Normally Distributed = Absolute Continuity? Let $(\Omega, \mathcal{F}, P)$ be a probability space.  A random variable $X: \Omega \to \mathbb{R}$ is said to have the standard normal distribution if it has the density $f:\mathbb{R} \to [0,\infty)$ given by
$$
f(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}.
$$
(A) More precisely, let $(\mathbb{R}, \mathcal{B})$ be the real line with the Borel sigma algebra and define the distribution measure $\mu : \mathcal{B} \to [0,1]$ by $\mu(B) = P(X^{-1}(B))$.  Let $\lambda: \mathcal{B} \to [0,\infty]$ be the Lebesgue measure on $\mathbb{R}$.  Then both $\mu$ and $\lambda$ are $\sigma$-finite measures on $(\mathbb{R}, \mathcal{B})$, and if $\mu << \lambda$ (not sure how to show this) then by the Radon-Nikodym theorem, there exists a non-negative Borel-measurable $f$ such that
$$
\mu(B) = \int_B f\, \rm{d}\lambda.
$$
We call $f$ the density, or Radon-Nikodym derivative, of $\mu$ w.r.t. $\lambda$.
(B) On the other hand, if we define the the measure $\mu$ on $(\mathbb{R}, \mathcal{B})$ by
$$
\mu(B) := \int_B f \, \rm{d} \lambda
$$
where again $\lambda$ is the Lebesgue measure on $(\mathbb{R}, \mathcal{B})$ and $f$ is as above, then $\mu$ is indeed a probability measure on $(\mathbb{R}, \mathcal{B})$ and $\mu << \lambda$.  However, I don't quite see how $\mu(B) = P(X^{-1}(B))$ is guaranteed now.
So, when we state that a random variable has a normal (or any) distribution, it seems like we really use method (B), where we define the density first, and then the distribution measure $\mu$ is whatever results from setting $\mu(B) = \int_B f \, \rm{d} \lambda$.  
Is this correct?  If so, how are we guaranteed that $\mu(B) = P(X^{-1}(B))$?  Also, is it generally true that $\mu << \lambda$, even though it doesn't seem necessary when specifying a distribution?
 A: In (A), assuming you're talking about that normal variable: Saying $\mu(B)=P(X^{-1}(B))$ is the definition of "$\mu$ is the distribution of $X$". And now the definition of "$X$ is normal with mean $0$ and variance $1$" is "if we let $\mu$ be the distribution of $X$, as defined above, then $\mu(B)=\int_B f\,d\lambda$, where $f$ is as above".
Regarding the second question, or what I think is the second question, regarding getting $X$ starting with a distribution function: Say $f\ge0$ on $\mathbb R$ and $\int f\,d\lambda=1$. Define $\mu$ by $\mu(B)=\int_B f\,d\lambda$; then $\mu$ is a probability measure on $\mathbb R$. At this point in (B) you ask a question about $X$; you can't answer because you haven't said what $X$ is!
So far we have a Borel probability measure $\mu$ on $\mathbb R$. So $(\mathbb R,\mathcal B,\mu)$ is a probability space. Now $X$ is a random variable on this probability space. That means $X:\mathbb R\to\mathbb R$; in fact $X$ is the random variable on this probability space defined by $$X(t)=t\quad(t\in\mathbb R).$$That's the definition of $X$. Definitions are good: It's now clear that $X^{-1}(B)=B$, so $$P(X^{-1}(B))=P(B)=\int_B f\,d\lambda.$$
No, probability distributions don't have to be absolutely continuous. Say $X=1$. The distribution is the measure $\mu=\delta_1$.
