Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
First prove that $\mu(f) \le \mu(|f|)$, then prove that $-\mu(f) \le \mu(|f|)$. – Qiaochu Yuan Jun 3 '11 at 10:00
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. – Martin Sleziak Jun 3 '11 at 10:27
@Martin Sleziak thanks, that's very helpful – Freeman Jun 3 '11 at 13:20
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 $f\geq 0$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.