Is there any difference between the notations $\int f(x)d\mu(x)$ and $\int f(x) \mu(dx)$? Suppose $\mu$ is a measure. Is there any difference in meaning between the notation
$$\int f(x)d\mu(x)$$
and the notation
$$\int f(x) \mu(dx)$$?
 A: Just to add to what has been said already -- the notations $\int f\,d\mu$, $\int f(x)\,\mu(dx)$ and $\int f(x)\,d\mu(x)$ are all very common and have identical meanings. There is also the even briefer notation $\mu(f)$, so you can consider the measure as something acting directly on a function. You can even skip the parentheses and just write $\mu f$, as used by Kallenberg, Foundations of Modern Probability. If you don't have any reason to explicitly write the variable of integration, then either of $\int f\,d\mu$, $\mu(f)$, $\mu f$ will do (although the first is probably the clearest for most people). If $\mu$ is a probability measure then $\mathbb{E}_\mu[f]$ is also common or, simply, $\mathbb{E}[f]$ (the expectation or expected value of $f$) if there is no confusion over which measure is being used. If you do need to write the variable, then I don't think that there is any real preference between $\mu(dx)$ and $d\mu(x)$. While the latter does seem a bit more consistent with the notation $\int f\,d\mu$, the former is often more convenient. This is just because sometimes you can be forced into this notation anyway when there is more than one variable, and it is nice to be consistent. For example a kernel $\mu(x,A)$ is a measurable function of the first variable, $x$, and is a measure in the second, $A$. You would then write $\int f(x,y)\,\mu(x,dy)$ for the integral, whereas $\int f(x,y)\,d\mu(x,y)$ would be confusing.
I just looked at the introductions to some of my probability textbooks and Revuz & Yor, Continuous Martingales and Brownian Motion, does explicitly mention several notations which are to be used interchangeably.

For a measure $m$ on $(E,\mathcal{E})$ and $f\in\mathcal{E}$, the integral of $f$ with respect to $m$, if it makes sense, will be denoted by any of the symbols
  $$
\int f\,dm,\ \int f(x)\,dm(x),\ \int f(x)\,m(dx),\ m(f),\ \langle m,f\rangle,
$$
  and in case $E$ is a subset of a euclidean space and $m$ is the Lebesgue measure, $\int f(x)\,dx$.

So, that's the four different notations mentioned above along with the additional one $\langle m,f\rangle$ which, I must say, I don't think is very common at all.
A: There is no difference in meaning. I grabbed five books at random from my shelf and four out of five use the form $\int f(x) \mu(dx)$, while one uses the form  $\int f(x) d\mu(x)$. The form suppressing the variable of integration $\int f d\mu$ is also very common. I have never seen $\int f(x) d\mu$, which seems to me a weird hybrid. I work in probability so there may be some bias.
A: The conventions I have seen are as follows. When there is no need to see the variable of integration, $\int f\, d\mu$ is used and is preferred.   There is no need to show a dummy variable you are not using.
When you need to see the varible of integration, as you need to see it here,
$$E(X^2) = \int_{-\infty}^\infty x^2 \, dF_X(x),$$
you put the variable of integration in parens after the integrator.  I have seen the $\mu(dx)$ notation on some older probability books.
A: At times, I find the $\mu(dx)$ notation to be quite intuitive. Informally, if we think of $dx$ as representing an infinitesimally small "chunk" of the real line, then $\mu(dx)$ is its measure.
For a formal example, let $F$ be right-continuous and increasing and $f$ continuous. Let $\mu$ be the Lebesgue-Stieltjes measure associated to $F$, that is, $\mu((a,b])=F(b)-F(a)$. Let $\{x_j\}_{j=1}^n$ be a partition of some interval $I$ and let $\Delta x_j = (x_{j-1},x_j]$. Although it is customary to let $\Delta x_j$ denote the length of this interval, in cases where we may apply different notions of length to the same interval, it may make more sense to simply let $\Delta x_j$ denote the interval itself.
In this case, we have
$$
\int_I f(x)\,\mu(dx) = \lim_{n\to\infty}\sum_{j=1}^n f(x_j)\mu(\Delta x_j),
$$
provided the mesh of the partition tends to zero. In this setting, the $\mu(dx)$ notation keeps both sides of the equality notationally consistent with one another.
That said, though, whether you choose to use $\mu(dx)$ or $d\mu(x)$, the meaning is the same.
