Absolute value & integrability

steven187

hello all

Iv been working on alot of integrability questions and im 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 realise 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

Homework Helper
Gold Member
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

Last edited:

quasar987

Homework Helper
Gold Member
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:

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

?

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