I've been reading through some exercises and solutions and came across a student's argument that seemed rather odd, though it was crucial for the proof and his answer got a full mark.
It was claimed that if $f \in {L_\infty }\left[ {0,1} \right]$ then the following set is of Lebesgue measure zero: $$E = \left\{ {x \in \left[ {0,1} \right]|f\left( x \right) \ge {{\left\| f \right\|}_\infty }} \right\}$$
I just figured that if $f\left( x \right) \equiv 1$ then $E = \left[ {0,1} \right]$ and then $m\left( E \right) = 1$.
So I'm quite baffled
You correctly demonstrated with a counterexample that $$ \left\{ {x \in \left[ {0,1} \right]|f\left( x \right) \ge {{\left\| f \right\|}_\infty }} \right\} $$ does not necessarily have measure zero. Correct would be that $$ E = \left\{ {x \in \left[ {0,1} \right]|f\left( x \right) \color{red}{>} {{\left\| f \right\|}_\infty }} \right\} $$ has measure zero, because $E$ is the countable union of the sets $$ E_n = \{ x \in [0, 1] \mid f(x) > \| f\|_\infty + \frac 1n \} $$ and each $E_n$ has measure zero according to the definition of $\| f\|_\infty $ as the essential supremum of $|f|$, see also L-infinity: $$ \|f\|_{\infty }\equiv \inf\{C\geq 0:|f(x)|\leq C{\text{ for almost every }}x\} $$