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For an integrable function $f$ on $(a,b)$, how would you prove $| \int (f)| \leq \int (|f|)?$

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First prove that $\mu(f) \le \mu(|f|)$, then prove that $-\mu(f) \le \mu(|f|)$. –  Qiaochu Yuan Jun 3 '11 at 10:00
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In my book, integration of arbitrary measurable functions is defined by the integration of non-negative measurable functions, that is, $\int f\,dx=\int f^+\,dx-\int f^-\,dx$ and $\int |f|\,dx=\int f^+\,dx+\int f^-\,dx$. So it is a natural consequence that $|\int f\,dx|\le\int |f|\,dx$ –  ziyuang Jun 3 '11 at 10:10
    
@ziyuang that slightly similar to what i'm trying to work with, but what is f^+ sorry? just a function that gives a larger integral? –  Freeman Jun 3 '11 at 10:22
    
$f^+$ and $f^-$ is more or less standard notation for $f^+(x)=\max\{0,f(x)\}$ and $f^-(x)=\max\{0,-f(x)\}$. Note that $f=f^+-f^-$ and $|f|=f^++f^-$. See e.g. en.wikipedia.org/wiki/Integral#Lebesgue_integral –  Martin Sleziak Jun 3 '11 at 10:27
    
@Martin Sleziak thanks, that's very helpful –  Freeman Jun 3 '11 at 13:20

1 Answer 1

up vote 4 down vote accepted

Let $f$ be a complex measurable function on $(a,b)$. Choose a complex number $\alpha$ such that $\left|\alpha\right|=1$ and $\alpha \int f=\left|\int f\right|$. We now compute that:

$\left|\int f\right|=\alpha \int f = \int \alpha f\leq \int \left|f\right|$.

Exercise 1: Why does the inequality hold in the above proof? (Hint: observe that the first equality shows that $\int \alpha f$ is a real number.)

Exercise 2: The proof above tells you under what conditions equality holds in the inequality $\left|\int f\right|\leq \int \left|f\right|$. State these conditions. (Hint: equality holds in the inequality given if and only if $\int \alpha f = \int \left|f\right|$.)

I think that it is worth explaining the intuition behind the above proof. The case when $f\geq 0$ is trivial. Let us consider the case $f<0$. In this case, we wish to show that $\left|\int f\right|\leq \int \left|f\right|$. The key point is to use the case $f\geq 0$; we can apply the case $f\geq 0$ to $-f$ and conclude that $\left|\int (-f)\right|\leq \int\left|(-f)\right|=\int (-f)$.

Exercise 3: Deduce that $\left|\int f\right|\leq \int \left|f\right|$ if $f<0$.

In the general case, the complex number $\alpha$ above acts in a similar way to the transference of the case $f<0$ to the case $f\geq 0$.

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This is the way I learned it too! (from Folland, I think) –  JavaMan Jun 3 '11 at 11:09

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