How can one show the inequality that $\|f+g\|_\infty ≤ \|f\|_\infty + \|g\|_\infty$?
Can we use same real number $a$ for both $f$ and $g$ ? i.e,
$$\|f\|_\infty = \text{ess} \sup(f)=\inf\{a\in\mathbb{R} : \mu\{x\in X : |f(x)| > a\}=0\}$$
$$\|g\|_\infty = \text{ess} \sup(f)=\inf\{a\in\mathbb{R} : \mu\{x\in X : |g(x)| > a\}=0\}$$
Best Answer
We have $|f(x) + g(x)| \leq |f(x)| + |g(x)|$ for every $x$, by the triangle inequality. Now $|f(x)| \leq \|f\|_\infty$ almost everywhere, and similarly $|g(x)| \leq \|g\|_\infty$ almost everywhere. Therefore, $$|f(x) + g(x)| \leq \|f\|_\infty + \|g\|_\infty$$ almost everywhere. Since the right hand side is an almost-everywhere upper bound for $|f(x) + g(x)|$, and $\|f + g\|_\infty$ is the infimum of all such almost-everywhere upper bounds, it follows that $$\|f + g\|_\infty \leq \|f\|_\infty + \|g\|_\infty$$
Edit to respond to the question raised in the comments:
@George: I mean that there is a set $N$ with $\mu(N) = 0$ such that $|f(x)| \leq \|f\|_\infty$ for all $x \in N^c$. This is an easy consequence of the definition of the essential supremum.
Indeed, if there is no such set $N$, then $|f(x)| > \|f\|_\infty$ on a set of positive measure. Therefore, at least one of the sets $\{x : |f(x)| > \|f\|_\infty + 1/n\}$ must have positive measure (where $n$ is a positive integer). But this contradicts the definition of $\|f\|_\infty$.