Proof that a random measure with orthogonal increments is a measure Let me first state what I mean by a random measure with orthogonal increments.

Definition: A random measure with orthogonal increments $Z$ is a collection $\left(Z(B): B \in \mathcal{B}_{(-\pi,\pi]}\right)$  of zero-mean, complex random variables $Z(B)$ indexed by the Borel sets $\mathcal{B}_{(-\pi,\pi]}$ defined on some probability space $(\Omega,\mathcal{U},P)$ such that for some finite measure $\mu$ on $(-\pi,\pi]$
  $$\mathrm{cov}(Z(B_1),Z(B_2))=\mu(B_1\cap B_2) \quad \forall{B_1,B_2}\in \mathcal{B}_{(-\pi,\pi]}$$

Honestly, I don't think I understand this definition. If $Z$ is a measure then it must be true that $Z(\emptyset) = 0$. How does this follow from the definition above exactly? Do I take $B_1 = B_2 = \emptyset$ and argue that given $\mathrm{cov}(Z(B_1),Z(B_2))=\mu(B_1\cap B_2) = 0$ and $E[Z(B)] = 0$ for any $B \in \mathcal{B}_{(-\pi,\pi]}$, $Z(\emptyset) = 0$ $\ P$-almost surely?
Next I would like to show that $Z(\cup_j B_j) = \sum_jZ(B_j)$ in mean square whenever $B_1,B_2,\ldots$ is a sequence of pairwise disjoint Borel sets. I am not sure what the mathematical formulation of this result is. My guess is
$$E[\lvert Z(\bigcup_j B_j) - \sum_jZ(B_j)\rvert^2] = 0$$
as this would imply  $Z(\cup_j B_j) = \sum_jZ(B_j)$ $\ P$-almost surely, which I think is the most you can expect from a random measure. (If it were an ordinary measure, we would insist on algebraic equality.)
Assuming that my guess is right I proceed. From the definition I have (after a few steps)
$$\mathrm{var}\left(Z\left(\bigcup_j B_j\right)\right) = \mathrm{var}\left(\sum_jZ(B_j)\right) $$
I don't know how to continue from this point on. I would really appreciate some help. Thanks.
 A: You essentially have proven the result.  I will restate and embellish upon your calculations to show that $Z$ satisfies the definition of a Vector Measure.  
Let $\mathcal{F}$ be the Borel field for the interval $(-\pi, \pi]$.  Let $X$ be the set of complex valued $L^2$, mean-zero, random variables on the probability space $(\Omega, \mathcal{U}, P)$.  We then have the function $$Z : \mathcal{F} \to X$$

$Z(\emptyset) = 0$ 

By definition $Z(\emptyset)$ is the random variable in $X$ s.t. 
$$\mathrm{cov}(Z(\emptyset), Z(\emptyset)) = \mu(\emptyset \cap \emptyset) = 0$$
The only function in $X$ that satisfies this (i.e. has $L^2$ norm of $0$) is the zero function.

Additivity: Suppose $B_1, B_2$ are disjoint.  Then
  $$Z(B_1 \cup B_2) = Z(B_1) + Z(B_2)$$

From the definition of $Z$ we know that 
$$\textrm{cov}(Z(B_1 \cup B_2), Z(B_i)) = \mu(B_i)$$
Therefore, by linearity of covariance, and additivity of $\mu$, and definition of $Z$:
\begin{eqnarray*}
\textrm{cov}(Z(B_1 \cup B_2), Z(B_1) + Z(B_2)) &=& \mu(B_1 \cup B_2) \\
&=& \sqrt{\mu(B_1 \cup B_2)} \sqrt{\mu(B_1) + \mu(B_2)}\\
&=& \sqrt{\textrm{var}(Z(B_1\cup B_2))}\sqrt{\textrm{var}(Z(B_1)) + 2\textrm{cov}(Z(B_1), Z(B_2)) + \textrm{var}(Z(B_2))} \\
 &=& \sqrt{\textrm{var}(Z(B_1\cup B_2))}\sqrt{\textrm{var}(Z(B_1) + Z(B_2))} \\
\end{eqnarray*}
This can only happen if $Z(B_1 \cup B_2)$ and $Z(B_1) + Z(B_2)$ are colinear.  Additionally, since they have the same norm, they must be equal.

Countable Additivity: Suppose $B_i \in \mathcal{F}$, disjoint, then
  $$Z(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty Z(B_i)$$
  where the right hand side converges wrt to the norm of $X$.

The approach we can take here is to show $S_n = \sum_{i=1}^n Z(B_i)$ is cauchy wrt the $L^2$ norm.
\begin{eqnarray*}
\int_\Omega \overline{(S_n - S_m)}(S_n - S_m) dP &=& \mathrm{Cov}(\sum_{i={m+1}}^nZ(B_i), \sum_{i={m+1}}^nZ(B_i)) \\
&=& \sum_{i={m+1}}^n \mathrm{Var}(Z(B_i)) \\
&=& \sum_{i={m+1}}^n \mu(B_i) \\
\end{eqnarray*}
Here we used $Z(B_i)$ are uncorrelated since the $B_i$ are disjoint.  Now since $\sum_{i=1}^\infty \mu(B_i) = \mu(\cup_{i = 1}^\infty B_i) < \infty$ (since $\mu$ is a finite measure) we know that the partial sums $\sum_{i=m}^n \mu(B_i)$ become arbitrarily small so we are done showing they are Cauchy.
Let $S$ be the limit of the Cauchy sequence $S_n = \sum_{i=1}^n Z(B_i) = Z(\cup_{i=1}^n B_i)$.  It remains to show that $S = Z(\cup_{i=1}^\infty B_i)$.
\begin{eqnarray*}
||Z(\cup_{i=1}^\infty B_i) - S||_2 &=& ||(Z(\cup_{i=1}^\infty B_i) - S_n) + (S_n - S)||_2 \\
&=& ||(Z(\cup_{i=n+1}^\infty B_i) + (S_n - S)||_2 \\
&\leq& ||(Z(\cup_{i=n+1}^\infty B_i)||_2 + ||S_n - S||_2 \\
&=& \mu(\cup_{i=n+1}^\infty B_i) + ||S_n - S||_2 \\
\end{eqnarray*}
As above, $\mu(\cup_{i=n+1}^\infty B_i)$ becomes arbitrarily small as $n$ becomes large.  Furthermore, $||S_n - S||_2$ does as well since $S_n$ converges to $S$.
