$ \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 $\alpha$
 A: 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.
A: 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$.
A: 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.
