The general idea behind proving the triangle inequality of the essential supremum norm.(Royden 4th edition on pg.138)

Question

The proof is given below:

But I do not understand the role of $E_{n}$ in the proof and how to create it. what criteria we are looking for in it? and why we create it?

Could anyone explains this for me, please?

Answer 1

By definition $||f||_{\infty}$ is the infimum of all essential upper bounds.

In the first part it is shown that this infimum is attained and, hence, is a minimum.

This is then helpful and (a bit implicitly) used to show that the sum of two essentially bounded functions is also essentially bounded. Or in other words: $f,g \in L^{\infty}(E)\Rightarrow f+g \in L^{\infty}(E)$.

To show that the infimum is attained it is necessary to show that there is a measurable subset $N \subset E$ with $|f(x)|\leq ||f||_{\infty}$ on $E\setminus N$ such that $m(N)= 0$.

Since $||f||_{\infty}$ is the infimum of all essential upper bounds, the author now approaches $||f||_{\infty}$ from above by invoking that $||f||_{\infty}+\frac 1n$ is an essential upper bound, hence, there are for each $n\in\mathbb{N}$ measurable subsets $E_n$ with $m(E_n) = 0$ such that $|f(x)|\leq ||f||_{\infty}+\frac 1n$ on $E\setminus E_n$.

Now, if $x \in E \setminus \bigcup_{n=1}^{\infty}E_n$ this means that $|f(x)| \leq ||f||_{\infty}+\frac 1n$ for all $n\in\mathbb{N}$ which is equivalent with $|f(x)| \leq ||f||_{\infty}$.

Since $E_{\infty}=\bigcup_{n=1}^{\infty}E_n$ is the at most countable union of sets of measure zero, we know that $m(E_{\infty}) = 0$. So, $||f||_{\infty}$ is the least essential upper bound.