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Let $A = \left \{x \in [0,1] \cap ({\mathbb{R}}- {\mathbb{Q}}) \left| \begin{array}{ll} {\textrm{in decimal expansion of } x, \textrm { say } 0.{a_1}{a_2}{a_3}\ldots \\ \textrm{first 1 (if exist) is followed by 2 }} \end{array} \right. \right \}$. What is the measure of $A$? I guess that it is 0. But don't know how to proceed. Please any help

bhim sen Chaudhary
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    This set has, in particular, $[0.12, 0.13) \cap (\mathbb{R} - \mathbb{Q})$ as a subset, so it can't have measure 0. – Duncan Ramage Jun 02 '17 at 20:24
  • ok I got. so what will be its measure. – bhim sen Chaudhary Jun 02 '17 at 20:26
  • The set of all rational numbers in [0, 1] has measure 0 so the set of all irrational numbers in that interval has measure 1. I don't know what you mean by "first 1 (if exist) is followed by 2 (if exists)". What do you mean by "1 (if exists)" and "2 (if exists)"? "1" and "2" certainly exist! – user247327 Jun 02 '17 at 20:31
  • If I understand it correctly, it's "all irrational numbers such that the first "1" in their decimal expansion is followed by a "2", provided a "1" exists in their decimal expansion". – Duncan Ramage Jun 02 '17 at 20:34
  • wherever first time 1 occur in decimal expansion it followed by 2. – bhim sen Chaudhary Jun 02 '17 at 20:34

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Consider the following disjoint sets \begin{eqnarray} A_i &=& \left\{x \in [0,1]\cap(\mathbb R -\mathbb Q)|\text{the first $1$ in x's decimal expansion exists and is followed by $i$}\right\}\\ B &=& \left\{x \in [0,1]\cap(\mathbb R -\mathbb Q)|\text{x has no $1$'s in its decimal expansion}\right\} \end{eqnarray} Your set is $A_2$. From symmetry it should be clear that $m[A_i] = m[A_j]$ for all $i,j$, and we also have $[0,1]\cap (\mathbb{R}-\mathbb{Q}) = (\cup_0^9A_i) \cup B$. Since $m[[0,1]\cap (\mathbb{R}-\mathbb{Q})] = 1$, we have $1 = 10 m[A_2] + m[B]$. From this answer, $m[B] = 0$, and thus we have $m[A_2] = 1/10$.

eyeballfrog
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