If $f, g \geq 0$ measurable, then $\int f-g = \int f -\int g$
Is this always true or does it need a special condition to hold?
I know if we have addition this holds ($\int f+g = \int f +\int g$)
if $f, g$ measurable, there exist two increasing sequences of simple functions that converges to $f$ and $g$ respectively and their sum is also an increasing sequence of simple functions that converges to $f +g$
This method wouldn't work when I have a difference of two functions, but I still saw this hold sometimes.
p.s. It's from the proof of dominated convergence theorem. This is the last few lines of the proof from the lecture.
$\lim \int|f-f_n|=0$
$|\int f- \int f_n|=|\int f-f_n| \leq \int|f-f_n| \rightarrow0$
therefore $\int f_n \rightarrow \int f$