Using the f(x) generated in the previous theorem
\begin{align*}
\mu & = E[X] \\
& = \sum_{x=0}^{\infty} x \cdot \frac{(\lambda T)^x}{x!} e^{-\lambda T}\\
& = \lambda T e^{-\lambda T} \sum_{x=1}^{\infty} \frac{(\lambda T)^{x-1}}{(x-1)!} \\
& = \lambda T e^{-\lambda T} \sum_{k=0}^{\infty} \frac{(\lambda T)^k}{k!} \\
& = \lambda T e^{-\lambda T} e^{\lambda T} \\
& = \lambda T
\end{align*}
which confirms the use of \(\mu\) in the original probability formula.