# $\mu(|f|\geq \alpha) = \frac{1}{\alpha} ||f||_1$

I'm having difficulty with this problem here:

Let $$f\in L^1(X,\mathcal{M},\mu)$$ with $$||f||_1 \neq 0$$.

Prove that there exists a unique $$\alpha$$ so that $$\mu(\lbrace|f|\geq \alpha\rbrace) = \frac{1}{\alpha} ||f||_1$$.

My first thought was to use the Hardy LittleWood Maximal Theorem which says that there exists $$C$$ so that

$$\mu(\lbrace Hf(x) >\alpha\rbrace) \leq \frac{C}{\alpha} ||f||_1$$

Unfortunately this theorem does not take me very far in this approach.

Any help is appreciated.

Edit: It seems the question was asking to show that there is at most one such $$\alpha$$

• you may want to check Chebyshev's inequality: en.wikipedia.org/wiki/… – daw Feb 14 '19 at 16:07
• @daw I don't think that will be useful because the proof of Markov's and Chebyshev's inequality follows from "truncation" argument. You may visualise it as chopping a rectangular piece of ginger from raw. The chopped piece will always be strictly smaller than the raw ginger, unless there's nothing to chop (i.e. $f$ is an indicator function). – GNUSupporter 8964民主女神 地下教會 Feb 14 '19 at 16:15
• Just to let you know the generally accepted policy here: If you ask a question that you didn't mean to ask, but end up with correct answers to the question you did ask, it is considered polite to ask your actual question in a new question instead of editing your first question. See here for more discussion on the policy. – K.Power Feb 14 '19 at 16:20
• To respond to the edit, one has to show that the given eqaulity holds iff $f$ is (up to a multiplicative constant) an indicator function of a measurable set (i.e. $1_A(x) = 1$ if $x \in A$, $0$ if $x \notin A$, $A \in \cal{M}$). The if-part is trivial. The only-if-part can be shown by assuming that $f$ is not a scalar multiple of an indicator function. Therefore, for all $k \in \Bbb{R}$, $A \in \cal{M}$, $f \ne 1_A$. Note that $\{|f|\ge a\} \in \cal{M}$, so once we've shown that a strict inequality, we're almost done, and the rest is trivial. – GNUSupporter 8964民主女神 地下教會 Feb 14 '19 at 16:49

The conclusion doesn't hold for $$f(x) = (1-|x|)_+ = \max(1-|x|,0)$$ and if $$\mu$$ is the Lebesgue measure. The shape of piecewise linear fucntion $$f$$ is a centred peak, so $$||f||_1 = 1$$. so that $$\mu(|f| \ge a) = 2(1-a) \ne \dfrac1a = \dfrac{||f||_1}{a}$$. By AM-GM inequality, $$2 \, a(1-a) \le 2 \left(\frac{a+(1-a)}{2}\right)^2 \le 2 \cdot \frac14 = \frac12 < 1,$$ so the last inequality holds.

Let $$X:=[0,1]$$ and $$\mu$$ the Lebesgue measure. Let $$f:X\to\mathbb{R}$$ be as $$f(x):=2$$ for $$x\in[0,1/2]$$ and $$f(x):=1$$ for $$x\in(1/2,1]$$. Furthermore let $$\alpha:=2$$

Then $$2\mu([0,1/2]) =1 < 1 +1/4 = \int_0^1 f(x) d\mu(x)$$ which is a counterexample of your equality.

However the proposition is true when you substitute $$'='$$ by $$'\le'$$:

$$\mu(\lbrace|f|\geq \alpha\rbrace) \le \frac{1}{\alpha} ||f||_1$$.

• The question asks to show the existence and uniqueness some $\alpha$, not to show equality for all $\alpha$. – guy Feb 14 '19 at 15:44
• Yes you are right. – Maksim Feb 14 '19 at 15:54

As others have said this is not necessarily true, but it is true that we are always guaranteed a $$\alpha>0$$ s.t $$\mu\{|f|\geq\alpha\}\leq \frac{\|f\|_1}{\alpha}.$$ Regardless of the measure space we know that $$\mu\{|f|\geq \|f\|_1\}\leq 1$$, so if we choose $$\alpha=\|f\|_1$$ the inequality will hold.