Let $\mu$ be a probability measure. More specifically, let $\mu$ be the measure associated with the uniform probability distribution on a half-open interval, say $[0,a).$
If I am not wrong, if we denote $\mathcal{B}$ as the Borel $\sigma$-algebra of $\mathbb{R}$, then the definition of $\mu$ is roughly something like:
$$\mu(A) = \dfrac{|A \cap [0,a)|}{a}, \quad \forall A \in \mathcal{B} $$ where $|\cdot|$ is the Lebesgue measure.
I am trying to connect this with the density of the uniform distribution which is $1/a$ on $[0,a)$.
My attempt, which may be not formal or even wrong, is to write $\mu$ as a function of single points such as: $$\mu(x) = \dfrac{|\{x \} \cap [0,a)|}{a}, \quad x \in \mathbb{R} $$ and then I would like to obtain: $\dfrac{\mathrm{d} \mu (x)}{\mathrm{d}x} = \begin{cases} 1/a, & \text{ if } x \in [0,a)\\ 0, & \text{ otherwise.} \end{cases}$
However I cannot just take the derivative like this as $\mu$ is a function of sets. Even if this works, I will need to show the derivative of $|\{ x \} \cap [0,a)|$ for $x \in [0,a)$ is equal to $1$.
I tried to see if I can use Lebesgue's density theorem.. I am not sure how to link these two concepts.
In short, how can I link the measure and the density function of the uniform distribution by differentiating the measure. Apologies if I am violating some formalities here.
Ps. I can always define the density of this measure and obtain $\mu(A)$ by integrating it over $A$, but I am trying to understand if reverse is a common practice, or if it is even possible at all.
The density is the derivative of the CDF $F(x) := \mu((-\infty, x])$. You can check that for the uniform distribution on $(0, a]$, the CDF is $F(x) = \begin{cases}0 & x < 0 \\ x/a & 0 \le x < a \\ 1 & x \ge a\end{cases}.$