TITLE: Connecting abstract probability theory with simple distributions
QUESTION [5 upvotes]: Suppose I have the probability space $(X,\mathcal{E},P)$ and a random variable
$u : X \to \mathbb{R}$. $u$ induces a probability $Q (B) = P (u^{-1}(B))~\forall B \in \mathcal{B}$, the Borel $\sigma$-algebra on $\mathbb{R}$ and we can consider a new probability space  $(\mathbb{R},\mathcal{B},Q)$.
Now, I am having trouble connecting this theory to practical applications. For example, can I let $u$ to be a Standard Normal Random Variable? If so, the expectation of $u$ is defined as $E[u] = \int\limits_X u dP$. How does this equal
$\int\limits_{\mathbb{R}} x f(x) dx$ where $f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$?
I guess that the induced measure has to be brought into play along with some Radon-Nikodym theorem to show this but I am not sure how to do so. 
Any help is greatly appreciated.
Thanks,
Phanindra

REPLY [10 votes]: The definition of induced measure can be expressed as equality of integrals
over the two probability spaces
$$\int_\Omega 1_B(u(\omega))\, P(d\omega) = \int_\mathbb{R} 1_B(x)\, Q(dx).$$
This equation is true for every Borel set $B$, and can be extended to all 
bounded Borel measurable functions $g$; that is, 
$$\int_\Omega g(u(\omega))\, P(d\omega) = \int_\mathbb{R} g(x)\, Q(dx).$$
Provided the mean of $u$ exists, you can use truncation and dominated convergence
to extend this to the unbounded function $g(x)=x$ and conclude that 
$$\int_\Omega u(\omega)\, P(d\omega) = \int_\mathbb{R} x\, Q(dx).$$
If the random variable $u$ has a density function $f$, then integration 
with respect to $Q(dx)$ can be replaced by integration with respect to 
$f(x)\,dx$ so that the mean is $E(u)=\int_\mathbb{R} x f(x)\,dx$.
This fact is sometimes expressed with the funny terminology: 
the law of the unconscious statistician.