Given a measure on a measurable set, how do I compute the Lebesgue integral of a given function? I'm trying to study Measure Theory on my own. I think I understood the theoretical framework of how measure theory should work, but I'm having troubles because of the lack of practical examples and right now I have the following doubt: if I'm given the measure and a function, how do I explicitly compute the integral?
Here's a simple example of what I mean.
Let's say I have $A \subset \mathbb{R}$ and my measure $\delta_1$ such that:
$$
      \delta_1(A) = 
      \begin{cases}
      1 & \text{if } 1 \in A \\
      0 & \text{if } 1 \notin A
    \end{cases}
$$
Let's say I take the function $f(x) = |x|$.
How do I compute the integral 
$$
\int_A f \, d\delta_1
$$
I do realise this is a very simple example, but what I don't understand is how, formally, my measure is supposed to act inside my integral.
Could you show me how to continue explicitly the equation?
$$\int_Af \, d\delta_1 = \cdots $$
Thanks in advance
 A: I answered the concept for a general measure here: How to calculate an integral given a measure?
I'll copy it below, then supplement with an example using the Dirac measure, like you are.
Recall the definition of the Lebesgue integral of $f$ on a measure space $(X,\mathfrak{F},\mu)$: If $f$ is non-negative, then
$$\int\limits_X f \, d\mu=\sup\left\lbrace\int\limits_X s\,d\mu:\ 0\leq s\leq f,\ s\text{ simple}\right\rbrace.$$ We define it on absolutely integrable functions which can be both positive and negative by breaking them into positive and negative parts $f^+=\max \{f,0\}$ and $f^-=-\min\{f,0\}$ (respectively), then applying the above to each piece. Also, if $f\in L^1$, there exists a sequence of simple functions $(s_n)$ converging to $f$ pointwise so that $$\int\limits_X f\,d\mu=\lim_{n\rightarrow\infty}\int\limits_X s_n \, d\mu.$$ To find this approximating sequence, we find monotone non-decreasing approximating sequences for the positive and negative parts of $f$ (you can find this construction in any textbook). So, we really just need to remember how to integrate simple functions: 
If $s(x)=\sum\limits_{j=1}^n c_j\chi_{A_j}(x),$ where $A_j\in\mathfrak{F},$ then $$\int\limits_X s \, d\mu = \sum\limits_{j=1}^n c_j\mu(A_j).$$ Putting this all together allows us to calculate the integral of $f\in L^1(X)$.
For the example, consider the Dirac measure for $x\in X$: $$\delta_x(A)=\chi_A(x)=\begin{cases} 0 & x\notin A,\\ 1 &x\in A. \end{cases}$$ We'll claim that $$\int f\, d\delta_x=f(x).$$ To see this, check on a simple function $f=\sum c_j\chi_{A_j}$ to get $$\int f\, d\delta_x=\sum c_j\delta_x(A_j)=\sum c_j\chi_{A_j}(x)=f(x).$$ In your particular example, you will need to insert a cut-off over the set $A$. So, it will depend on whether or not $1\in A$. Since $|x|$ is equal to $1$ for $x=1$, the answer will be $\chi_A(1)$.
(To be slicker in justifying the expression for integrating with respect to $\delta_x$: for a fixed $x$, the constant function $g$ defined by $g(y)=f(x)$ is equal to $f$ $\delta_x$-a.e., as the set $\{y: f(x)=g(y)\neq f(y)\}$ does not contain $x$.
Hence, their integrals are the same, and the integral of $g$ is just $\delta_x(\text{whole space}) f(x)=f(x).$)
