Absolute value & integrability

[steven187]

hello all

Iv been working on a lot of integrability questions and I am having trouble with this problem
let f be integrable on [a,b] then show that |f| is integrable and that

$$|\int_{a}^{b}f|\le \int_{a}^{b}|f|$$

now this is what i know

$$\int_{a}^{b^U}f =\int_{a_{L}}^{b}f= \int_{a}^{b}f$$

$$U(f,P)-L(f,P)<\epsilon$$

and

$$|f(x)|\le M \forall x\epsilon [a,b]$$ is there anything else i can gain from a function being integrable on a closed interval?

muchly appreciated if someone could tell me where to start and some directions? I realize that it is only through practice that i will be able to know where to start and where to go from there, please help

thank you

steven

 

[quasar987]

Hi,

You need to show that for a given partition P, S(|f|,P) - s(|f|,P) $\leq$ S(f,P) - s(f,P).

It is easy: use the definition of s(,) and S(,) and work the three different cases for a given interval in the partition: 1) f(x) is stricly < 0 for all x in that interval. 2) f(x) is stricly > 0 for all x in that interval. 3) f(x) is < 0 for some x and > 0 for some other x in that interval.

 

[steven187]

hello all

this is what i have done so far, i hope it is correct, i have shown that
$$U(|f|,P)-L(|f|,P)<\epsilon$$
and so |f| is integrable that wasnt a problem
then since -|f(x)|<=f(x)<=|f(x)| for all x an element of [a,b]
then we integrate the whole inequality to get
$$-\int_{a}^{b}|f(x)| \le\int_{a}^{b}f(x)\le\int_{a}^{b}|f(x)|$$
and hence
$$|\int_{a}^{b}f|\le \int_{a}^{b}|f|$$

In terms of the above method about proving the 3 different cases i got pretty confused going down that path, some further details would be helpful

steven

 

[quasar987]

How about simply invoquing the caracterisation of the integral

$$\int_{a}^{b}f(x)dx = \lim_{|p|\rightarrow 0}\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1})$$

and the triangle inequality:

$$\forall x,y \in \mathbb{R}, \ |x+y| \leq |x|+|y|$$

?