:
Let $B_t$ be Brownian motion and fix $t_0\ge 0$. Prove that $$\bar{B_t}:=B_{t_0+t}-B_{t_0};\quad t\ge 0$$ is a Brownian motion.

I try to prove it by definition. In the book, Brownian motion is defined as
$$P^x(B_{t_1}\in F_1, \cdots, B_{t_k}\in F_k) = \\ \int\limits_{F_1 \times \cdots \times F_k}p(t_1, x, x_1)\cdots p(t_k-t_{k-1}, x_{k-1}, x_k)dx_1 \ldots dx_k, $$
 where
$$p(t,x,y) = (2\pi t)^{-n/2}\cdot \exp(-\frac{|x-y|^2}{2t})$$
Hence for simplicity, we only need to show $$P^0(\bar{B}_{t_1}\in F_1, \bar{B}_{t_2}\in F_2) =  \int\limits_{F_1 \times F_2}p(t_1, 0, x_2) p(t_2-t_1, x_3, x_2)dx_2 dx_3$$
Suppse $B_t$ starts at $x_0$. By definition of $\bar{B_t}$
$P^0(\bar{B}_{t_1}\in F_1,\bar{B}_{t_2}\in F_2)=P^0(B_{t_0+t_1}-B_{t_0}\in F_1, B_{t_0+t_2}-B_{t_0}\in F_2)\tag{1}$
Then I don't know how to handle it. I know I should use  $B_t$'s finite dimensional distribution. So if I can write $(1)$ as 
$$P^{x_0}[\bigcup_{x_1\in\mathbb{R}^n} \left(B_{t_0}=x_1,B_{t_0+t_1}\in F_1+x_1, B_{t_0+t_2}\in F_2+x_1\right)]\tag{2}\\=\int\limits_{\mathbb{R}^n\times (F_1+x_1)  \times (F_2+x_1)}p(t_0, x_0, x_1)p(t_1, x_1, x_2) p(t_2-t_1, x_2, x_3)dx_1 dx_2 dx_3 $$
Then apply change of variable formula and Fubini's theorem, I can get the result.
But I don't know how to justify $(2)$ and the last equality. I think it should be true because it's the conditional probability. But the book didn't introduce this concept and since $\mathbb{R}^n$ is uncountable, I don't know how this is valid in terms of measure theory.
 A: Why aren't you applying the characterization:  "Any continuous real-valued process $(X_t)$ that is zero-mean Gaussian process with covariance $\mathrm{cov}  (X_t,X_s)=t\wedge s$ is a Brownian motion" ? (Rogers and Williams, page 4).
I write $B'_t$ for your new process. It is a Gaussian process. 
Then $B'_t$ has continuous sample paths (since $B_t$ does). Also
$$E(B'_t) = E(B_{t_o+t}) - E(B_{t_0}) = 0. $$
Finally for $s < t$
\begin{align}
\mathrm{cov}(B'_t,B'_s) &= E((B_{t_o+t}-B_{t_o})(B_{t_o+s}-B_{t_o}))\\
&= \min(t_0+t, t_o+s) + \min(t_0,t_0) - \min(t_0+s,t_0) + \min(t_0+t,t_0)\\
&= (t_0 +s) + t_0 - t_0 -t_0 = s = \min(s,t).
\end{align}
In the second line I have expanded the brackets, used linearity of expectation, and the characterization applied to $(B_t)$.
A: First of all, recall that $\{B_{t_{k+1}}-B_{t_k}\}_{k\in \mathcal{Z}^+}$ is a collection of independent normal random variables with mean $0$. Thus, it is clear that
\begin{align*}
P(\tilde{B_t}\in F) =\int_{F} p(t,0,x)dx
\end{align*}
since 
\begin{align*}
variance=\begin{bmatrix}
-1&1
\end{bmatrix}
 \begin{bmatrix}
 t_0&t_0\\
 t_0&t+t_0
 \end{bmatrix}
\begin{bmatrix}
-1\\1
\end{bmatrix}=t.
\end{align*}
Now, suppose that $k\in\mathbb{Z}^+$ be given. Then observe that
\begin{align*}
&P(\tilde{B}_{t_1}\in F_1 ,\dots, \tilde{B}_{t_k}\in F_k )=\int_{F_1} p(t_1,0,x_1)P(\tilde{B}_{t_1}=x_1,\tilde{B}_{t_2}\in F_2 ,\dots, \tilde{B}_{t_k}\in F_k )dx_1\\
&=\int_{F_1} p(t_1,0,x_1)P(B_{t_0}={B}_{t_1+t_0}-x_1,\tilde{B}_{t_2}\in F_2 ,\dots, \tilde{B}_{t_k}\in F_k )dx_1\\
&=\int_{F_1} p(t_1,0,x_1)P({B}_{t_2+t_0}-B_{t_1+t_0}+x_1\in F_2 ,\dots, {B}_{t_k+t_0}-B_{t_0}+x_1\in F_k )dx_1\\
&=\int_{F_1\times F_2} p(t_1,0,x_1)p(t_2-t_1,x_1,x_2)P({B}_{t_2+t_0}-B_{t_1+t_0}+x_1=x_2,{B}_{t_3+t_0}-B_{t_1+t_0}+x_1\in F_2 ,\dots, {B}_{t_k+t_0}-B_{t_0}+x_1\in F_k )dx_1dx_2\\
&=\int_{F_1\times F_2} p(t_1,0,x_1)p(t_2-t_1,x_1,x_2)P(-B_{t_1+t_0}+x_1=x_2-{B}_{t_2+t_0},{B}_{t_3+t_0}-B_{t_1+t_0}+x_1\in F_2 ,\dots, {B}_{t_k+t_0}-B_{t_0}+x_1\in F_k )dx_1dx_2\\
&=\int_{F_1\times F_2} p(t_1,0,x_1)p(t_2-t_1,x_1,x_2)P({B}_{t_3+t_0}+x_2-{B}_{t_2+t_0}\in F_3 ,\dots, {B}_{t_k+t_0}+x_2-{B}_{t_2+t_0}\in F_k )dx_1dx_2\\
&=\cdots \\
&=\int_{F_1\times \cdots \times F_k} p(t_1,0,x_1)p(t_2-t_1,x_1,x_2)\cdots p(t_k-t_{k-1},x_{k-1},x_k)dx_1\cdots dx_k
\end{align*}
since for each $i\in \{1,2,\dots, k\}$, 
\begin{align*}
\text{ the variance of }B_{t_{k+1}}-B_{t_k}+x_k=\begin{bmatrix}
-1&1
\end{bmatrix}
 \begin{bmatrix}
 t_{k}+t_0&t_{k}+t_0\\
 t_{k}+t_0&t_{k+1}+t_0
 \end{bmatrix}
\begin{bmatrix}
-1\\1
\end{bmatrix}=t_{k+1}-t_k.
\end{align*}
and the mean of $B_{t_{k+1}}-B_{t_k}+x_k$ is $x_k$.
Since $k\in \mathbb{Z}^+$ was arbitrary, we are done.
