real analysis - Set derived from definition of $Vert f Vert_infty$

Someone told me that the set

$B_n := \{x \in X : \vert f(x) \vert > \Vert f \Vert_\infty - \frac{1}{n}\}$ for $n \in \mathbb{N}$

(where $B_n$ has finite positive measure), is derived from the definition of the essential sup norm

$\Vert f \Vert_\infty = inf\{ a \ge 0 : \mu( \{ \vert f(x) \vert > a \} ) = 0 \}$,

but I really don't see why this is true, or even the intuition behind it. Could someone shed some light on this?

Thank you.

Answer

Suppose that $\mu(B_{n})=0$. That would imply that the $a$ in the definition of $\Vert f\Vert_{\infty}$ satisfies $a\leq \Vert f\Vert_{\infty}-\frac{1}{n}$. Since we assumed that $\mu(\{|f(x)|>\Vert f\Vert_{\infty}-1/n\})=0$. But this is non-sense since it implies

\begin{align*}
\Vert f \Vert_{\infty} &\leq \Vert f\Vert_{\infty}-\frac{1}{n} \\
\Rightarrow 0 &\leq -\frac{1}{n}
\end{align*}