I Showing equivalence of two definitions of essential supremum

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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\}##.
 
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I agree with your proof.


I think the sets are equal because if you pick some a such that f is always smaller than a (which is the problem region for you attempt at a bijection), you can just arbitrarily define like, g(x)=f(x) if x is not zero, g(0)=a
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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