Any set of positive Lebesgue measure, have a Lebesgue measurable PROPER subset with the same measure? 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!
 A: 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  $$m(A)=m(B\cup \{a\})=m(B)+m(\{a\})=m(B)+0=m(B).$$
A: 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.
A: 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 $f$.
