Rewriting the numerator,
 \begin{eqnarray*}
\sum_{i=1}^n (X_{T'_i}-\hat{\mu}T'_i)^2 &=& \sum_{i=1}^n\bigl(
 (X_{T'_i}-\mu T'_i) +(\mu-\hat{\mu})T'_i\bigr)^2 \\
 &=& \sum_{i=1}^n(X_{T'_i}-\mu T'_i)^2
    + 2 \sum_{i=1}^n (X_{T'_i}-\mu T'_i)(\mu-\hat{\mu})T'_i
    \\
    & & + \sum_{i=1}^n \bigl( (\mu-\hat{\mu})T'_i\bigr) ^2.
 \end{eqnarray*}
By the previous lemma and the Law of Large Numbers, the first term
of the sum converges to $n \sigma^2E(T_1)$. We have shown that the
denominator converges to $n E(T_1)$. Therefore to prove
$\hat{\sigma}^2$ is strongly consistent it is sufficient to show,
 \[
(\mu-\hat{\mu})^2 \sum_{i=1}^n  {T'_i}^2 / \sum_{i=1}^n T'_i
\goesto 0
\]
and
\[
 \sum_{i=1}^n (X_{T'_i}-\mu T'_i)(\mu-\hat{\mu})T'_i
 / \sum_{i=1}^n T'_i  \goesto 0
 \]
 with probability 1.

We have shown that $E({T_1}^2)$ exists and is finite, and that
$\mu-\hat{\mu}\goesto 0$ with probability 1, so the first of these
limits is proven.

The second limit is shown using the Cauchy-Schwarz inequality,
\begin{eqnarray*}
0 &\le& \left ( \sum_{i=1}^n (X_{T'_i}-\mu
T'_i)(\mu-\hat{\mu})T'_i  / \sum_{i=1}^n T'_i \right) ^2
\\
&\le& \left( \sum_{i=1}^n (X_{T'_i}-\mu T'_i) / \sum_{i=1}^n T'_i
\right ) \left ( \sum_{i=1}^n(\mu-\hat{\mu})T'_i  / \sum_{i=1}^n
T'_i\right)
\\ &\goesto& \sigma^2 \cdot 0 = 0
\end{eqnarray*}
with probability 1.
