Why is $\mu(A) = \int_E \textbf{1}_A (x)\mu(dx)$, where $(E, \mathcal{E}, \mu)$ is a probability space and $A \in \mathcal{E}$? I'm not sure how to interpret the notation. It looks like we calculate the value of the measure for (small) intervals on $A$ and sum them up. But for that to make sense, these small intervals must be non-overlapping, which the notation doesn't seem to specify.
So could someone please give an intuitive explanation of the notation and/or reference where it's possible to read more about it?
This is part of the definition of Lebesgue's integral. If $(E, \mathcal{E}, \mu)$ is a measure space, for a measurable set $A \in \mathcal{E}$, the integral $\int_E \mathbf{1}_A(x) \, \mathrm{d}\mu(x)$ is defined the be $\mu(A)$, which is the measure of the set $A$.
For the intuition, I think it is a good idea to compare Riemann and Lebesgue's integrals, and to compare the interpretation. I will not be very rigorous here, I'm just seeking an intuition.
Recall that we can interpret a Riemann integral $\int_a^b f(x) \mathrm{d} x$ as the area between the curve traced by $f(x)$ and the $x$-axis, seeing that as an infinite sum of areas of infinitely thin rectangles bounded by the $x$-axis and the curve made by $f$. A Lebegue's integral $\int_A f(x) \, \mathrm{d}\mu(x)$, where $A \in \mathcal{E}$, can be interpreted in a similar way, but with two important differences.
First, the set $A$ is not in general an interval, so the integral cannot really be interpreted as an infinite sum of areas of infinitely small rectangles. Nevertheless, you can still interpret an infinite sum, but of products of the value of $f$ on small measurable sets, which could be more abstract objects. But that's not the important difference where the intuition is different.
The most important difference to see in the interpretation (at least in my opinion) is that with Lebesgue's integral, a different "weight" is given to each of these small sets, which is defined by the measure. In a Riemann integral $\int_a^b f(x) \mathrm{d}x$, we split the interval $[a,b]$ in small intervals $[a_n,b_n]$, and then "sum up" products of the form $f(x) (b_n - a_n)$, which are areas of rectangles of lengths $b_n - a_n$ and of height given by a value $f(x)$ for $x \in [a_n,b_n]$. In a Lebesgue integral $\int_A f \,\mathrm{d}\mu$, we can interpret that we split the set $A$ into small sets $A_n$, but then we sum up products of the form $f(x) \mu(A_n)$, so the measure $\mu$ gives a different weight to every set $A_n$. If $\mu$ is the Lebesgue measure and $A$ is an interval, each $A_n$ is a small interval and its measure is its length, but it can be far more general than that. Even if $A$ is an interval, if the measure is different, than the weights attributed to each small interval $A_n$ will be different, so the result will not be the same.
Now for your specific example, since the function $\mathbf{1}_A$ is $1$ on $A$ and $0$ outside $A$, the integral $\int_E \mathbf{1}_A(x) \,\mathrm{d}\mu(x)$ is simply $\int_A \mathrm{d}\mu(x)$. For the intuition, recall that a similar Riemann integral $\int_a^b \mathrm{d}x$ gives $b - a$, which is the length of the interval $[a,b]$ on which the integral is done. Similarly, the integral $\int_A \mathrm{d}\mu(x)$ should give the "length" of the set $A$, but here the integral is done with respect to the measure $\mu$. Hence, the result is $\mu(A)$, which is the "length" of the set $A$ as measured by the measure $\mu$.