Is it $\mu(dx)$ or $d\mu(x)$? or they are equal?

Question

In probability theory, I have seen two forms of an integral. Let $\mu$ be a Borel measure and $f$ is a function.

What is the difference between the following two forms: \begin{eqnarray} \int_{\mathbb{R}^d} f(x) \mu(dx) \end{eqnarray} and \begin{eqnarray} \int_{\mathbb{R}^d} f(x) d\mu(x) \end{eqnarray}

Please give some references for your answer.

Answer 1

Since the other answers didn't provide a reference and you asked for one in the comments, a reference can be found in Schillings Measures Integrals and Martingales (ISBN 9780521615259), Page 77. There they state: In case we need to exhibit the integration variable, we write $$ \int u d \mu=\int u(x) \mu(d x)=\int u(x) d \mu(x) $$

Note that for a distribution of a random variable: $$\mathbb P_X(A)= \mathbb P ( X \in A)$$ Schilling also writes $$\int u(X(\omega)) \mathbb P(d \omega)= \int u(s) \mathbb P (X \in ds)$$ for suitable function $u$.

Answer 2

Let me start by saying that unfortunately I have no references at hand.

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Both are notations for the same thing: the integral of function $f$ with respect to measure $\mu$.

Which to use is a matter of taste.

Personally I prefer: $$\cdots\mu(dx)$$ because somehow a measurement of the infinitesimal small $dx$ takes place.

In the special case where $\lambda$ denotes the Lebesgue measure on $\mathbb R$ you could say that we have the equality: $$\lambda(dx)=dx$$

i.e. the measure of $dx$ equals $dx$ itself.

If $\mu$ is also a measure on $\mathbb R$ and this with a density $f$ wrt the Lebesgue measure then we can state:$$\mu(dx)=f(x)\lambda(dx)=f(x)dx$$

Answer 3

The $\mu(dx)$ notation is convenient in stochastic processes. If $X$ is a (for simplicity time-homogeneous) Markov process, then instead of a transition matrix (as in Markov chains) one uses transition functions.

Ignoring some technical measurability conditions, a transition function is defined by the property $$\mathbb{P}(X_t \in A \mid X_s = x) = \int_A P_t(x, dy)$$ Here we assume that $P_t(x, \cdot)$ is a probability measure. The formula above means that if $\mu = P_t(x, \cdot)$, then $$\mathbb{P}(X_t \in A \mid X_s = x) = \int_A \mu(dy)$$

It would be really clumsy to write $$\mathbb{P}(X_t \in A \mid X_s = x) = \int_A dP_t(x, \cdot)$$