Consider the random variable $X$ defined on the probability space $(\Omega, \mathcal{F}, P)$ distributed as a uniform on $[0,1]$.
The probability distribution function of $X$ is defined as a map $$ p:\mathcal{B}(\mathbb{R})\rightarrow [0,1] $$ such that, for any $E\in \mathcal{B}(\mathbb{R})$, $$ p(E):=\mathbb{P}(X^{-1}(E)) $$ $p$ is a probability measure for $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ and $(\mathbb{R}, \mathcal{B}(\mathbb{R}), p)$ is a probability space.
Let $\mu$ be the Lebesgue measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$.
Why $p(E)=\mu(E)$ for any $E\subseteq [0,1]$? Is that by definition?
My attempt which I think is wrong is that: the probability density function of $X$ is $$ f(t)=\begin{cases} 1 \text{ if $t$ $\in$ $[0,1]$}\\ 0 \text{ otherwise} \end{cases} $$ We know that $f$ is the probability density function of $X$ with respect to $\mu$ meaning that $$ f:=\frac{dp}{d\mu} $$ Is this somehow related with having $p(E)=\mu(E)$ for any $E\subseteq [0,1]$?
This is more or less the definition of uniform distribution. Properties we certainly expect from a uniform (on $[0,1]$) random variable $X$ are that we want $\Bbb P([0,1])=1$ and $\Bbb P([a,b])=\Bbb P([a+c,b+c])$ whenever $0\le a\le b\le b+c\le 1$. Together with additivity, this already leads to $\Bbb P(X\in E)=\mu(E\cap[0,1])$.