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Any set of positive Lebesgue measure, have a Lebesgue measurable proper subset with the same measure?

For example: if $A$ is a measurable set, with $m(A) > 0$,

I can say that $x \in A$ (since $A$ is not empty because $m(A) > 0$), then can I say that $B = A\setminus \{x\}$ is a measurable subset of $A$ and $m(B) = m(A)$ (since $m({x}) = 0$)?

I want to use this "claim" to solve a part of a problem that I have to show that if $\int_{B} f = 0$ for any measurable subset $B$ of $A$, then $\int_{A}f = 0 = \int_{B}f$ to eventually show that $f = 0$ a.e on $A$.

Is this claim true?

By the way, I also know the Inner approximation: If $E$ is a measurable, then for any $\epsilon > 0$, exits $F \subset E$ such that $m(E\setminus F) < \epsilon$. and $F$ is closed and the similar one for open sets (Outter approximation).

Did the previous theorem says that $m(E\setminus F) = 0$?

Thank you in advance!

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    $\begingroup$ Firstly, it looks like you're specifically thinking about the Lebesgue measure - is this correct? The property that for all $E$ measurable and for all $\epsilon > 0$, there exists a closed $F \subset E$ such that $m(E \backslash F) < \epsilon$ is the inner regularity property of the Lebesgue measure. $\endgroup$ – Kenny Wong Aug 22 '17 at 13:20
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    $\begingroup$ Secondly, are you really asking whether any set of positive measure has a proper subset of the same measure? If we're using the Lebesgue measure, then this true because a set with one element is guaranteed to be null - as you pointed out. $\endgroup$ – Kenny Wong Aug 22 '17 at 13:21
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    $\begingroup$ As for your actual problem, I would personally use the fact that if $f$ is non-negative and $\int_B f = 0$, then $f = 0$ a.e. on $B$. We could then consider taking $B = \{ x : f(x) \geq 0\}$ and $B = \{ x : f(x) \leq 0\}$. $\endgroup$ – Kenny Wong Aug 22 '17 at 13:23
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    $\begingroup$ Yes, for the reason you pointed out. The only set that doesn't contain a point is the empty set. And the empty set is null. $\endgroup$ – Kenny Wong Aug 22 '17 at 13:26
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    $\begingroup$ But I'm afraid I'm missing something... How does this claim help with showing that $\int_B f = 0$ for all measurable $B \subset A$ implies that $f = 0$ a.e. on $A$? $\endgroup$ – Kenny Wong Aug 22 '17 at 13:27
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Singletons (one-member subsets) are Lebesgue sets of measure $0.$ The complement $D^c$ of a Lebesgue set $D$ is a Lebesgue set. If $D,E$ are Lebesgue sets then so are $D\cap E$ and $D\cup E.$ If $D,E$ are disjoint Lebesgue sets then $M(D\cup E)=m(D)+m(E).$

If $a\in A$ where $A$ is $any$ non-empty Lebesgue set then $B=A$ \ $\{a\}$ is a proper subset of $A$. And $B= A\cap (\{a\})^c$ is a Lebesgue set which is disjoint from $\{a\}.$ So $$m(A)=m(B\cup \{a\})=m(B)+m(\{a\})=m(B)+0=m(B).$$

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    $\begingroup$ Countable sets are Lebesgue sets of measure $0.$ So if $A$ is a Lebesgue set of positive measure then $A$ is uncountable and we can remove any countable subset from $A$ and still have a Lebesgue set with the same measure. $\endgroup$ – DanielWainfleet Aug 23 '17 at 4:32
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Yes, your claim is true.

However, you do not need that specific property in your application: it is not specified that $B$ is a proper subset of $A$, therefore chosing $B=A$ is allowed.

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  • $\begingroup$ I was thinking about that, but this is a qualifying problem and I think they make a mistake because that one will be trivial. lol $\endgroup$ – Richard Clare Aug 22 '17 at 13:36
  • $\begingroup$ This will mean that any set with a possitive Lebesgue measure, has an uncountable number of singleton? $\endgroup$ – Richard Clare Aug 22 '17 at 13:41
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    $\begingroup$ yes, because it is an uncountable set $\endgroup$ – supinf Aug 22 '17 at 13:49
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Let it be that $A$ is measurable and $m(B)=0$ for every measurable set $B\subseteq A$.

Then evidently $\int g1_A=0$ for every step function $g$.

Consequently $\int f1_A=0$ for every nonnegative measurable function $f$.

This because $\int f1_A$ is by definition the suprememum of the values $\int g1_A$ where $g$ is a step function with $g\leq f$.

This can be expanded to measurable functions $f$.

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