Since, $ X_{T_1} -\mu \, T_1 =\sum\sigma Z_j, $ it is sufficient
to show that $E\bigl((\sum Z_j)^2\bigr) =E(T_1)$.
\begin{eqnarray*}
    E\left[\left( \sum_{j=1}^{T_1} Z_j \right)^2 \right] &=&
    \sum_{k\ge 1} \int_{\{T_1=k\}}\left(\sum_{j=1}^k Z_j\right)^2
    \, d{\cal P} \\
    &=& \sum_{k\ge 1}\sum_{i=1}^k\int_{\{T_1=k\}} Z_i^2\,d{\cal P}
      +2\sum_{k\ge 1}\sum_{i=1}^{k-1}\sum_{j=i+1}^k
      \int_{\{T_1=k\}} Z_i Z_j\,d{\cal P}
\end{eqnarray*}
Consider the first term on the right-hand side,
\begin{eqnarray*}
\sum_{k\ge 1}\sum_{i=1}^k\int_{\{T_1=k\}} Z_i^2\,d{\cal P} &=&
\sum_{i\ge 1}\sum_{k\ge i}\int_{\{T_1=k\}} Z_i^2\,d{\cal P} \\
&=& \sum_{i\ge 1}\int_{\{T_1\ge i\}}Z_i^2 \, d{\cal P} \\
&=& \sum_{i\ge 1} \left( E(Z_i^2) - \int_{\{T_1< i\}}Z_i^2\,d{\cal P}  \right) \\
&=& \sum_{i\ge 1} E(Z_i^2)( 1 - P(T_1<i))
\end{eqnarray*}
because $\{T_1<i\}\in  {\cal F}_{i-1}$ and is independent of
$Z_j$. Noting that $E(Z_i^2)=1$, we have that this first term is
$\sum P(T_1\ge i) = E(T_1)$.

The second term is rearranged and shown to equal zero,
\begin{eqnarray*}
2\sum_{k\ge 1}\sum_{i=1}^{k-1}\sum_{j=i+1}^k
      \int_{\{T_1=k\}} Z_i Z_j\,d{\cal P} &=&
      2 \sum_{i\ge 1}\sum_{j\ge i+1}\sum_{k\ge j}\int_{\{T_1=k\}}
      Z_i Z_j \, d{\cal P} \\
      &=& 2 \sum_{i\ge 1}\sum_{j\ge i+1}\int_{\{T_1\ge j\}}
      Z_iZ_j\, d{\cal P} \\
      &=&2 \sum_{i\ge 1}\sum_{j\ge i+1}\left(
          E(Z_iZ_j) - \int_{\{T_1< j\}}Z_i Z_j\,d{\cal P}
      \right)
\end{eqnarray*}
For $i\not=j$, $E(Z_iZ_j)=E(Z_i)E(Z_j)=0$. Also, since
$I_{\{T_1<j\}}Z_i$ is independent of $Z_j$ when $i<j$, the
integral is zero.