Thanks to tomasz, user251257 and Zoran Loncarevic I present my solution, which solves the stated question as a corollary.
Proposition. Let $(X,\mathcal{A},\mu)$ be a measure space and $f \in \mathcal{M}$.
Then the function $\nu: \mathcal{A} \to [0,\infty]$ defined by
$$\nu(A) := \int_A f d\mu$$ is a measure on $\mathcal{A}$.
For calculating an integral with respect to above measure, there is a nice proposition.
Proposition. Let $(X,\mathcal{A},\mu)$ be a measure space and $f,g \in
\mathcal{M}$. With the terminology established above it holds that $$\int_X g d\nu = \int_X fg d\mu$$
Proof. Assume $g := \sum_{i = 1}^n a_i \chi_{A_i} \in \Sigma^+$. Then
$$\int_X g d\nu = \sum_{i = 1}^n a_i \nu(A_i) = \sum_{i = 1}^n a_i \int_{A_i} f d\mu = \int_X f\left(\sum_{i = 1}^n a_i\chi_{A_i}\right) d\mu = \int_X fg d\mu$$
Now let $g \in \mathcal{M}^+$. Then we find a sequence $(\varphi_n)_{n \in \mathbb{N}}$ of functions in $\Sigma^+$ such that $\varphi_n \nearrow g$ pointwise. Hence
$$\int_X g d\nu = \lim_{n \to\infty} \int_X \varphi_n d\nu = \lim_{n \to \infty} \int_X f\varphi_n d\mu = \int_X fg d\mu$$
by monotone convergence since $f\varphi_n \nearrow fg$. Lastly let $g \in \mathcal{M}$. Write $g = g^+ - g^-$. Thus
$$\int_X g d\nu = \int_Xg^+ d\nu - \int_X g^- d\nu = \int_X fg^+ d\mu - \int_X fg^- d\mu = \int_X fg d\mu$$
Corollary. We have $$\int_{- \infty}^\infty x^2 d\mu = \frac{2}{a^3}$$
Proof. By above proposition we have that $$\int_{- \infty}^\infty x^2 d\mu = \int_0^\infty x^2 e^{-ax}dx$$ Twice partial integration yields $$\begin{align}\int_0^\infty x^2 e^{-ax}dx &= -\frac{1}{a}x^2e^{-ax}\Big\vert_0^\infty + \frac{2}{a}\int_0^\infty xe^{-ax}dx\\
&= \frac{2}{a}\int_0^\infty xe^{-ax}dx\\
&= \frac{2}{a}\left[-\frac{1}{a}xe^{-ax}\Big\vert_0^\infty + \frac{1}{a}\int_0^\infty e^{-ax}dx \right]\\
&= -\frac{2}{a^3}e^{-ax}\Big\vert_0^\infty\\
&= \frac{2}{a^3}
\end{align}$$
