Bartle's Elements of Integration Exercise 6.T

Question

Let$(X,\mathcal{M},\mu)$be a measure space. Bartle defines $$\|f\|_\infty = \inf\{S(N);N\in\mathcal{M}, \mu(N)=0\}$$ where $$S(N) = \sup \{|f(x)|;x\not\in N\}$$ How do I show $|f(x)|\leq \|f\|_\infty$ a.e.? Also, that this is the infimum of real constants with this property, that is, if $A<\|f\|_\infty$, then $\exists E\in\mathcal{M}$ with $\mu(E)>0$ such that $|f(x)|>A,\,\forall x\in\ E$.
I've actually seen this as the definition of $\|\cdot\|_\infty$ in some other books, but I can't see how to prove they are equivalent. I've tried proving it by contradiction but I seem to get nowhere. Any tips?

Answer 1

Take a a sequence $(N_n)_n$ in $\mathcal{M}$ with $\mu(N_n) = 0$ for all $n$ such that $S(N_n) \searrow \Vert f\Vert_\infty$. Then define $$N := \bigcup_n N_n$$

Then $\mu(N) \leq \sum_n \mu(N_n) =0$ and $\{x \in X: |f(x)| >\Vert f \Vert_\infty\} \subseteq N$, for if $|f(x)| > \Vert f \Vert_\infty$, there is $n$ with $|f(x)| > S(N_n) > \Vert f \Vert_\infty$ which implies $x \in N_n$. Thus $\{x \in X: |f(x)| > \Vert f \Vert_\infty\}$ is contained in a set of measure $0$ and thus also has measure $0$, from which the statement follows.

Answer 2

Answer for the second question: Let $A <||f||_{\infty}$. Let $E=\{x: |f(x)| >A\}$. If $\mu (E) =0$ then $|f(x)| \leq A$ for $x \notin E$ so $S(E) \leq A$. But $\|f\|_{\infty} \leq S(E) \leq A$ so we have a contradiction. Hence $\mu (E) >0$.