simple functions let  $m_f(A)=\int_{A} f dm$
suppose that f is a simple function that is $f= \sum c_{i} 1_{A_{i}}$
describe f.
this is part of a problem where i had to prove that $m_f(A)$ is a measure and i had also to describe it when f= the characteristic function.
the only issue i have in the case of simple functions is that when it takes negative values how do i integrate.
i know that for non negative simple functions $\int_{A}\sum c_{i} 1_{A_{i}}= \sum c_{i} m(A_{i})$. however i am not sure  if this fact still holds when the function takes negative values .
 A: You can only ensure that $m$ is a measure if $f$ is a non-negative measurable function. Otherwise suppose as in your case, if you take a simple function, $f=\sum c_i1_{A_i}$ and $c_n$ is negative for some $n$, then $m(A_n)<0$, which implies $m$ is not a measure.
However, for a measurable function $f$, $m(A)=\int_A fd\mu$ defines a set function that is countably additive and satisfies $m(\emptyset)=0$.
A: $m_f$ will only be a measure for non-negative functions $f$ (Else, if $c_i = -t$, $t>0$ then $m_f(A_i) = -t\ m(A_i)<0$).
As to your question, the integral in measure theory is defined this way: Take $f = f_+ - f_-$ where $f_+ = \max(f,0)$ and $f_- = \max(-f, 0)$ then
$$\int_A f d\mu := \int_Af_+ d\mu - \int_A f_- d\mu$$

In your special case if
$$f = \sum_{i=1}^N c_i 1_{A_i}$$
where $\mu(A_i) < \infty \quad \forall i$ and $I_+ := \{i : c_i > 0\}$, $I_- := \{i : c_i < 0\}$ and thus
$$f_+ = \sum_{i\in I_+} c_i 1_{A_i}, \qquad f_- = \sum_{i\in I_-} -c_i 1_{A_i}$$
And by definition
$$\int f d\mu = \int \sum_{i\in I_+} c_i 1_{A_i} d\mu - \int \sum_{i\in I_-} -c_i 1_{A_i} d\mu = \sum_{i=1}^N c_i \int 1_{A_i} d\mu = \sum_{i=1}^N c_i \mu(A_i)$$
