I'm studying $L^p$ spaces for the first time. I am following, among others, the book by Stein & Shakarchi Functional Analysis. Introduction to Further Topics in Analysis. Before the proof of the famous theorem they introduce the following lemma (p.14).
Lemma 4.2 Suppose $1\leq p,q\leq\infty$ are conjugate exponents.
(i) If $g\in L^q$, then $\displaystyle\|g\|_{L^q}=\sup_{\|f\|_{L^p}\leq1}\left|\int fg\right|$.
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In the proof of the above lemma for the case $p=1$ and $q=\infty$ they say:
Let $\epsilon>0$, and $E$ a set of finite positive measure, where $|g(x)|\geq\|g\|_{L^\infty}-\epsilon$. (Such a set exists by the definition of $\|g\|_{L^\infty}$ and the fact that the measure $\mu$ is $\sigma$-finite.)
I have a couple of doubts with what is said in the previous paragraph.
Doubt 1. When they say $|g(x)|\geq\|g\|_{L^\infty}-\epsilon$, should I understand
$$\forall x\in E:\ |g(x)|\geq\|g\|_{L^\infty}-\epsilon?$$
Doubt 2. I'm failing to see why such set $E$ must exists. Suppose we are talking about $L^1(X,\mu)$ space. I know that if the measure $\mu$ is $\sigma$-finite there exists a sequence of subsets $(X_n)$ such that
$$X=\bigcup_{n=1}^\infty X_n,\quad\mu(X_n)<\infty,\ \forall n\in\mathbb{N}.$$
I also know that if $g\in L^\infty(X,\mu)$ then
$$\|g\|_{L^\infty}=\mathrm{ess}\,\sup g<\infty\Longrightarrow\mu\{g>\mathrm{ess}\,\sup g\}=0.$$
I think I should choose the set $E$ with the help of said sequence $(X_n)$, but I don't see how to do it.
Best Answer
Let $\epsilon >0$ and $E=\{x: |g(x)| >\|g\|_{\infty} -\epsilon\}$. It is easy to that this set is measurable. If this set has measure $0$ then $|g| \leq \|g\|_{\infty} -\epsilon$ almost everywhere. But this contradicts the definition of $\|g\|_{\infty}$ as the smallest number $M$ such that $|g(x)| \leq M$ almost everywhere so we must have $\mu (E) >0$. Of course, $|g(x)| >\|g\|_{\infty} -\epsilon$ for every $x \in E$.