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I would like to know the intuitive explanation with simple examples about the exact meaning of essential sup and inf. What I understand about sup, that it is the largest value of a subset that is a set of a larger set. The same goes for inf but it is the smallest value of a subset that is a set of a bigger set. How do we see essential in this case?

Thank you

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    $\begingroup$ "value of a subset that is a set of a larger set" is meaningless. $\endgroup$ Sep 28 '16 at 2:11
  • $\begingroup$ Could you then explain why it is meaningless? $\endgroup$
    – Elekko
    Sep 28 '16 at 6:11
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In measure theory, sets of measure zero are often treated as sets that do not matter (in some sense). It makes sense to get a definition of supremum where you can disregard some elements.

Consider a measure space $ ( \Omega, \Sigma, \mu) $, when working in such spaces you want to understand everything through the measure $ \mu$. Say we have a real function $f: \Omega \to \mathbb{R} $ defined in this measure space. The function might well be unbounded but it may happen that the points where the function grows too much may not be important to us.

Imagine there is a constant $c$ such that $ \mu(\{x \in \Omega, |f(x)| > c \}) = 0 $. That means that our measure doesn't pay much attention to the set of points where $ |f| > c $. So in that sense $ |f| \leq c \: \: a.e. $, meaning that $c$ is an essential upper bound of $f$. Taking the infimum of all the real numbers with such property, you would get the essential supremum.

So the essential supremum is the least (smallest) essential upper bound.

This means that for the means of your measure $ \mu$ you can think of $f$ as bounded by this value (even though it may be unbounded).

Think about the real numbers with the Lebesgue measure $(\lambda)$.
Then the function: $$ f(x) = \begin{cases} x & x \in \mathbb{Q} \\ 3 & x \in \mathbb{R}\setminus \mathbb{Q} \end{cases} $$ is essentially bounded by $4$. Because $ \{x \in \mathbb{R}, |f(x)| > 4 \} \subseteq \mathbb{Q} $ and of course $ \lambda(\mathbb{Q}) = 0 $. So because our measure doesn't give much importance to the set of rational numbers, the essential supremum is considered only over the other set.

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  • $\begingroup$ This is a pretty nice explanation. Thank you! $\endgroup$
    – Charith
    Apr 14 at 14:45

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