\(\newcommand\notesclassalt[2]{\if\smallformat1#2\else#1\fi} \newcommand\lectureend[1]{% \noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}} \marginpar{\raggedleft#1}% } \newcommand\isdefinition[1]{\textbf{#1}} \newcommand\f\frac \newcommand\intd{\mathop{}\!\mathrm{d}} \newcommand\ind[1]{1_{#1}} \newcommand\seq\coloneqq \newcommand\qes{=:} \newcommand\loc{\mathrm{loc}} \newcommand\A{\mathcal A} \newcommand\R{\mathcal R} \newcommand\Q{\mathcal Q} \newcommand\B{\mathcal B} \renewcommand\P{\mathcal P} \newcommand\spt[1]{\mathrm{spt}(#1)} \newcommand\var{\mathop{\mathrm{var}}} \newcommand\diam{\mathop{\mathrm{diam}}} \newcommand\dist{\mathop{\mathrm{dist}}} \newcommand\dmo{\mathrm d} \newcommand\dm[2]{\dmo(#1,#2)} \newcommand\tb[1]{\mathop{\partial{}}{#1}} \newcommand\mb[1]{\mathop{\partial_*}{#1}} \newcommand\mbe[1]{\mb{(#1)}} \newcommand\mbb[1]{\mb{\Bigl(#1\Bigr)}} \newcommand\tbe[1]{\tb{(#1)}} \newcommand\tbb[1]{\tb{\Bigl(#1\Bigr)}} \newcommand\rb[1]{\mathop{\partial^*}{#1}} \newcommand\tc[1]{\overline{#1}} \newcommand\mc[1]{\tc{#1}^*} \newcommand\ti[1]{\mathring{#1}} \newcommand\tia[1]{#1^{\mathrm o}} \newcommand\tim[1]{\tia{(#1)}} \newcommand\tib[1]{\tia{\bigl(#1\bigr)}} \newcommand\mi[1]{\ti{#1}^*} \newcommand\mia[1]{#1^{\mathrm o*}} \newcommand\mim[1]{\mia{(#1)}} \newcommand\mib[1]{\mia{\bigl(#1\bigr)}} \newcommand\smg{\mathcal{H}} \newcommand\smo{\smg^{d-1}} \newcommand\sm[1]{\smg^{d-1}(#1)} \newcommand\smb[1]{\smg^{d-1}\Bigl(#1\Bigr)} \newcommand\smi[1]{\smg^0(#1)} \newcommand\smib[1]{\smg^0\Bigl(#1\Bigr)} \newcommand\lmr[1]{|#1|} \newcommand\lmg{\mathcal{L}} \newcommand\lmog{\lmg_*} \newcommand\lm[1]{\lmg(#1)} \newcommand\lmb[1]{\lmg\Bigl(#1\Bigr)} \newcommand\lmib[1]{\lmio\Bigl(#1\Bigr)} \newcommand\lmo[1]{\lmog(#1)} \newcommand\lmob[1]{\lmog\Bigl(#1\Bigr)} \newcommand{\mres}{\!\mathbin{\vrule height 1.6ex depth 0pt width 0.13ex\vrule height 0.13ex depth 0pt width 1.1ex}\!} \newcommand\restrictzero[2]{#1\,\mres#2} \newcommand\D{\mathbb D} \newcommand\Dq{\mathcal D} \newcommand\M{\mathcal M} \)
Is Lebesgue measure a Radon measure? Is Lebesgue outer measure a Radon outer measure?
Solution: Yes.
Do the annuli \(A_n=\tc {B(x,n)}\setminus B(x,n-1)\) form a decomposition, i.e. is it true that \(Ω=A_1∪A_2∪…\) and for any \(n\neq k\) we have \(A_n∩A_k=∅\)?
Solution: Almost, but no. It is true that \(Ω=A_1∪A_2∪…\), but \(A_n∩A_{n+1}=\tb {B(x,n)}\), and \(\tb {B(x,n)}\neq ∅\) for certain metric spaces such as \(ℝ^d\) with the Euclidean metric.
Is it true, that for any \(E⊂Ω\ni x\) and any Radon measure \(μ\) that \(μ(E\setminus B(x,n))→0\) as \(n→∞\)?
Solution: No, take \(E=ℝ^d\) and \(μ=\lmg \). It is true however if \(μ(E)<∞\) as a consequence of the "easure continuity"Lemma 1.2.7.
Let \(μ\) be the counting measure on \(ℝ\), i.e. \(μ(E)\) equals the number of elements in \(E\) if \(E\) is finite, and \(μ(E)=∞\) otherwise. Is \(μ\) a Radon measure? Is \(μ\) a Radon outer measure?
Solution: All sets \(E⊂ℝ\) are measurable, in particular Borel sets, but \(μ\) is not a Radon (outer) measure because compact sets do not have finite \(μ\)-measure.
Does ?? hold for the counting measure?
Solution: No. If \(E\) is closed and nonempty then for any open \(U\supset E\) the set \(U\setminus E\) is open and nonempty and thus infinite. If \(E\) is open and nonenmpty and \(C\subset E\) is closed then the same is true for \(E\setminus C\).
Is it true that for any measurable \(A⊂ℝ^d\) with \(\lm A<∞\) and \(ε>0\) exists a compact set \(K⊂A\) with \(\lm {A\setminus K}<ε\)? What, if \(\lm A=∞\)?
Solution: No, if \(\lm A=∞\), take \(A=ℝ^d\).
If \(\lm A<∞\), then yes. By ?? there exist a closed set \(C⊂A\) with \(\lm {A\setminus C}<ε/2\) and \(\lm C\leq \lm A<∞\). Thus by the "easure continuity"lemma there exists an \(n∈ℕ\) such that \(K\coloneqq C∩\tc {B(0,n)}\), which is compact, satisfies \(\lm {A\setminus K}=\lm {A\setminus C}+\lm {C\setminus K}<ε\).
Let \(E⊂Ω\) be measurable. Is the characteristic function given by
Solution: Yes, because for any \(a∈ℝ\) we have \(\{f<a\}∈\{∅,ℝ^d\setminus A,ℝ^d\}\), all of which are measurable.
Are polynomials Lebesgue measurable functions?
Solution: Yes, because they are continuous.