# Integration Theorem

• Dec 15th 2009, 03:09 PM
slevvio
Integration Theorem
I have this in my notes:

Define continuous functions on [a,b] by $f_+ (t) = max(f(t),0)$ and $f_- (t) = max(-f(t),0).$ Then $f(t) = f_+ (t) - f_- (t)$ and $|f(t)| = f_+ (t) + f_- (t).$. The result follows from linearity and the triangle inequality by the following calculation:

$|\int_a^b f(t)dt | = | \int_a^b (f_+ (t) - f_- (t))dt | \le \int_a^b f_+(t)dt + \int_a^b f_- (t) dt= \int_a^b |f(t)| dt$.

I understand that $| \int_a^b (f_+ (t) - f_- (t))dt | \le |\int_a^b f_+(t)dt| + |\int_a^b f_- (t) dt|$ but not why $| \int_a^b (f_+ (t) - f_- (t))dt | \le \int_a^b f_+(t)dt + \int_a^b f_- (t) dt$.

Any help in this will be much appreciated thank you :)
• Dec 16th 2009, 03:09 AM
HallsofIvy
It uses the fact that $\left|\int f(x)dx\right|\le \int |f(x)|dx$.
• Dec 16th 2009, 03:15 AM
slevvio
What I have written is the proof for that fact but its one of the steps that I don't understand :(
• Dec 16th 2009, 05:20 AM
HallsofIvy
Oh, of course, I completely misread.

Do you see that if f(x) is non-negative on (a,b) then $\left|\int_a^b f(x)dx\right|= \int_a^b f(x)dx= \int_a^b |f(x)| dx$?

Thats true because both f(x) and the integral are never negative so you can just "discard" the absolute values.

Now, by its definition, $f_+(x)$ is non-negative. By its definition, $f_-(x)$ is non-positive so is $-f_-(x)$ is non-negative.

Clearly, $f(x)= f_+(x)+ f_-(x)$ while $|f(x)|= f_+(x)- f_-(x)$. If that's not clear, think about what happens for each x if (1) $f(x)\ge 0$, (2) $f(x)< 0$.

Now, $\int_a^b |f(x)|dx= \int_a^b (f_+(x)- f_-(x))dx= \int_a^b f_+(x)dx- \int_a^b f_-(x) dx$ while $\left|\int_a^b f(x)dx\right|= \left|\int_a^b f^+(x)dx+ \int_a^b f_-(x)dx\right|$.

Finally, use the fact that, for numbers x and y, $|x- y|\le |x|+ |y|$.
• Dec 21st 2009, 03:40 AM
slevvio
thank you very much