1
$\begingroup$

Let $(X,\Sigma_x,\mu)$ be a measurable space ($X$-the set, $\Sigma_x$ - is a $\sigma$ algebra over X, and $\mu$ is a positive measure).

Let $f\in L^1(\mu)$ (meaning $\int_X|f|d\mu<\infty$), $f:X\to \mathbb{C}$.

How do I prove $|\int_X{}fd\mu|\leq\int_X{}|f|d\mu$?

my thoughts:

Denote $z=\int_Xfd\mu$, let $a\in\mathbb{C}$ be such that $az=|z|$ and then:

$|\int_X{}fd\mu|=|z|=az=a\int_X{}fd\mu$

but...then what?...

any help? :)

2 Answers 2

1

Hope this helps:

Proposition: If $f\in L^{1}$, then $\left|\int f\right| \leq \int |f|$.

Proof - If $f$ is real then $$\left|\int f\right| = \left|\int f^{+} - \int f^{-}\right| \leq \int f^{+} + \int f^{-} = \int |f|$$ If $f$ is complex-valued and $\int f\neq 0$, then $\int f = \left|\int f\right|e^{-i\theta}$. So $$\left|\int f\right| = e^{-i\theta}\int f = \int e^{-i\theta}f$$ In particular, $\int e^{-i\theta}f$ is real, so we have \begin{align*} \left|\int f\right| = Re\int e^{-i\theta}f = \int Re(e^{-i\theta} f) &\leq \int |Re(e^{-i\theta}f)|\\ &\leq \int |e^{-i\theta} f|\\ &= \int |f| \end{align*} Some of the notation above is from Real Analysis by Folland, if you have any questions please let me know.

2

In spirit, this is just the triangle inequality, and that's how we'll prove it. First assume $f$ is real valued. Remember that

$\int f d\mu = \sup_g \int g d\mu$

where the supremum is taken over all simple functions $g \leq f$, i.e. all $g$ of the form

$g(x) = \sum_j c_j 1_{A_j}$

for some finitely many real numbers $c_j$ and measurable sets $A_j$. The inequality holds for any such $g$, since

$\Big| \int g d\mu \Big| = \Big| \sum_j c_j \mu(A_j) \Big| \leq \sum_j |c_j|\mu(A_j) = \int |g| d\mu,$

by the triangle inequality. Thus

$\Big| \int f d\mu \Big| \leq \sup_g \Big| \int g d\mu \Big| \leq \sup_g \int |g| d\mu = \int |f| d\mu.$

Note that the assumption $f \in L^1(\mu)$ is unnecessary, since in the case that $f$ is not integrable, the right hand side is infinite, so the inequality holds trivially.

If $f$ is complex valued, then take $\alpha \in \mathbb{C}$ so that $\alpha \int f \in \mathbb{R}$, and write

$\Big| \int f d\mu \Big| = \int Re(\alpha f) d\mu \leq \int |Re(\alpha f)| d\mu \leq \int |\alpha f| d\mu = \int |f| d\mu$.

Edit: There are a couple of not-completely-trivial facts I've used. It would be good to check these to make sure you understand what's going on!

  • 0
    but how can you say something like $g\leq f$ when f is a complex function? – 2017-02-17
  • 0
    Good point! This works for real valued functions: now I think you can get it to work by rotating, as you suggested. I think Hermes' suggestion leads to an easier proof though... – 2017-02-17