property of Lebesgue integral If $f$ and $g$ are nonnegative Lebesgue measurable functions, then we know that 
$\int (f+g) d\lambda = \int f d \lambda + \int g d \lambda $. Given the difinition of integral of an arbitrary Lebesgue measurable function, that is $ \int f d\lambda = \int f^+ d\lambda - \int f^- d\lambda$, how do you prove $\int (f_1 - f_2) d\lambda = \int f_1 d\lambda - \int f_2 d\lambda $ for nonnegative Lebesgue integrable functions $f_1$ and $f_2$ ?
That $f_1 - f_2$ is Lebesgue integrable is easily seen. Now, I can show that $\int (-f) d\lambda = -\int f d\lambda$, as $\int (-f) d\lambda = \int (-f)^+ d\lambda - \int (-f)^- d\lambda = \int f^- d\lambda - \int f^+ d\lambda = -\int fd\lambda$. Any ideas how to proceed?
 A: So I came up with a way to do it:
The idea is to split the domain of integration into part on which $f_1 - f_2$ is positive and part on which it is negative, and use the identity
$$ \int |f_1 - f_2| = \int (f_1 -f_2)^+ + \int (f_1-f_2)^- $$
More precisely we let $E = \{x: f_1(x) \ge f_2(x) \}$, so $\int_E |f_1 - f_2| = \int_E (f_1 - f_2) $. Now since $f_1 - f_2$ is nonnegative we can apply the argument of drhap to deduce that  $\int_E (f_1 - f_2) = \int_E f_1 - \int_E f_2$.
Similarly, $\int_{E^c} |f_1 - f_2| = \int_{E^c} -(f_1-f_2) = \int_{E^c} f_2 -\int_{E^c}f_1$. From what I already proved, the negative sign can be taken out of the second integral. Now if we combine these we'll get the result.
A: Use the usual trick... split up the cases where $f_1-f_2$ is positive and negative and use your original result, then combine terms properly.
Namely, consider $(f_1-f_2)^+$ and $(f_1-f_2)^-$ now what exactly is 
$$\int (f_1-f_2)^+ - \int(f_1-f_2)^-$$
?
Now use linearity etc.
A: Let $h=f+g$ and notice that $$h^+-h^-=f^+-f^-+g^+-g^-.$$ This means $$h^++f^-+g^- = h^-+f^++g^+,$$ and since both sides have constant sign, $$\int h^++ \int f^-+ \int g^- = \int h^-+ \int f^++ \int g^+.$$ Now collect terms in a suitable way.
This proof is taken from W. Rudin's Real and complex analysis.
