Exact meaning of Lebesgue integral's domain As an economist trying to learn probability theory, I have become quite confused regarding some of the essentials of Lebesgue integration. 
Say we are working with a Lebesgue integral for a non-negative measurable function $f$ on a measure space $(X,E,u)$ with the usual definition:
$$\int_{X}^{}f\,du$$
Intuitively, I understand the above to be the area under the graph of $f(X)$, where the x-axis is "represented" by the measure $u$. Let's say we wanted to calculate a sub-section of this area. It is then common to see something ala:
$$\int_{X}^{}f(x)1_{A}\,du=\int_{A}^{}f(x)\,du$$
where $A \in E$. But what does $\int_{A}^{}f(x)\,du$ actually mean? Is this the integral on $f(x)$ defined on a new measure space $(A,E,u)$? I am especially confused by this, when we start delving into probability theory, where the above integrals would equate to $E[X]$. Here it does not seem possible to me to change the underlying probability space, as I understand $(\Omega, \mathbb{F}, \mathbb{P})$ to be somewhat "fixed".
Any intuition or explanation of above would be much appreciated!
 A: Fix a measure space $(X,\mathcal M,\mu)$ and let $A \in \mathcal M$.
You can define a new measure space $(A,\mathcal M_A,\mu_A)$ by defining $$\mathcal M_A = \{E \cap A \mid E \in \mathcal M\}$$ and $$\mu_A(E) = \mu(A \cap E), \quad E \in \mathcal M_A.$$
If $f : X \to \mathbb R$, we can define $f|_A : A \to \mathbb R$ by restriction: $f|_A(x) = f(x)$ for all $x \in A$.
It is just a matter of checking definitions to see that $\mathcal M_A$ is a $\sigma$-algebra, $\mu_A$ is measure on $(A,\mathcal M_A)$, $f|_A$ is $\mu_A$-measurable if and only if $f \cdot 1_A$ is $\mu$-measurable, and that $$\int_X f \cdot 1_A \, d\mu = \int_A f|_A \, d\mu_A$$ whenever either integral exists.
The common value is denoted $\displaystyle \int_A f \, d\mu$, and is almost universally defined to be equal to the left-hand side of the equality above.
A: You must keep in mind that the lebesgue integral for a positive fonction is the superior limit of the integrals of the inferior simples functions (https://en.wikipedia.org/wiki/Simple_function).
When you lebesgue integrate a simple function like $f(x)=\sum_{\alpha_i}\alpha_i\mathbb{1}_{A_i}$ the integral on all whole domain, $\cup_{i}A_i$ is $\sum_{\alpha_i}\alpha_i\mu(A_i)$ where $A_i$ is the domain where $f$ is $\alpha_i$ on, i.e. $A_i=f^{-1}(\alpha_i)$. So X is all the value where f is defined. If you think of integrate on $A$ think that you are integrating all the values $\alpha_i$ of $f$ where $f^{-1}(\alpha_i)\in A$
for example, you are integrating the simple function that is $1$ in $[0,1]$ $2$ in $[1,2]$ 3 in $[2,3]$ etc. 
you want to integrate it on $A=[1,3]$, you take the image $f(\mathbb{R})=\{1,2,3,..\}$ is $f^{-1}(1)$ in $A$ ? No, pass it, is $f^{-1}(2)$ in $A$ yes, to you add to your sum $2\times \mu(f^{-1}(2))=2\times1$ and do the same for $f^{-1}(3)$ etc. 
Now that you are working on positive function but not necessarily simple, you know for sure, because you function is mesurable that for $B$ in the tribue of arrival, you have $f^{-1}(B)$ in the starting tribue, if $f^{-1}(B)$ in $A$, you count it etc. 
I hope it gives you a hint, and better comprehension of the lebesgue integral.
