Graduate&CompetitionMathematics

{
  "id": "6a87b547d2762196b6bf3514",
  "subject": "Mathematics",
  "level": "Graduate",
  "qtype": "Proof",
  "question": "Let $\\mu$ be a finite Borel measure on $[0,1]$, and let $\\mu_s$ denote its singular part with respect to Lebesgue measure $m$ in the Lebesgue decomposition $\\mu=\\mu_{ac}+\\mu_s$. Prove that there exists a dense $G_\\delta$ set $G\\subseteq[0,1]$ such that $m(G)=0$ and $\\mu_s(G)=\\mu_s([0,1])$.",
  "options": null,
  "answer": "See explanation",
  "solution": "By singularity of $\\mu_s$ with respect to $m$, choose a Borel set $A\\subseteq[0,1]$ such that $m(A)=0$ and $\\mu_s(A)=\\mu_s([0,1])$ (if $\\mu_s=0$, take $A=\\emptyset$). For each $n$, by outer regularity of Lebesgue measure, choose a relatively open set $U_n\\subseteq[0,1]$ containing $A$ with $m(U_n)<1/(3n)$; if $A=\\emptyset$, take $U_n=\\emptyset$. Also choose a relatively open dense set $W_n\\subseteq[0,1]$ with $m(W_n)<1/(3n)$, for example a union of sufficiently small open intervals around a countable dense subset of $[0,1]$. Set $V_n=U_n\\cup W_n$. Then each $V_n$ is relatively open and dense in $[0,1]$, and $m(V_n)\\le m(U_n)+m(W_n)<2/(3n)$. Let $G=\\bigcap_{n=1}^\\infty V_n$. By the Baire category theorem, $G$ is dense in $[0,1]$ and is a $G_\\delta$ set. Since $A\\subseteq U_n\\subseteq V_n$ for every $n$, $A\\subseteq G$, so $\\mu_s(G)\\ge \\mu_s(A)=\\mu_s([0,1])$, hence equality. Finally, $m(G)\\le m(V_n)<2/(3n)$ for every $n$, so $m(G)=0$."
}