Continuous distributions I've just started with various types of continuous distributions and the very first one is the uniform continuous distribution. One can justify that the function given as its pdf is indeed A pdf, but how does one justify that pdf is indeed THE pdf of the distribution. One should then first define its cdf to be the improper integral of that function from negative infinity till t, for each t in R. But here I run into another problem. Is such a definition possible, without any inconsistencies with the axioms of probability? 
I've learned that the pdf is unique as it is also defined to be continuous on its support. Thus using the Fundamental Theorem of Calculus, one can see its uniqueness. 
 A: If $F$ is the distribution of a random variable $X$, i.e. $$F(x) = \mathbb P(X\leqslant x)$$ for $x\in\mathbb R$, and $F$ is continuous (i.e. $X$ is a continuous random variable), then a probability density function of $X$ is any function $f$ satisfying
$$\mathbb P(X\in B) = \int_B f(x)\mathsf dx $$
for all Borel sets $B$. This function need not be unique; for example, for the uniform $[0,1]$ distribution, we have
$$ F(x) = \begin{cases} 0,& x\leqslant 0\\ x,& 0<x<1\\ 1,& x\geqslant 1.\end{cases}$$
One density would be
$$ f(x) = 1_{(0,1)}(x).$$ 
But we could also use the density
$$g(x) = 1_{[0,1]}(x), $$
as $f=g$ a.e. and so $$\int_B f(x)\mathsf dx = \int_B g(x)\mathsf dx$$ for any Borel set $B$.
A: A pdf is not unique. In measure theoretic probability you find that the pdf is determined only up to a set of measure zero (on the line, which intuitively means it has no length). If you alter the pdf on a set of measure zero, you obtain a new pdf, which has the same corresponding cdf. 
This means that for example $1_{(0,1)}$ and $1_{[0,1]}$ are both valid pdfs of the variable which is uniform on $[0,1]$.
