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 $f(x)\mathrm{d}x$?
Best Answer
$\mu (B)=\int \chi_{x \in B} d\mu(x)$ 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.