Meaning of $\int_E {f(x) \mu(dx)}?$ Suppose $f$ is a measurable real-valued function defined on a measure space $(E, X , \mu)$. What is the meaning of the RHS of the following integral

$$\int_E{f d\mu} = \int_E {f(x) \mu(dx)}?$$

I understand that LHS means 'integrate $f$ with respect to the measure $\mu$'. However, I fail to understand RHS. 
Remark: The integral above is taken from here, under 'Construction - Integration'.
 A: The notation $\mu(dx)$ seems to come from the Lebesgue integration
$$
\sum f(\xi_i)\mu([x_i,x_{i+1}])
$$
where after taking the limit (plus translation invariance) the symbol $\mu(dx)$ appears. 
The alternative notation $d\mu(x)$ looks more like to arrive from the Riemann-Stieltjes approach 
$$
\sum f(\xi_i)(g(x_{i+1})-g(x_i))
$$
where the limit gives $dg(x)$.
Another way to see the relation between those two notations (assuming everything to exist)
$$
d\mu(x)=\mu'(x)dx=\tilde\mu(dx).
$$
A: Suppose you're doing integration with the intuitive approach!
Then in Riemann Integral, you multiply the height of function in a very small part of horizontal line, called $dx$ !
But In Lebesgue Integral, The horizontal line must not be the real line, similarly the size of that small part ($dx$) is $\mu(dx)$ or in different notation $d\mu(x)$. Unfortunately, Lebesgue Integral doesn't have a standard notation as Riemann's Integral.
This is notation helps you, as well, to work with multiple integrals with different variables.
A: The RHS just emphasizes that $f$ and $\mu$ are functions on $X$. Sometimes $\int_E f d\mu$ is written $\int_E f(x) d\mu(x)$ as well. 
A: My understanding is the following, based in part by reading of this book, e.g. the Change of variables formula.
For Lebesgue integrals, it is customary to use $d\mu$ to signify the use of the measure $\mu$ in the integration. This is reflected in $\int_{\Omega} f d\mu$ of the LHS. 
A random variable $X$ is a measurable map from $(\Omega, \mathcal{F})$ to $(R, \mathcal{R})$. To make things explicit, let's write $X$ as $h(\omega), \omega\in \Omega$.
 Then we can write
$$ \int_{\Omega} f d\mu = \int_{\Omega} f(h(\omega)) d\mu.$$
Note a random variable induces a new measure on Borel sets in $\mathcal{R}$. Let's denote the new measure as $\nu$. Then $\nu(A) = \mu(h^{-1}(A))$, for $A \in \mathcal{R}$. The $\mu(dx)$ in the RHS really means $d\nu$, i.e. $$ \mu(dx) \equiv d\nu \equiv d\left( \mu \circ h^{-1} \right).$$
Hence
$$ \int_R f \mu(dx) \equiv \int_R f(x) d\nu.$$
Note $dx$ is conventionally reserved for $d\lambda$ where $\lambda(\cdot)$ is the Lebesgue measure on Borel sets. The reasons for using $\mu(dx)$ instead of $d\nu$ that I can think of:


*

*to avoid defining a new symbol $\nu$;

*to signify a new measure induced by applying $\mu$ over (inverse of) Borel sets, i.e. $\mu \circ h^{-1}$.


Note I silently changed your integration domains of $E$ to $\Omega$ and $R$ to make things correct. Finally your equation really is
$$ \int_{\Omega} f(h(\omega)) d\mu =  \int_R f(x) d\nu.$$
The equation means one can carry out integrations over $R$, the domain of random variables, instead of over the the original sample space $\Omega$ for computational convenience, and the results are guaranteed to be the same as long as the induced measure over $(R, \mathcal{R})$, denoted by $\mu(dx)$, is used.
Note in the above formulation, the LHS integrates over $\Omega$. This is good in theory, but bad for practical computation. And the RHS integrates over $R$ which is much more amenable to computation. And in practice, we often have the density function for $\nu$ with respect to the Lebesgue measure, i.e. $\frac{d\nu}{dx} = g(x)$, then the RHS becomes
$$ \int_R f(x)g(x)dx $$
that everyone recognizes. 
