an $L^\infty$ norm equal to a supremum My question arose while studying an article which finds the $K$-functional for the pair of spaces $L^1,L^\infty$, so it's related to interpolation theory, but I think it can be solved with some $\inf,\sup$ manipulations. I'm sorry if the tags aren't correct.
Consider $(X,\Sigma, \mu)$ an arbitrary measure space. For each measurable function $f: X \to \Bbb{C}$ and $\alpha \geq 0$ define 
$$ f_*(\alpha)=\mu(\{ x \in X : |f(x)|>\alpha \}),$$
the distribution function of $f$.
For any measurable function $f:X \to \Bbb{C}$ for which there exists $\alpha>0$ with $f_*(\alpha)<\infty$, define $f^* :(0,\infty) \to [0,\infty)$ by
$$ f^*(t)=\inf \{ y>0 :f_*(y) \leq t\}.$$
It can be proved from the definitions that for every $t>0$ we have
$$ f^*(f_*(t))\leq t,\ f_*(f^*(t))\leq t.$$ 
Moreover, the function $f^*$ is continuous from the right. Both of the functions $f_*,f^*$ are non increasing. 
What I need to prove is that 

$$\sup_{t>0} f^*(t)= \| f\|_\infty,$$ for every function $f \in L^\infty$.

The inequality $\leq $ is straight from the definition. The other one I can't seem to get, and the text says that it's pretty hard, but with "a little more effort" it can be done. I haven't succeeded.
 A: For the inequality $\ge$ you'll have to show:
$$
  \forall \varepsilon>0 \,\exists\, t>0: f^*(t) \ge \|f\|_\infty - \varepsilon =: c_\varepsilon.
$$
By the definition of $f^*$, the inequality $f^*(t) \ge c_\varepsilon$ means:
$$
  \text{for all $y>0$ satisfying}\ f_*(y)\le t\  \text{one has}\ y \ge c_\varepsilon.
$$
By contraposition, the latter is equivalent to
$$
  \forall y\in(0,c_\varepsilon): f_*(y)>t.
$$
Thus, given $\varepsilon>0$, you have to find $t>0$ such that the latter holds. This is easy: Take any $t \in (0,f_*(c_\varepsilon))\,$ (and use that $f_*(y) \ge f_*(c_\varepsilon)$ for $y\in(0,c_\varepsilon)$).
A: Well, it is quite confusing notationwise, but let's just go through it one step at a time:
Suppose $\sup_{t > 0} f^\ast(t) < \|f\|_\infty$. Then there is $\epsilon > 0$ such that for all $t > 0$ we have
$$ f^\ast(t) \le \|f\| - 2\epsilon $$
This means by definition 
$$\forall t > 0: \quad \inf\{y > 0| f_\ast(y) \le t\} \le \|f\| - 2\epsilon$$
Therefore $\forall t > 0$ there exists $y > 0$ such that $f_\ast(y) \le t$ and $y\le \|f\| - \epsilon$.
By definition this means that for all $t> 0$ there is a $y$ satisfying $\|f\|- \epsilon \ge y > 0$ such that 
$$\mu(\{x \in X |\, |f(x)| > y \}) \le t $$
But then, in particular we get that for all $t>0$ we have
$$\mu(\{x \in X |\, |f(x)| > \|f\|_\infty -\epsilon \}) \le t $$
Since this is true independent of $t$, letting $t \to 0$, we get a contradiction to the definition of $\|f\| = \|f\|_\infty$.
