Is it $\mu(dx)$ or $d\mu(x)$? or they are equal? 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. 
 A: Let me start by saying that unfortunately I have no references at hand.

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:$$\mu(dx)=f(x)\lambda(dx)=f(x)dx$$
A: 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 $u$.
A: 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 
$$\mathbb{P}(X_t \in A \mid X_s = x) = \int_A dP_t(x, \cdot)$$
