If $f\in L^{\infty}(E)$.
How $$S_1=\sup\left\{M:\mu(\left\{x\in E:|f(x)|\ge M \right\}) \ne 0 \right\} $$ and $$S_2=\inf\left\{M:\mu(\left\{x\in E:|f(x)|\gt M \right\}) = 0 \right\} $$ are equivalent?
This is Herald Hanche-Olsen's answer
If you put $$\begin{aligned} f_1(M)&=\mu(\{x\in E\colon |f(x)|\ge > M\}),\\ f_2(M)&=\mu(\{x\in E\colon |f(x)|> M\}), \end{aligned} $$ then both $f_1$ and $f_2$ are (non-strictly) decreasing functions. Furthermore, $f_1\ge f_2$, while if $M_1>M2$, then $f_1(M_1)\le > f_2(M_2)$. The result follows easily from these observations. (Intuitively, $f_1$ and $f_2$ are practically the same function; they differ only at discontinuities.
In this answer i'm not getting notion of introduction of infimum and supremum over $f_1$ and $f_2$?Also,if $f_1$ & $f_2$ are practically the same function then how the infimum of one can be equal to the supremum of other?
If you put $$\begin{aligned} f_1(M)&=\mu(\{x\in E\colon |f(x)|\ge M\}),\\ f_2(M)&=\mu(\{x\in E\colon |f(x)|> M\}), \end{aligned} $$ then both $f_1$ and $f_2$ are (non-strictly) decreasing functions. Furthermore, $f_1\ge f_2$, while if $M_1>M2$, then $f_1(M_1)\le f_2(M_2)$. The result follows easily from these observations. (Intuitively, $f_1$ and $f_2$ are practically the same function; they differ only at discontinuities.