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Let $A$ be the subset of $[0,1]$ and $A$ is the union of all open intervals $(a_i,b_i)$ such that each rational number in $(0,1)$ is contained in some $(a_i,b_i)$ and $\sum _{i=1}^ \infty (b_i-a_i) < 1$,

1) how do I show $Bd(A)=[0,1]-A$? 2) Now using this taking $U=A$ how do I prove that if $f=\mathcal X_U$(characteristic function of $U$) except on a set of measure zero. Then $f$ is not integrable on $[0,1]$

Okay, for 1) I see that no points in $A$ is $Bd(A)$ so $Bd(A) \subseteq [0,1]-A$. Now for $x \in [0,1]-A$ what will be the form of $x$? I think $x$ will be the irrational number s.t $x \notin (a_i,b_i)$ $\forall i$. Now for every open $U$ s.t $x \in U$, $U \cap \Bbb Q \cap (0,1) \neq \phi$. so it is the $Bd(A)$, hence we are done...(Right??)

2) Now for the second part what we know that the function is integrable iff it is discontinuous on a set of measure zero. AND the function $f=\mathcal X_U:[0,1] \to \Bbb R $ is integrable iff $Bd(U)$ has measure $0$.

So what I shall do next??

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  • $\begingroup$ Your $A$ could be $(0,1)$ so I think you can not prove the nonmeasurability of $\chi_U$. $\endgroup$ – Hanul Jeon Sep 13 '15 at 11:17
  • $\begingroup$ Sorry $\sum _{i=1}^ \infty (b_i-a_i) < 1$. This condition I forgot to include... I have included now.. $\endgroup$ – user152715 Sep 13 '15 at 11:48
  • $\begingroup$ Your proof for 1) is OK. For 2) show your function is discontinuous at least on this boundary, and this boundary has nonzero measure. $\endgroup$ – GEdgar Sep 13 '15 at 12:24
  • $\begingroup$ I know something like this has to be done but what is the process?? $\endgroup$ – user152715 Sep 13 '15 at 13:18

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