# Density of a probability measure that is not absolutely continuous wrt Lebesgue measure

Let $$\mu$$ be a probability measure defined over a Borel $$\sigma$$-algebra of $$\mathbb R$$. Let $$B$$ be a Borel set. A paper I am reading uses that the probability $$\mu$$ assigns to $$B$$ is:

$$\displaystyle \int_{\mathbb R} \chi_{x \in B} \mathrm{d}\mu(x)$$ where $$\chi$$ is the indicator function.

Does this imply that $$\mu$$ is absolutely continuous with respect to the Lebesgue measure and it has a density? The reason I am confused is that the paper does not assume this, however by the Radon-Nikodym theorem, we know that a density function $$f(x)$$ such that $$\mu(B) = \int_{B} f(x) \mathrm{d}x$$ exists only if $$\mu$$ is absolutely continuous. However, even if we do not have such $$f(x)$$, can we have

$$\mu(B) = \displaystyle \int_{B} \mathrm{d}\mu(x)$$ Or is $$\mathrm{d}\mu(x)$$ the same as $$f(x)\mathrm{d}x$$?

$$\mu (B)=\int \chi_{x \in B} d\mu(x)$$ follows from the definition of integral of a simple function. ($$\chi_{x \in B}$$, usually written as $$\chi_B$$ is a simple function). There is no absolute continuity involved here.
• Thanks for your answer. Sorry for not being clear, I know that indicator is a simple function. My main confusion: is $\mathrm{d}\mu(x)$ well-defined for any probability measure $\mu$? Jun 21, 2021 at 9:03
• $\int f(x)d\mu(x)$ is just another notation for $\int fd\mu$ @independentvariable Jun 21, 2021 at 9:05