- #1
psie
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- TL;DR Summary
- I've encountered the following two definitions of essential supremum and was wondering if someone could check if my "proof" of equivalence of the two definitions is correct.
Assume ##f: X\to\mathbb R## to be a measurable function on a measure space ##(X,\mathcal A,\mu)##. The first definition is ##\operatorname*{ess\,sup}\limits_X f=\inf A##, where $$A=\{a\in\mathbb R: \mu\{x\in X:f(x)>a\}=0\}$$ and the second is ##\operatorname*{ess\,sup}\limits_X f=\inf B## where $$B= \left\{\sup_X g:g=f\ \text{pointwise a.e.}\right\}.$$.
First, it is easily seen that if ##h=f## a.e., then ##\mu\{x\in X: f(x) > \sup_X h\} = 0##, so ##B\subset A##, which shows that ##\inf B\geq \inf A##. Now, I can't show the other inclusion of the sets and I suspect these sets are not necessarily equal, but what I can show is that if ##a\in A##, then ##h:=\min\{f,a\}=f## a.e. and ##\sup_X h\leq a## (*). So for ##a\in A##, we can find a ##b\in B## such that ##b\leq a##. Thus ##\inf A\geq \inf B##.
Any thoughts on this?
(*) ##h:=\min\{f,a\}## means ##h(x):=\min\{f(x),a\}##.
First, it is easily seen that if ##h=f## a.e., then ##\mu\{x\in X: f(x) > \sup_X h\} = 0##, so ##B\subset A##, which shows that ##\inf B\geq \inf A##. Now, I can't show the other inclusion of the sets and I suspect these sets are not necessarily equal, but what I can show is that if ##a\in A##, then ##h:=\min\{f,a\}=f## a.e. and ##\sup_X h\leq a## (*). So for ##a\in A##, we can find a ##b\in B## such that ##b\leq a##. Thus ##\inf A\geq \inf B##.
Any thoughts on this?
(*) ##h:=\min\{f,a\}## means ##h(x):=\min\{f(x),a\}##.