
Integration Theorem
I have this in my notes:
Define continuous functions on [a,b] by $\displaystyle f_+ (t) = max(f(t),0) $ and $\displaystyle f_ (t) = max(f(t),0). $ Then $\displaystyle f(t) = f_+ (t)  f_ (t) $ and $\displaystyle f(t) = f_+ (t) + f_ (t). $. The result follows from linearity and the triangle inequality by the following calculation:
$\displaystyle \int_a^b f(t)dt  =  \int_a^b (f_+ (t)  f_ (t))dt  \le \int_a^b f_+(t)dt + \int_a^b f_ (t) dt= \int_a^b f(t) dt $.
I understand that $\displaystyle  \int_a^b (f_+ (t)  f_ (t))dt  \le \int_a^b f_+(t)dt + \int_a^b f_ (t) dt $ but not why $\displaystyle  \int_a^b (f_+ (t)  f_ (t))dt  \le \int_a^b f_+(t)dt + \int_a^b f_ (t) dt $.
Any help in this will be much appreciated thank you :)

It uses the fact that $\displaystyle \left\int f(x)dx\right\le \int f(x)dx$.

What I have written is the proof for that fact but its one of the steps that I don't understand :(

Oh, of course, I completely misread.
Do you see that if f(x) is nonnegative on (a,b) then $\displaystyle \left\int_a^b f(x)dx\right= \int_a^b f(x)dx= \int_a^b f(x) dx$?
Thats true because both f(x) and the integral are never negative so you can just "discard" the absolute values.
Now, by its definition, $\displaystyle f_+(x)$ is nonnegative. By its definition, $\displaystyle f_(x)$ is nonpositive so is $\displaystyle f_(x)$ is nonnegative.
Clearly, $\displaystyle f(x)= f_+(x)+ f_(x)$ while $\displaystyle f(x)= f_+(x) f_(x)$. If that's not clear, think about what happens for each x if (1)$\displaystyle f(x)\ge 0$, (2) $\displaystyle f(x)< 0$.
Now, $\displaystyle \int_a^b f(x)dx= \int_a^b (f_+(x) f_(x))dx= \int_a^b f_+(x)dx \int_a^b f_(x) dx$ while $\displaystyle \left\int_a^b f(x)dx\right= \left\int_a^b f^+(x)dx+ \int_a^b f_(x)dx\right$.
Finally, use the fact that, for numbers x and y, $\displaystyle x y\le x+ y$.
