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12h
comment If every borel measurable function continuous in compact metric space then metric space is finite
Welcome to Stackexchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
1d
comment What did Whitehead and Russell's “Principia Mathematica” achieve?
@ncmathsadist In that case you should discuss the aims of the Principia and how they were shown to be infeasible.
2d
answered If a subset of metric space $(X,d)$ like $S$ is closed and bounded, does it imply that $X$ is totally bounded?
2d
answered Question on proving tight sequences.
2d
comment Question on proving tight sequences.
You would have to show that for every $\epsilon>0$, there exists such $M$.The compact set in the definition of tightness would then be $[-M,M]$.
May
1
comment Construction of a Borel subset $E$ of $\Bbb R$ such that both $E$ and $E^c$ has positive “density” everywhere.
This has been asked before here.
Apr
30
comment General property regarding outer measure for a nested sequence of sets (measurable or not).
@DavidC.Ullrich Get elected.
Apr
30
comment General property regarding outer measure for a nested sequence of sets (measurable or not).
@DavidC.Ullrich Thanks for the comment,it should be correct now.
Apr
30
comment General property regarding outer measure for a nested sequence of sets (measurable or not).
@stalker2133 That $(F_n)$ is a Partition means that every point in $[0,1]$ lies in some of the $F_n$ and is therefore not in $E_n$. Since $\bigcap_n E_n$ is the set of points that are in every $E_n$, this intersection is empty and contains no point at all. If you make such statements, please prove them. You haven't given a counterexample an there is none.
Apr
29
comment General property regarding outer measure for a nested sequence of sets (measurable or not).
I actually looked at Theorem 1.15 in that book, it says something entirely different.
Apr
29
comment General property regarding outer measure for a nested sequence of sets (measurable or not).
That cannot be true. It is possible to partition $[0,1]$ into a continuum of disjoint sets with outer measure $1$ (see here), so one can partition $[0,1]$ also into a sequence $(F_n)$ of disjoints non-measurable subsets of $[0,1]$ with outer measure $1$ each. Letting $E_n=[0,1]\backslash\bigcup_{i=1}^nF_n$ gives you a counterexample.
Apr
28
comment Show if $f(x)$ is measurable then $f(x+a)$ is also measurable
Yes, every Borel set is Lebesgue measurable. But the converse is not true. It seems your definition is that $\{x\in\mathbb{R}:f(x)<r\}$ has to be Lebesgue measurable for each $r\in\mathbb{R}$?
Apr
28
comment Show if $f(x)$ is measurable then $f(x+a)$ is also measurable
What does measurability of $f$ mean here? Is $f^{-1}(B)$ a Borel set for $B$ Borel or only guaranteed to be Lebesgue measurable?
Apr
26
answered Problem with proof of the upper hemicontinuity of correspondence
Apr
26
comment Is $\omega-1$ finite?
How do you define $\omega-1$?
Apr
24
comment Existence of a locally essentially unbounded integrable function
Here is a very easy solution: Take your favorite nonnegative integrable function $f$ and change it on rational numbers $q$ so that $f(q)$ is the denominator of the representation of $q$ as a fraction in lowest terms.
Apr
24
answered Induce measure between topological spaces.
Jan
28
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23
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Nov
20
awarded  Enlightened