It is now more convenient to work with increments $\mu + \sigma
Z_i$ with respect to normalized random variables $Z_i$.
\begin{eqnarray*}
   E(X_{T_1}) &=&   E\left (
                   \sum_{j=1}^{T_1} \mu + \sigma Z_j
                       \right ) \\
     &=&  \sum_{k\ge 1} \int_{\{T_1=k\}} \sum_{j=1}^k ( \mu +
     \sigma Z_j ) \, d{\cal P} \\
     &=& \sum_{k\ge 1} k \mu P(T_1=k) + \sigma \sum_{k\ge 1}
          \sum_{j=1}^k \int_{\{T_1=k\}} Z_j \, d {\cal P} \\
\end{eqnarray*}
The first term on the right-hand side is obviously $\mu E(T_1)$.
Rearranging the sum of the second term on the right hand side,
\begin{eqnarray*}
\sigma \sum_{k\ge 1}
          \sum_{j=1}^k \int_{\{T_1=k\}} Z_j \, d {\cal P}
          &=&
          \sigma \sum_{j\ge 1}\sum_{k \ge j}
             \int_{\{T_1=k\}} Z_j \, d {\cal P} \\
          &=& \sigma \sum_{j\ge 1}\int_{\{T_1\ge j\}} Z_j\,d {\cal
          P}\\
          &=& \sigma \sum_{j\ge 1}\left( E(Z_j) -
          \int_{\{T_1<j\}}Z_j \,d {\cal P}\right)
\end{eqnarray*}
Since $Z_j$ and $\{T_1<j\}$ are independent,
\begin{eqnarray*}
 E(Z_j) -
          \int_{\{T_1<j\}}Z_j \,d {\cal P} &=& E(Z_j)(1 -
          P(T_1<j)) = 0
\end{eqnarray*}
because $E(Z_j)=0$. We conclude that the second term is zero.
