 # I am a graduate student of Mathematics.In the book Measure,Integration and Real Analysis by Sheldon abbracciopj 2022-06-30 Answered
I am a graduate student of Mathematics.In the book Measure,Integration and Real Analysis by Sheldon Axler there is a question that asks the reader to show that there exists no measure space $\left(X,\mathcal{S},\mu \right)$ such that $\left\{\mu \left(E\right):E\in \mathcal{S}\right\}=\left[0,1\right)$.I am not sure how to do it.I was thinking of taking a nested sequence $\left\{{E}_{n}\right\}$ of sets in $\mathcal{S}$ such that $\mu \left({E}_{n}\right)=1-1/n$ but I don't think that will work.Can someone give me any clue?
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If $\left(X,\mathcal{S},\mu \right)$ denotes a measure space then $\mathcal{S}$ denotes a $\sigma$-algebra on $X$ so that by definition $X\in \mathcal{S}$ and consequently:
$\mu \left(X\right)\in \left\{\mu \left(E\right)\mid E\in \mathcal{S}\right\}$
This with $\mu \left(E\right)\le \mu \left(X\right)$ for every $E\in \mathcal{S}$.
So the set $\left\{\mu \left(E\right)\mid E\in \mathcal{S}\right\}$ has a largest element in $\mu \left(X\right)$.
However evidently the set $\left[0,1\right)$ does not have a largest element and we conclude that the two sets do not coincide.

We have step-by-step solutions for your answer! kokoszzm
Use your idea along with the property of "continuity from below" that all measures have. You can show that an event has measure one in this way.

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