I'm trying to better understand the measure theoretic definition of a probability density function, but I believe I'm making a mistake somewhere if someone could clarify. From what I currently gather, let $(\Omega,F,\mathbb{P})$ be a probability space and let $X:\Omega\rightarrow \mathbb{R}$ be a real valued random variable. Then, the distribution of $X$ is the push-forward measure
$$
X_*\mathbb{P}(E)=\mathbb{P}(X^{-1}(E))=\mathbb{P}(X\in E)
$$
where $E\subseteq \mathbb{R}$ is some measurable set in $\mathbb{R}$. The cumulative distribution function of $X$ is then
$$
F(a) = \mathbb{P}(X\in (-\infty,a]) = \mathbb{P}(X^{-1}(-\infty,a])
$$
To obtain the probability density function, we assume that $\mathbb{P}<<\lambda$ where $\lambda$ is the Lebesgue measure and then note
$$
F(a)=\mathbb{P}(X^{-1}(-\infty,a]) = \int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} = \int_{X^{-1}(-\infty,a]} \frac{d\mathbb{P}}{d \lambda} d\lambda
$$
where $\frac{d\mathbb{P}}{d \lambda}$ is the Radon-Nikodym derivative of $\mathbb{P}$ with respect to $\lambda$. We then set $f=\frac{d\mathbb{P}}{d \lambda}$ and call $f$ the probability density function.
Alright, so I know there's an error here since I really want
$$
F(a) = \int_{-\infty}^a f d\lambda
$$
but instead I have
$$
\int_{X^{-1}(-\infty,a]} f d\lambda
$$
Basically, there's $X^{-1}$ here and I don't think there should be, so what's the error in what I'm doing with Radon-Nikodym? Second, I'm still not entirely sure how to define $\mathbb{P}$ concretely. Say we're working with a normal distribution, I know that
$$
f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$
and
$$
F(x) = \frac{1}{2}\left[1+\textrm{erf}\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]
$$
In this case, what would $\mathbb{P}$ be?
Thanks for the help in advance.
Best Answer
You claim the following change of variables: $$\int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} \overset{?}{~=~} \int_{\color{crimson}{X^{-1}(-\infty,a]}} \frac{d\mathbb{P}}{d \lambda} d\lambda$$
However, you should change the integration interval when you changed the variable.
$$\int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} ~=~ \int_{\color{navy}{(-\infty,a]}} \frac{d\mathbb{P}}{d \lambda} d\lambda$$
That is all.