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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 $f(x)\mathrm{d}x$?

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$\mu (B)=\int \chi_{x \in B} d\mu(x)$ 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.

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  • $\begingroup$ 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 $\mu$? $\endgroup$ Jun 21, 2021 at 9:03
  • $\begingroup$ $\int f(x)d\mu(x)$ is just another notation for $\int fd\mu$ @independentvariable $\endgroup$ Jun 21, 2021 at 9:05

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