Integration with respect to $\mu(E) := \int_E\chi_{[0,\infty)}(x) e^{-ax} dx$ Having a semester of abstract measure theory behind me, it is still hard to solve actual exercises. Recently I stumbled upon a very interesting one, which I have not encountered before. Consider the emasure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),dx)$ where $\mathcal{B}(\mathbb{R})$ denotes the usual Borel $\sigma$-algebra and $dx$ the Lebesgue measure. For a constant $a > 0$ define the measure $\mu$ on $\mathcal{B}(\mathbb{R})$ by $$\mu(E) := \int_E\chi_{[0,\infty)}(x) e^{-ax} dx$$

Now the task is to calculate $$\int_{\mathbb{R}}x^2 d\mu$$

My first (and natural) idea was to approximate $x^2$ by simple functions and then find a monotone sequence $\varphi_n$ which converges pointwise to $x^2$. I mean this how one calculates the integral by definition. I think this gets kind of complicated if one does not find the right simple functions. Has anyone a better idea? 
 A: 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 $$\begin{align}\int_0^\infty x^2 e^{-ax}dx &= -\frac{1}{a}x^2e^{-ax}\Big\vert_0^\infty + \frac{2}{a}\int_0^\infty xe^{-ax}dx\\
&= \frac{2}{a}\int_0^\infty xe^{-ax}dx\\
&= \frac{2}{a}\left[-\frac{1}{a}xe^{-ax}\Big\vert_0^\infty + \frac{1}{a}\int_0^\infty e^{-ax}dx \right]\\
&= -\frac{2}{a^3}e^{-ax}\Big\vert_0^\infty\\
&= \frac{2}{a^3}
\end{align}$$
