# Can an essential supreme/infimum norm be defined in terms of supreume and infimum?

From Wikipedia:

the essential supremum of $f$, ess sup $f,$ is defined by $$\mathrm{ess } \sup f=\inf \{a \in \mathbb{R}: \mu(\{x: f(x) > a\}) = 0\}\,$$ if the set $\{a \in \mathbb{R}: \mu(\{x: f(x) > a\}) = 0\}$ of essential upper bounds is not empty, and ess sup $f = +∞$ otherwise.

Exactly in the same way one defines the essential infimum as the largest essential lower bound, that is, $$\mathrm{ess } \inf f=\sup \{b \in \mathbb{R}: \mu(\{x: f(x) < b\}) = 0\}\,$$ if the set of essential lower bounds is not empty, and as $−∞$ otherwise.

I was wondering if they can be defined equivalently in terms of $\sup$ and $\inf$ respectively?

For example, $$\mathrm{ess } \sup f=\inf \{\sup(f(A)): \forall \text{ measurable subset } A, \mu(A^c) = 0\}\,$$ and $$\mathrm{ess } \inf f=\sup \{\inf(f(A)): \forall \text{ measurable subset } A, \mu(A^c) = 0\}\, ?$$

Thanks and regards!

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In fact, the two definitions you show are the same.

We assume that $f:\mathbb{R}\to\mathbb{R}$ is measurable. Let $A=\{\sup f(S): S\subset \mathbb{R} \mbox{ measurable}, \mu(S^c)=0\}$ and $B=\{a: \mu(\{x: f(x)>a\})=0\}$.

Claim: $\inf A =\inf B$.

Proof: Suppose $p\in A$ and $p=\sup f(S)$ for some measurable set $S$ such that $\mu(S^c)=0$. Then $\{x: f(x)>p\}\subset S^c$, so $\mu(\{x: f(x)>p\})=0$ and $p\in B$. Thus $A\subset B$ and $\inf A\ge \inf B$.

On the other hand, suppose $a\in B$. Then $\mu(\{x: f(x)>a\})=0$ so the set $S=\{x: f(x)\le a\}$ is one of the defining subset in the definition of $A$. Clearly $\sup f(S)\le a$. This proves that $\inf A\le \inf B$.

The proof for the essential inf is similar.

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Thanks, TCL! I was wondering if you could help me with this related question? – Tim May 12 '13 at 6:28