Measure Theory: Sets

Fall 2021.3 #real_analysis/qual/work

Recall that a set \(E \subset \mathbb{R}^{d}\) is measurable if for every \(c>0\) there is an open set \(U \subseteq {\mathbf{R}}^d\) such that \(m^{*}(U \setminus E)<\epsilon\).

  • Prove that if \(E\) is measurable then for all \(\epsilon>0\) there exists an elementary \(\operatorname{set} F\), such that \(m(E \Delta F)<\epsilon\).

    Here \(m(E)\) denotes the Lebesgue measure of \(E\), a set \(F\) is called elementary if it is a finite union of rectangles and \(E \Delta F\) denotes the symmetric difference of the sets \(E\) and \(F\).

  • Let \(E \subset \mathbb{R}\) be a measurable set, such that \(0<m(E)<\infty\). Use part (a) to show that \begin{align*} \lim _{n \rightarrow \infty} \int_{E} \sin (n t) d t=0 \end{align*}

Spring 2020.2 #real_analysis/qual/completed

Let \(m_*\) denote the Lebesgue outer measure on \({\mathbf{R}}\).

a.. Prove that for every \(E\subseteq {\mathbf{R}}\) there exists a Borel set \(B\) containing \(E\) such that \begin{align*} m_*(B) = m_*(E) .\end{align*}

b.. Prove that if \(E\subseteq {\mathbf{R}}\) has the property that \begin{align*} m_*(A) = m_*(A\bigcap E) + m_*(A\bigcap E^c) \end{align*} for every set \(A\subseteq {\mathbf{R}}\), then there exists a Borel set \(B\subseteq {\mathbf{R}}\) such that \(E = B\setminus N\) with \(m_*(N) = 0\).

Be sure to address the case when \(m_*(E) = \infty\).

  • Definition of outer measure: \begin{align*} m_*(E) = \inf_{\left\{{Q_j}\right\} \rightrightarrows E} \sum {\left\lvert {Q_j} \right\rvert} \end{align*} where \(\left\{{Q_j}\right\}\) is a countable collection of closed cubes.
  • Break \({\mathbf{R}}\) into \({\textstyle\coprod}_{n\in {\mathbf{Z}}} [n, n+1)\), each with finite measure.
  • Theorem: \(m_*(Q) = {\left\lvert {Q} \right\rvert}\) for \(Q\) a closed cube (i.e. the outer measure equals the volume).

  • \(m_*(Q) \leq {\left\lvert {Q} \right\rvert}\):

  • Since \(Q\subseteq Q\), \(Q\rightrightarrows Q\) and \(m_*(Q) \leq {\left\lvert {Q} \right\rvert}\) since \(m_*\) is an infimum over such coverings.

  • \({\left\lvert {Q} \right\rvert} \leq m_*(Q)\):

  • Fix \({\varepsilon}> 0\).

  • Let \(\left\{{Q_i}\right\}_{i=1}^\infty \rightrightarrows Q\) be arbitrary, it suffices to show that \begin{align*}{\left\lvert {Q} \right\rvert} \leq \qty{\sum_{i=1}^\infty {\left\lvert {Q_i} \right\rvert}} + {\varepsilon}.\end{align*}

  • Pick open cubes \(S_i\) such that \(Q_i\subseteq S_i\) and \({\left\lvert {Q_i} \right\rvert} \leq {\left\lvert {S_i} \right\rvert} \leq (1+{\varepsilon}){\left\lvert {Q_i} \right\rvert}\).

  • Then \(\left\{{S_i}\right\} \rightrightarrows Q\), so by compactness of \(Q\) pick a finite subcover with \(N\) elements.

  • Note \begin{align*} Q \subseteq \bigcup_{i=1}^N S_i \implies {\left\lvert {Q} \right\rvert} \leq \sum_{i=1}^N {\left\lvert {S_i} \right\rvert} \leq \sum_{i=1}^N (1+{\varepsilon}) {\left\lvert {Q_j} \right\rvert} \leq (1+{\varepsilon})\sum_{i=1}^\infty {\left\lvert {Q_i } \right\rvert} .\end{align*}

  • Taking an infimum over coverings on the RHS preserves the inequality, so \begin{align*}{\left\lvert {Q} \right\rvert} \leq (1+{\varepsilon}) m_*(Q)\end{align*}

  • Take \({\varepsilon}\to 0\) to obtain final inequality.

  • If \(m_*(E) = \infty\), then take \(B = {\mathbf{R}}^n\) since \(m({\mathbf{R}}^n) = \infty\).

  • Suppose \(N \coloneqq m_*(E) < \infty\).

  • Since \(m_*(E)\) is an infimum, by definition, for every \({\varepsilon}> 0\) there exists a covering by closed cubes \(\left\{{Q_i({\varepsilon})}\right\}_{i=1}^\infty \rightrightarrows E\) depending on \({\varepsilon}\) such that \begin{align*} \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon})} \right\rvert} < N + {\varepsilon} .\end{align*}

  • For each fixed \(n\), set \({\varepsilon}_n = {1\over n}\) to produce such a covering \(\left\{{Q_i({\varepsilon}_n)}\right\}_{i=1}^\infty\) and set \(B_n \coloneqq\bigcup_{i=1}^\infty Q_i({\varepsilon}_n)\).

  • The outer measure of cubes is equal to the sum of their volumes, so \begin{align*} m_*(B_n) = \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon}_n)} \right\rvert} < N + {\varepsilon}_n = N + {1\over n} .\end{align*}

  • Now set \(B \coloneqq\bigcap_{n=1}^\infty B_n\).

    • Since \(E\subseteq B_n\) for every \(n\), \(E\subseteq B\)
    • Since \(B\) is a countable intersection of countable unions of closed sets, \(B\) is Borel.
    • Since \(B_n \subseteq B\) for every \(n\), we can apply subadditivity to obtain the inequality \begin{align*} E \subseteq B \subseteq B_n \implies N \leq m_*(B) \leq m_*(B_n) < N + {1\over n} {\quad \operatorname{for all} \quad} n\in {\mathbf{Z}}^{\geq 1} .\end{align*}
  • This forces \(m_*(E) = m_*(B)\).

Suppose \(m_*(E) < \infty\).

  • By (a), find a Borel set \(B\supseteq E\) such that \(m_*(B) = m_*(E)\)
  • Note that \(E\subseteq B \implies B\bigcap E = E\) and \(B\bigcap E^c = B\setminus E\).
  • By assumption, \begin{align*} m_*(B) &= m_*(B\bigcap E) + m_*(B\bigcap E^c) \\ m_*(E) &= m_*(E) + m_*(B\setminus E) \\ m_*(E) - m_*(E) &= m_*(B\setminus E) \qquad\qquad\text{since } m_*(E) < \infty \\ \implies m_*(B\setminus E) &= 0 .\end{align*}
  • So take \(N = B\setminus E\); this shows \(m_*(N) = 0\) and \(E = B\setminus (B\setminus E) = B\setminus N\).

If \(m_*(E) = \infty\):

  • Apply result to \(E_R\coloneqq E \bigcap[R, R+1)^n \subset {\mathbf{R}}^n\) for \(R\in {\mathbf{Z}}\), so \(E = {\textstyle\coprod}_R E_R\)
  • Obtain \(B_R, N_R\) such that \(E_R = B_R \setminus N_R\), \(m_*(E_R) = m_*(B_R)\), and \(m_*(N_R) = 0\).
  • Note that
    • \(B\coloneqq\bigcup_R B_R\) is a union of Borel sets and thus still Borel
    • \(E = \bigcup_R E_R\)
    • \(N\coloneqq B\setminus E\)
    • \(N' \coloneqq\bigcup_R N_R\) is a union of null sets and thus still null
  • Since \(E_R \subset B_R\) for every \(R\), we have \(E\subset B\)
  • We can compute \begin{align*} N = B\setminus E = \qty{ \bigcup_R B_R } \setminus \qty{\bigcup_R E_R } \subseteq \bigcup_R \qty{B_R\setminus E_R} = \bigcup_R N_R \coloneqq N' \end{align*} where \(m_*(N') = 0\) since \(N'\) is null, and thus subadditivity forces \(m_*(N) = 0\).

Fall 2019.3. #real_analysis/qual/completed

Let \((X, \mathcal B, \mu)\) be a measure space with \(\mu(X) = 1\) and \(\{B_n\}_{n=1}^\infty\) be a sequence of \(\mathcal B\)-measurable subsets of \(X\), and \begin{align*} B \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}x\in B_n \text{ for infinitely many } n}\right\}. \end{align*}

  • Argue that \(B\) is also a \(\mathcal{B} {\hbox{-}}\)measurable subset of \(X\).

  • Prove that if \(\sum_{n=1}^\infty \mu(B_n) < \infty\) then \(\mu(B)= 0\).

  • Prove that if \(\sum_{n=1}^\infty \mu(B_n) = \infty\) and the sequence of set complements \(\left\{{B_n^c}\right\}_{n=1}^\infty\) satisfies \begin{align*} \mu\left(\bigcap_{n=k}^{K} B_{n}^{c}\right)=\prod_{n=k}^{K}\left(1-\mu\left(B_{n}\right)\right) \end{align*} for all positive integers \(k\) and \(K\) with \(k < K\), then \(\mu(B) = 1\).

Hint: Use the fact that \(1 - x ≤ e^{-x}\) for all \(x\).

  • Borel-Cantelli: for a sequence of sets \(X_n\), \begin{align*} \left\{{x {~\mathrel{\Big\vert}~}x\in X_n \text{ for infinitely many $n$} }\right\} &= \bigcap_{N\geq 1} \bigcup_{n\geq N} X_n = \limsup_n X_n \\ \left\{{x {~\mathrel{\Big\vert}~}x\in X_n \text{ for all but finitely many $n$} }\right\} &= \bigcup_{N\geq 1} \bigcap_{n\geq N} X_n = \liminf X_n .\end{align*}

  • Properties of logs and exponentials: \begin{align*} \prod_n e^{x_n} = e^{\Sigma_n x_n} \quad\text{and} \quad \sum_n \log(x_n) = \log\left(\prod_n x_n\right) .\end{align*}

  • Tails of convergent sums vanish.

  • Continuity of measure: \(B_n \searrow B\) and \(\mu(B_0)<\infty\) implies \(\lim_n \mu(B_n) = \mu(B)\), and \(B_n\nearrow B \implies \lim_n \mu(B_n) = \mu(B)\).

  • The Borel \(\sigma{\hbox{-}}\)algebra is closed under countable unions/intersections/complements,
  • \(B = \limsup_n B_n = \cap_{N\geq 1} \cup_{n\geq N} B_n\) is an intersection of unions of measurable sets.

  • Tails of convergent sums vanish, so \begin{align*} \sum_{n\geq N} \mu(B_n) \xrightarrow{N\to\infty} 0 .\end{align*}
  • Also, \begin{align*} B_M \coloneqq\bigcap_{N = 1}^M \bigcup_{n\geq N} B_n \searrow B .\end{align*}
  • A computation: \begin{align*} \mu(B) &\coloneqq\mu\left(\bigcap_{N\geq 1} \bigcup_{n\geq N} B_n\right) \\ &\leq \mu\left( \bigcup_{n\geq N} B_n \right) && \forall N \\ &\leq \sum_{n\geq N} \mu(B_n) && \forall N \\ &\overset{N\to\infty}\longrightarrow 0 ,\end{align*} where we’ve used that we’re intersecting over fewer sets and this can only increase measure.

  • Since \(\mu(X) = 1\), in order to show \(\mu(B) = 1\) it suffices to show \(\mu(X\setminus B) = 0\).

  • A computation: \begin{align*} \mu(B^c) &= \mu\qty{ \qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty B_n }^c }\\ &= \mu\qty{ \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty B_n^c } \\ &\leq \sum_{N=1}^\infty \mu\qty{ \bigcap_{n=N}^\infty B_n^c } \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} \mu\qty{ \bigcap_{n=N}^K B_n^c } && \text{continuity of measure from above} \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} \prod_{n=N}^K \qty{1 - \mu(B_n)} && \text{by assumption} \\ &\leq \sum_{N=1}^\infty \lim_{K\to\infty} \prod_{n=N}^K e^{-\mu(B_n)} && \text{by hint} \\ &= \sum_{N=1}^\infty \lim_{K\to\infty} e^{-\sum_{n=N}^K \mu(B_n)} \\ &= \sum_{N=1}^\infty e^{-\lim_{K\to\infty} \sum_{n=N}^K \mu(B_n)} && \text{by continuity of } f(x) = e^x \\ &= \sum_{N=1}^\infty e^{-\sum_{n=N}^\infty \mu(B_n)} \\ &= \sum_{N=1}^\infty 0 \\ &= 0 .\end{align*}

  • Here we’ve used that every tail of a divergent sum is divergent: if \(\sum_{n=1}^\infty a_n \to \infty\) then for every \(N\), the tail \(\sum_{n=N}^\infty a_n \to \infty\) as well.

  • We’ve also use that if \(b_n\to \infty\) then \(e^{-b_n} \to 0\).

Spring 2019.2 #real_analysis/qual/completed

Let \(\mathcal B\) denote the set of all Borel subsets of \({\mathbf{R}}\) and \(\mu : \mathcal B \to [0, \infty)\) denote a finite Borel measure on \({\mathbf{R}}\).

  • Prove that if \(\{F_k\}\) is a sequence of Borel sets for which \(F_k \supseteq F_{k+1}\) for all \(k\), then \begin{align*} \lim _{k \rightarrow \infty} \mu\left(F_{k}\right)=\mu\left(\bigcap_{k=1}^{\infty} F_{k}\right) \end{align*}

  • Suppose \(\mu\) has the property that \(\mu (E) = 0\) for every \(E \in \mathcal B\) with Lebesgue measure \(m(E) = 0\). Prove that for every \(\epsilon > 0\) there exists \(\delta > 0\) so that if \(E \in \mathcal B\) with \(m(E) < δ\), then \(\mu(E) < ε\).

  • Proof of continuity of measure.
  • Using limsup/liminf sets (intersections of unions and vice-versa) and (sub)additivity to bound measures.
    • Control over lower bound: use tails of convergent sums
    • Control over upper bound: use rapidly converging coefficients like \(\sum 1/2^n\)
  • Convergent sums have vanishing tails.
  • Intersecting over more sets can only lose measure, taking a union over more can only gain measure.
  • Similarly intersecting over fewer sets can only gain measure, and taking a union over fewer sets can only lose measure.

Use a limsup or liminf of sets and continuity of measure. Note that choosing a limsup vs a liminf is fiddly – for one choice, you can only get one of the bounds you need, for the other choice you can get both.

  • Observation: \(\mu\) finite means \(\mu(E) < \infty\) for all \(E \in\mathcal{B}\), which we’ll need in several places.

  • Prove a more general statement: for any measure \(\mu\), \begin{align*} \mu(F_1) < \infty,\, F_k \searrow F \implies \lim_{k\to\infty}\mu(F_k) = \mu(F) ,\end{align*} where \(F_k \searrow F\) means \(F_1 \supseteq F_2 \supseteq \cdots\) with \(\bigcap_{k=1}^\infty F_k = F\).

    • Note that \(\mu(F)\) makes sense: each \(F_k \in \mathcal{B}\), which is a \(\sigma{\hbox{-}}\)algebra and closed under countable intersections.
  • Take disjoint annuli by setting \(E_k \coloneqq F_k \setminus F_{k+1}\)

  • Funny step: write \begin{align*} F_1 = F {\textstyle\coprod}\coprod_{k=1}^{\infty} E_k .\end{align*}

    • This is because \(x\in F_1\) iff \(x\) is in every \(F_k\), so in \(F\), or
    • \(x\not \in F_1\) but \(x\in F_2\), noting incidentally \(x\in F_3, F_4,\cdots\), or,
    • \(x\not\in F_2\) but \(x\in F_3\), and so on.
  • Now take measures, and note that we get a telescoping sum: \begin{align*} \mu(F_1) &= \mu(F) + \sum_{k=1}^\infty \mu(E_k) \\ &= \mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(E_k) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k \setminus F_{k+1} ) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k) - \mu(F_{k+1} ) \hspace{5em}\text{to be justified}\\ &= \mu(F) + \lim_{N\to\infty} [ (\mu(F_1) - \mu(F_2)) + (\mu(F_2) - \mu(F_3)) + \cdots \\ & \hspace{8em} + (\mu(F_{N-1}) - \mu(F_N)) + (\mu(F_N) - \mu(F_{N+1})) ] \\ \\ &= \mu(F) + \lim_{N\to\infty} \mu(F_1) - \mu(F_{N+1}) \\ &= \mu(F) + \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) .\end{align*}

  • Justifying the measure subtraction: the general statement is that for any pair of sets \(A\subseteq X\), \(\mu(X\setminus A) = \mu(X) - \mu(A)\) when \(\mu(A) < \infty\): \begin{align*} X &= A {\textstyle\coprod}(X\setminus A) \\ \implies \mu(X) &= \mu(A) + \mu(X\setminus A) && \text{countable additivity} \\ \implies \mu(X) -\mu(A) &= \mu(X\setminus A) && \text{if } \mu(A) < \infty .\end{align*}

  • Now use that \(\mu(F_1)<\infty\) to justify subtracting it from both sides: \begin{align*} \mu(F_1) &= \mu(F) + \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) \\ \implies 0 &= \mu(F_1) - \lim_{N\to\infty} \mu(F_{N+1}) \\ \lim_{N\to\infty} \mu(F_{N+1}) &= \mu(F_1) .\end{align*}

  • Now use that \(\lim_{N\to\infty}\mu(F_{N+1}) = \lim_{N\to\infty} \mu(F_N)\) to conclude.

  • Toward a contradiction, negate the implication: there exists an \({\varepsilon}>0\) such that for all \(\delta\), there exists an \(E\in \mathcal{B}\) \begin{align*} m(E) < \delta && \text{but} \hspace{4em} \mu(E) > {\varepsilon} .\end{align*}

    • Goal: produce a set \(A\) with \(m(A)= 0\) but \(\mu(A)\neq 0\).
  • Take a sequence \(\delta_n = \alpha(n)\), some function to be determined later, produce sets \(E_n\) with \begin{align*} m(E_n) < \delta_n && \text{but} \hspace{4em} \mu(E_n) > {\varepsilon}\quad \forall n .\end{align*}

  • Set \begin{align*} A_M \coloneqq\bigcap_{N=1}^M \bigcup_{n=N}^\infty E_n \coloneqq\bigcap_{N=1}^M F_N \hspace{4em} F_N \coloneqq\bigcup_{n=N}^\infty E_n .\end{align*}

    • Observation: \(F_N \supseteq F_{N+1}\) for all \(N\), since the right-hand side involves taking a union over fewer sets.
    • Notation: define \begin{align*} A_\infty \coloneqq\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n .\end{align*}
  • Bounding the Lebesgue measure \(m\) from above: \begin{align*} m(A_\infty) &\coloneqq m\qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n } \\ &\leq m\qty{ \bigcup_{n=N}^\infty E_n } && \forall N \\ &\leq \sum_{n=N}^\infty m(E_n) && \forall N \quad \text{by countable subadditivity} \\ &\leq \sum_{n=N}^\infty \alpha(n) \\ \\ &\overset{N\to\infty}\longrightarrow 0 ,\end{align*} where we’ve used that intersecting over fewer sets (i.e. none) can only increase measure in the first bound.

    • We have control over the sequence \(\alpha(n)\), so we can choose it to be summable so that the tails converge to zero as rapidly as we’d like.
    • So e.g. for any \({\varepsilon}_1 >0\), we can choose \(\alpha(n) \coloneqq{\varepsilon}_1/2^n\), then \begin{align*} \sum_{n=N}^\infty \alpha(n) &\leq \sum_{n=1}^\infty {{\varepsilon}_1 \over 2^n} = {\varepsilon}_1 \to 0 .\end{align*}
  • Bounding the \(\mu\) measure from below: \begin{align*} \mu(A_\infty) &\coloneqq \mu\qty{\bigcap_{N=1}^\infty F_N} \\ &= \lim_{N\to\infty} \mu(F_N) && \text{by part (1) }\\ &= \lim_{N\to\infty} \mu\qty{ \bigcup_{n=N}^\infty E_n } \\ &\geq \lim_{N\to\infty} \mu(E_N ) \\ &\geq \lim_{N\to\infty} {\varepsilon}\\ &= {\varepsilon}\\ &>0 ,\end{align*} where we’ve used that taking a union over fewer sets can only make the measure smaller.

Fall 2018.2 #real_analysis/qual/work

Let \(E\subset {\mathbf{R}}\) be a Lebesgue measurable set. Show that there is a Borel set \(B \subset E\) such that \(m(E\setminus B) = 0\).

\todo[inline]{Move this to review notes to clean things up.}
\todo[inline]{What a mess, redo!!}

  • Definition of measurability: there exists an open \(O\supset E\) such that \(m_*(O\setminus E) < {\varepsilon}\) for all \({\varepsilon}> 0\).
  • Theorem: \(E\) is Lebesgue measurable iff there exists a closed set \(F\subseteq E\) such that \(m_*(E\setminus F) < {\varepsilon}\) for all \({\varepsilon}>0\).
  • Every \(F_\sigma, G_\delta\) is Borel.
  • Claim: \(E\) is measurable \(\iff\) for every \(\varepsilon\) there exist \(F_\varepsilon \subset E \subset G_\varepsilon\) with \(F_\varepsilon\) closed and \(G_\varepsilon\) open and \(m(G_\varepsilon \setminus E)< \varepsilon\) and \(m(E\setminus F_\varepsilon) < \varepsilon\).
    • Proof: existence of \(G_{\varepsilon}\) is the definition of measurability.
    • Existence of \(F_{\varepsilon}\): ?
  • Claim: \(E\) is measurable \(\implies\) there exists an open \(O\supseteq E\) such that \(m(O\setminus E) = 0\).
    • Since \(E\) is measurable, for each \(n\in {\mathbb{N}}\) choose \(G_n \supseteq E\) such that \(m_*(G_n\setminus E) < {1\over n}\).
    • Set \(O_N \coloneqq\bigcap_{n=1}^N G_n\) and \(O\coloneqq\bigcap_{n=1}^\infty G_n\).
    • Suppose \(E\) is bounded.
      • Note \(O_N \searrow O\) and \(m_*(O_1) < \infty\) if \(E\) is bounded, since in this case \begin{align*} m_*(G_n\setminus E) = m_*(G_1) - m_*(E) < 1 \iff m_*(G_1) < m_*(E) + {1\over n} < \infty .\end{align*}
      • Note \(O_N \setminus E \searrow O \setminus E\) since \(O_N\setminus E \coloneqq O_N \bigcap E^c \supseteq O_{N+1} \bigcap E^c\) for all \(N\), and again \(m_*(O_1 \setminus E) < \infty\).
      • So it’s valid to apply continuity of measure from above: \begin{align*} m_*(O\setminus E) &= \lim_{N\to\infty} m_*(O_N\setminus E) \\ &\leq \lim_{N\to \infty} m_*(G_N\setminus E) \\ &= \lim_{N\to\infty} {1\over N} = 0 ,\end{align*} where the inequality uses subadditivity on \(\bigcap_{n=1}^N G_n \subseteq G_N\)
    • Suppose \(E\) is unbounded.
      • Write \(E^k = E \bigcap[k, k+1]^d \subset {\mathbf{R}}^d\) as the intersection of \(E\) with an annulus, and note that \(E = {\textstyle\coprod}_{k\in {\mathbb{N}}} E_k\).
      • Each \(E_k\) is bounded, so apply the previous case to obtain \(O_k \supseteq E_k\) with \(m(O_k\setminus E_k) = 0\).
      • So write \(O_k = E_k {\textstyle\coprod}N_k\) where \(N_k \coloneqq O_k \setminus E_k\) is a null set.
      • Define \(O = \bigcup_{k\in {\mathbb{N}}} O_k\), note that \(E\subseteq O\).
      • Now note \begin{align*} O\setminus E &= \qty{{\textstyle\coprod}_k O_k}\setminus \qty{{\textstyle\coprod}_K E_k} \\ &\subseteq {\textstyle\coprod}_k \qty{O_k \setminus E_k} \\ \implies m_*(O\setminus E) &\leq m_*\qty{{\textstyle\coprod}\qty{O_k \setminus E_k} } = 0 ,\end{align*} since any countable union of null sets is again null.
    • So \(O\supseteq E\) with \(m(O\setminus E) = 0\).
  • Theorem: since \(E\) is measurable, \(E^c\) is measurable
    • Proof: It suffices to write \(E^c\) as the union of two measurable sets, \(E^c = S \bigcup(E^c - S)\), where \(S\) is to be determined.
    • We’ll produce an \(S\) such that \(m_*(E^c - S) = 0\) and use the fact that any subset of a null set is measurable.
    • Since \(E\) is measurable, for every \({\varepsilon}> 0\) there exists an open \({\mathcal{O}}_{\varepsilon}\supseteq E\) such that \(m_*({\mathcal{O}}_{\varepsilon}\setminus E) < {\varepsilon}\).
    • Take the sequence \(\left\{{{\varepsilon}_n \coloneqq{1\over n}}\right\}\) to produce a sequence of sets \({\mathcal{O}}_n\).
    • Note that each \({\mathcal{O}}_n^c\) is closed and \begin{align*} {\mathcal{O}}_n \supseteq E \iff {\mathcal{O}}_n^c \subseteq E^c .\end{align*}
    • Set \(S \coloneqq\bigcup_n {\mathcal{O}}_n^c\), which is a union of closed sets, thus an \(F_\sigma\) set, thus Borel, thus measurable.
    • Note that \(S\subseteq E^c\) since each \({\mathcal{O}}_n \subseteq E^c\).
    • Note that \begin{align*} E^c\setminus S &\coloneqq E^c \setminus \qty{\bigcup_{n=1}^\infty {\mathcal{O}}_n^c} \\ &\coloneqq E^c \bigcap\qty{\bigcup_{n=1}^\infty {\mathcal{O}}_n^c}^c \quad\text{definition of set minus} \\ &= E^c \bigcap\qty{\bigcap_{n=1}^\infty {\mathcal{O}}_n}^c \quad \text{De Morgan's law}\\ &= E^c \bigcup\qty{\bigcap_{n=1}^\infty {\mathcal{O}}_n} \\ &\coloneqq\qty{ \bigcap_{n=1}^\infty {\mathcal{O}}_n} \setminus E \\ & \subseteq {\mathcal{O}}_N \setminus E \quad \text{for every } N\in {\mathbb{N}} .\end{align*}
    • Then by subadditivity, \begin{align*} m_*(E^c\setminus S) \leq m_*({\mathcal{O}}_N \setminus E) \leq {1\over N} \quad \forall N \implies m_*(E^c\setminus S) = 0 .\end{align*}
    • Thus \(E^c\setminus S\) is measurable.

  • Since \(E\) is measurable, \(E^c\) is measurable.
  • Since \(E^c\) is measurable exists an open \(O\supseteq E^c\) such that \(m(O\setminus E^c) = 0\).
  • Set \(B \coloneqq O^c\), then \(O\supseteq E^c \iff {\mathcal{O}}^c \subseteq E \iff B\subseteq E\).
  • Computing measures yields \begin{align*} E\setminus B \coloneqq E\setminus {\mathcal{O}}^c \coloneqq E\bigcap({\mathcal{O}}^c)^c = E\bigcap{\mathcal{O}}= {\mathcal{O}}\bigcap(E^c)^c \coloneqq{\mathcal{O}}\setminus E^c ,\end{align*} thus \(m(E\setminus B) = m({\mathcal{O}}\setminus E^c) = 0\).
  • Since \({\mathcal{O}}\) is open, \(B\) is closed and thus Borel.

d.irect Proof (Todo)

\todo[inline]{Try to construct the set.}

Spring 2018.1 #real_analysis/qual/completed

Define \begin{align*} E:=\left\{x \in \mathbb{R}:\left|x-\frac{p}{q}\right|

Prove that \(m(E) = 0\).

  • Borel-Cantelli: If \(\left\{{E_k}\right\}_{k\in{\mathbf{Z}}}\subset 2^{\mathbf{R}}\) is a countable collection of Lebesgue measurable sets with \(\sum_{k\in {\mathbf{Z}}} m(E_k) < \infty\), then almost every \(x\in {\mathbf{R}}\) is in at most finitely many \(E_k\).
    • Equivalently (?), \(m(\limsup_{k\to\infty} E_k) = 0\), where \(\limsup_{k\to\infty} E_k = \bigcap_{k=1}^\infty \bigcup_{j\geq k} E_j\), the elements which are in \(E_k\) for infinitely many \(k\).

  • Strategy: Borel-Cantelli.

  • We’ll show that \(m(E) \bigcap[n, n+1] = 0\) for all \(n\in {\mathbf{Z}}\); then the result follows from \begin{align*} m(E) = m \qty{\bigcup_{n\in {\mathbf{Z}}} E \bigcap[n, n+1]} \leq \sum_{n=1}^\infty m(E \bigcap[n, n+1]) = 0 .\end{align*}

  • By translation invariance of measure, it suffices to show \(m(E \bigcap[0, 1]) = 0\).

    • So WLOG, replace \(E\) with \(E\bigcap[0, 1]\).
  • Define \begin{align*} E_j \coloneqq\left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}\ \exists p\in {\mathbf{Z}}^{\geq 0} \text{ s.t. } {\left\lvert {x - \frac{p}{j} } \right\rvert} < \frac 1 {j^3}}\right\} .\end{align*}

    • Note that \(E_j \subseteq {\textstyle\coprod}_{p\in {\mathbf{Z}}^{\geq 0}} B_{j^{-3}}\qty{p\over j}\), i.e. a union over integers \(p\) of intervals of radius \(1/j^3\) around the points \(p/j\). Since \(1/j^3 < 1/j\), this union is in fact disjoint.
  • Importantly, note that \begin{align*} \limsup_{j\to\infty} E_j \coloneqq\bigcap_{n=1}^\infty \bigcup_{j=n}^\infty E_j = E \end{align*}


    \begin{align*} x \in \limsup_j E_j &\iff x \in E_j \text{ for infinitely many } j \\ &\iff \text{ there are infinitely many $j$ for which there exist a $p$ such that } {\left\lvert {x - {p\over j}} \right\rvert} < j^{-3} \\ &\iff \text{ there are infinitely many such pairs $p, j$} \\ &\iff x\in E .\end{align*}

  • Intersecting with \([0, 1]\), we can write \(E_j\) as a union of intervals: \begin{align*} E_j =& \qty{0, {j^{-3}}} \quad {\textstyle\coprod}\quad B_{j^{-3}}\qty{1\over j} {\textstyle\coprod} B_{j^{-3}}\qty{2\over j} {\textstyle\coprod} \cdots {\textstyle\coprod} B_{j^{-3}}\qty{j-1\over j} \quad {\textstyle\coprod}\quad (1 - {j^{-3}}, 1) ,\end{align*} where we’ve separated out the “boundary” terms to emphasize that they are balls about \(0\) and \(1\) intersected with \([0, 1]\).

  • Since \(E_j\) is a union of open sets, it is Borel and thus Lebesgue measurable.

  • Computing the measure of \(E_j\):

    • For a fixed \(j\), there are exactly \(j+1\) possible choices for a numerator (\(0, 1, \cdots, j\)), thus there are exactly \(j+1\) sets appearing in the above decomposition.

    • The first and last intervals are length \(1 \over j^3\)

    • The remaining \((j+1)-2 = j-1\) intervals are twice this length, \(2 \over j^3\)

    • Thus \begin{align*} m(E_j) = 2 \qty{1 \over j^3} + (j-1) \qty{2 \over j^3} = {2 \over j^2} \end{align*}

  • Note that \begin{align*} \sum_{j\in {\mathbb{N}}} m(E_j) = 2\sum_{j\in {\mathbb{N}}} \frac 1 {j^2} < \infty ,\end{align*} which converges by the \(p{\hbox{-}}\)test for sums.

  • But then \begin{align*} m(E) &= m(\limsup_j E_j) \\ &= m(\bigcap_{n\in {\mathbb{N}}} \bigcup_{j\geq n} E_j) \\ &\leq m(\bigcup_{j\geq N} E_j) \quad\text{for every } N \\ &\leq \sum_{j\geq N} m(E_j) \\ &\overset{N\to\infty}\to 0 \quad\text{} .\end{align*}

  • Thus \(E\) is measurable as a subset of a null set and \(m(E) = 0\).

Fall 2017.2 #real_analysis/qual/completed

Let \(f(x) = x^2\) and \(E \subset [0, \infty) \coloneqq{\mathbf{R}}^+\).

  • Show that \begin{align*} m^*(E) = 0 \iff m^*(f(E)) = 0. \end{align*}

  • Deduce that the map

\begin{align*} \phi: \mathcal{L}({\mathbf{R}}^+) &\to \mathcal{L}({\mathbf{R}}^+) \\ E &\mapsto f(E) \end{align*} is a bijection from the class of Lebesgue measurable sets of \([0, \infty)\) to itself.

\todo[inline]{Walk through.}


It suffices to consider the bounded case, i.e. \(E \subseteq B_M(0)\) for some \(M\). Then write \(E_n = B_n(0) \bigcap E\) and apply the theorem to \(E_n\), and by subadditivity, \(m^*(E) = m^*(\bigcup_n E_n) \leq \sum_n m^*(E_n) = 0\).

Lemma: \(f(x) = x^2, f^{-1}(x) = \sqrt{x}\) are Lipschitz on any compact subset of \([0, \infty)\).

Proof: Let \(g = f\) or \(f^{-1}\). Then \(g\in C^1([0, M])\) for any \(M\), so \(g\) is differentiable and \(g'\) is continuous. Since \(g'\) is continuous on a compact interval, it is bounded, so \({\left\lvert {g'(x)} \right\rvert} \leq L\) for all \(x\). Applying the MVT, \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = f'(c) {\left\lvert {x-y} \right\rvert} \leq L {\left\lvert {x-y} \right\rvert} .\end{align*}

Lemma: If \(g\) is Lipschitz on \({\mathbf{R}}^n\), then \(m(E) = 0 \implies m(g(E)) = 0\).

Proof: If \(g\) is Lipschitz, then \begin{align*} g(B_r(x)) \subseteq B_{Lr}(x) ,\end{align*} which is a dilated ball/cube, and so \begin{align*} m^*(B_{Lr}(x)) \leq L^n \cdot m^*(B_{r}(x)) .\end{align*}

Now choose \(\left\{{Q_j}\right\} \rightrightarrows E\); then \(\left\{{g(Q_j)}\right\} \rightrightarrows g(E)\).

By the above observation, \begin{align*} {\left\lvert {g(Q_j)} \right\rvert} \leq L^n {\left\lvert {Q_j} \right\rvert} ,\end{align*}

and so \begin{align*} m^*(g(E)) \leq \sum_j {\left\lvert {g(Q_j)} \right\rvert} \leq \sum_j L^n {\left\lvert {Q_j} \right\rvert} = L^n \sum_j {\left\lvert {Q_j} \right\rvert} \to 0 .\end{align*}

Now just take \(g(x) = x^2\) for one direction, and \(g(x) = f^{-1}(x) = \sqrt{x}\) for the other.

Lemma: \(E\) is measurable iff \(E = K {\textstyle\coprod}N\) for some \(K\) compact, \(N\) null.

Write \(E = K {\textstyle\coprod}N\) where \(K\) is compact and \(N\) is null.

Then \(\phi^{-1}(E) = \phi^{-1}(K {\textstyle\coprod}N) = \phi^{-1}(K) {\textstyle\coprod}\phi^{-1}(N)\).

Since \(\phi^{-1}(N)\) is null by part (a) and \(\phi^{-1}(K)\) is the preimage of a compact set under a continuous map and thus compact, \(\phi^{-1}(E) = K' {\textstyle\coprod}N'\) where \(K'\) is compact and \(N'\) is null, so \(\phi^{-1}(E)\) is measurable.

So \(\phi\) is a measurable function, and thus yields a well-defined map \(\mathcal L({\mathbf{R}}) \to \mathcal L({\mathbf{R}})\) since it preserves measurable sets. Restricting to \([0, \infty)\), \(f\) is bijection, and thus so is \(\phi\).

Spring 2017.1 #real_analysis/qual/completed

Let \(K\) be the set of numbers in \([0, 1]\) whose decimal expansions do not use the digit \(4\).

We use the convention that when a decimal number ends with 4 but all other digits are different from 4, we replace the digit \(4\) with \(399\cdots\). For example, \(0.8754 = 0.8753999\cdots\).

Show that \(K\) is a compact, nowhere dense set without isolated points, and find the Lebesgue measure \(m(K)\).

  • Definition: \(A\) is nowhere dense \(\iff\) every interval \(I\) contains a subinterval \(S \subseteq A^c\).
    • Equivalently, the interior of the closure is empty, \(\qty{\overline{K}}^\circ = \emptyset\).


Claim: \(K\) is compact.

  • It suffices to show that \(K^c \coloneqq[0, 1]\setminus K\) is open; Then \(K\) will be a closed and bounded subset of \({\mathbf{R}}\) and thus compact by Heine-Borel.

  • Strategy: write \(K^c\) as the union of open balls (since these form a basis for the Euclidean topology on \({\mathbf{R}}\)).

    • Do this by showing every point \(x\in K^c\) is an interior point, i.e. \(x\) admits a neighborhood \(N_x\) such that \(N_x \subseteq K^c\).
  • Identify \(K^c\) as the set of real numbers in \([0, 1]\) whose decimal expansion does contain a 4.

    • We will show that there exists a neighborhood small enough such that all points in it contain a \(4\) in their decimal expansions.
  • Let \(x\in K^c\), suppose a 4 occurs as the \(k\)th digit, and write \begin{align*} x = 0.d_1 d_2 \cdots d_{k-1}~ 4 ~d_{k+1}\cdots = \qty{\sum_{j=1}^k d_j 10^{-j}} + \qty{4\cdot 10^{-k}} + \qty{\sum_{j=k+1}^\infty d_j 10^{-j}} .\end{align*}

  • Set \(r_x < 10^{-k}\) and let \(y \in [0, 1] \bigcap B_{r_x}(x)\) be arbitrary and write \begin{align*} y = \sum_{j=1}^\infty c_j 10^{-j} .\end{align*}

  • Thus \({\left\lvert {x-y} \right\rvert} < r_x < 10^{-k}\), and the first \(k\) digits of \(x\) and \(y\) must agree:

    • We first compute the difference: \begin{align*} x - y &= \sum_{i=1}^\infty d_j 10^{-j} - \sum_{i=1}^\infty c_j 10^{-j} = \sum_{i=1}^\infty \qty{d_j - c_j} 10^{-j} \\ \end{align*}
    • Thus (claim) \begin{align*} {\left\lvert {x-y} \right\rvert} &\leq \sum_{j=1}^\infty {\left\lvert {d_j - c_j} \right\rvert} 10^j < 10^{-k} \iff {\left\lvert {d_j - c_j} \right\rvert} = 0 \quad \forall j\leq k .\end{align*}
    • Otherwise we can note that any term \({\left\lvert {d_j - c_j} \right\rvert}\geq 1\) and there is a contribution to \({\left\lvert {x-y} \right\rvert}\) of at least \(1\cdot 10^{-j}\) for some \(j < k\), whereas \begin{align*} j < k \iff 10^{-j} > 10^{-k} ,\end{align*} a contradiction.
  • This means that for all \(j \leq k\) we have \(d_j = c_j\), and in particular \(d_k = 4 = c_k\), so \(y\) has a 4 in its decimal expansion.

  • But then \(K^c = \bigcup_x B_{r_x}(x)\) is a union of open sets and thus open.

Claim: \(K\) is nowhere dense and \(m(K) = 0\):

  • Strategy: Show \(\qty{\overline{K}}^\circ = \emptyset\).

  • Since \(K\) is closed, \(\overline{K} = K\), so it suffices to show that \(K\) does not properly contain any interval.

  • It suffices to show \(m(K^c) = 1\), since this implies \(m(K) = 0\) and since any interval has strictly positive measure, this will mean \(K\) can not contain an interval.

  • As in the construction of the Cantor set, let

    • \(K_0\) denote \([0, 1]\) with 1 interval \(\left({4 \over 10}, {5 \over 10} \right)\) of length \(1 \over 10\) deleted, so \begin{align*}m(K_0^c) = {1\over 10}.\end{align*}
    • \(K_1\) denote \(K_0\) with 9 intervals \(\left({1 \over 100}, {5\over 100}\right), ~\left({14 \over 100}, {15 \over 100}\right), \cdots \left({94\over 100}, {95 \over 100}\right)\) of length \({1 \over 100}\) deleted, so \begin{align*}m(K_1^c) = {1\over 10} + {9 \over 100}.\end{align*}
    • \(K_n\) denote \(K_{n-1}\) with \(9^{n}\) such intervals of length \(1 \over 10^{n+1}\) deleted, so \begin{align*}m(K_n^c) = {1\over 10} + {9 \over 100} + \cdots + {9^{n} \over 10^{n+1}}.\end{align*}
  • Then compute \begin{align*} m(K^c) = \sum_{j=0}^\infty {9^n \over 10^{n+1} } = {1\over 10} \sum_{j=0}^\infty \qty{9\over 10}^n = {1 \over 10} \qty{ {1 \over 1 - {9 \over 10 } } } = 1. \end{align*}

Claim: \(K\) has no isolated points:

  • A point \(x\in K\) is isolated iff there there is an open ball \(B_r(x)\) containing \(x\) such that \(B_r(x) \subsetneq K^c\).

    • So every point in this ball should have a 4 in its decimal expansion.
  • Strategy: show that if \(x\in K\), every neighborhood of \(x\) intersects \(K\).

  • Note that \(m(K_n) = \left( \frac 9 {10} \right)^n \overset{n\to\infty}\to 0\)

  • Also note that we deleted open intervals, and the endpoints of these intervals are never deleted.

    • Thus endpoints of deleted intervals are elements of \(K\).
  • Fix \(x\). Then for every \(\varepsilon\), by the Archimedean property of \({\mathbf{R}}\), choose \(n\) such that \(\left( \frac 9 {10} \right)^n < \varepsilon\).

  • Then there is an endpoint \(x_n\) of some deleted interval \(I_n\) satisfying \begin{align*}{\left\lvert {x - x_n} \right\rvert} \leq \left( \frac 9 {10} \right)^n < {\varepsilon}.\end{align*}

  • So every ball containing \(x\) contains some endpoint of a removed interval, and thus an element of \(K\).

Spring 2017.2 #real_analysis/qual/completed

  • Let \(\mu\) be a measure on a measurable space \((X, \mathcal M)\) and \(f\) a positive measurable function.

Define a measure \(\lambda\) by \begin{align*} \lambda(E):=\int_{E} f ~d \mu, \quad E \in \mathcal{M} \end{align*}

Show that for \(g\) any positive measurable function, \begin{align*} \int_{X} g ~d \lambda=\int_{X} f g ~d \mu \end{align*}

  • Let \(E \subset {\mathbf{R}}\) be a measurable set such that \begin{align*} \int_{E} x^{2} ~d m=0. \end{align*} Show that \(m(E) = 0\).

  • Absolute continuity of measures: \(\lambda \ll \mu \iff E\in\mathcal{M}, \mu(E) = 0 \implies \lambda(E) = 0\).
  • Radon-Nikodym: if \(\lambda \ll \mu\), then there exists a measurable function \({\frac{\partial \lambda}{\partial \mu}\,} \coloneqq f\) where \(\lambda(E) = \int_E f \,d\mu\).
  • Chebyshev’s inequality: \begin{align*} A_c \coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \geq c }\right\} \implies \mu(A_c) \leq c^{-p} \int_{A_c} {\left\lvert {f} \right\rvert}^p \,d\mu \quad \forall 0 < p < \infty .\end{align*}

  • Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym

  • Claim: \(\lambda \ll \mu\), i.e. \(\mu(E) = 0 \implies \lambda(E) = 0\).

    • Note that if this holds, by Radon-Nikodym, \(f = {\frac{\partial \lambda}{\partial \mu}\,} \implies d\lambda = f d\mu\), which would yield \begin{align*} \int g ~d\lambda = \int g f ~d\mu .\end{align*}
  • So let \(E\) be measurable and suppose \(\mu(E) = 0\).

  • Then \begin{align*} \lambda(E) \coloneqq\int_E f ~d\mu &= \lim_{n\to\infty} \left\{{\int_E s_n \,d\mu {~\mathrel{\Big\vert}~}s_n \coloneqq\sum_{j=1}^\infty c_j \mu(E_j),\, s_n \nearrow f}\right\} \end{align*} where we take a sequence of simple functions increasing to \(f\).

  • But since each \(E_j \subseteq E\), we must have \(\mu(E_j) = 0\) for any such \(E_j\), so every such \(s_n\) must be zero and thus \(\lambda(E) = 0\).

\todo[inline]{What is the final step in this approximation?}
  • Set \(g(x) = x^2\), note that \(g\) is positive and measurable.

  • By part (a), there exists a positive \(f\) such that for any \(E\subseteq {\mathbf{R}}\), \begin{align*} \int_E g ~dm = \int_E gf ~d\mu \end{align*}

    • The LHS is zero by assumption and thus so is the RHS.

    • \(m \ll \mu\) by construction.

    • Note that \(gf\) is positive.

  • Define \(A_k = \left\{{x\in X {~\mathrel{\Big\vert}~}gf \cdot \chi_E > {1 \over k} }\right\}\), for \(k\in {\mathbf{Z}}^{\geq 0}\)

  • Then by Chebyshev with \(p=1\), for every \(k\) we have

\begin{align*} \mu(A_k) \leq k \int_E gf ~d\mu = 0 \end{align*}

  • Then noting that \(A_k \searrow A \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}gf\cdot \chi_E(x) > 0}\right\}\), we have \(\mu(A) = 0\).

  • Since \(gf\) is positive, we have \begin{align*} x\in E \iff gf\chi_E(x) > 0 \iff x\in A \end{align*} so \(E = A\) and \(\mu(E) = \mu(A)\).

  • But \(m \ll \mu\) and \(\mu(E) = 0\), so we can conclude that \(m(E) = 0\).

Fall 2016.4 #real_analysis/qual/completed

Let \((X, \mathcal M, \mu)\) be a measure space and suppose \(\left\{{E_n}\right\} \subset \mathcal M\) satisfies \begin{align*} \lim _{n \rightarrow \infty} \mu\left(X \backslash E_{n}\right)=0. \end{align*}

Define \begin{align*} G \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}x\in E_n \text{ for only finitely many } n}\right\}. \end{align*}

Show that \(G \in \mathcal M\) and \(\mu(G) = 0\).

\todo[inline]{Add concepts.}

  • Claim: \(G\in {\mathcal{M}}\).

    • Claim: \begin{align*} G = \qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n}^c = \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c .\end{align*}

      • This follows because \(x\) is in the RHS \(\iff\) \(x\in E_n^c\) for all but finitely many \(n\) \(\iff\) \(x\in E_n\) for at most finitely many \(n\).
    • But \({\mathcal{M}}\) is a \(\sigma{\hbox{-}}\)algebra, and this shows \(G\) is obtained by countable unions/intersections/complements of measurable sets, so \(G\in {\mathcal{M}}\).

  • Claim: \(\mu(G) = 0\).

    • We have \begin{align*} \mu(G) &= \mu\qty{\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu \qty{\bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu(E_M^c) \\ &\coloneqq\sum_{N=1}^\infty \mu(X\setminus E_N) \\ &\overset{N\to\infty}\to 0 .\end{align*}
\todo[inline]{Last step seems wrong!}

Spring 2016.3 #real_analysis/qual/work

Let \(f\) be Lebesgue measurable on \({\mathbf{R}}\) and \(E \subset {\mathbf{R}}\) be measurable such that \begin{align*} 0

Show that for every \(0 < t < 1\), there exists a measurable set \(E_t \subset E\) such that \begin{align*} \int_{E_{t}} f(x) d x=t A. \end{align*}

Spring 2016.5 #real_analysis/qual/work

Let \((X, \mathcal M, \mu)\) be a measure space. For \(f\in L^1(\mu)\) and \(\lambda > 0\), define \begin{align*} \phi(\lambda)=\mu(\{x \in X | f(x)>\lambda\}) \quad \text { and } \quad \psi(\lambda)=\mu(\{x \in X | f(x)<-\lambda\}) \end{align*}

Show that \(\phi, \psi\) are Borel measurable and \begin{align*} \int_{X}|f| ~d \mu=\int_{0}^{\infty}[\phi(\lambda)+\psi(\lambda)] ~d \lambda \end{align*}

Spring 2016.2 #real_analysis/qual/work

Let \(0 < \lambda < 1\) and construct a Cantor set \(C_\lambda\) by successively removing middle intervals of length \(\lambda\).

Prove that \(m(C_\lambda) = 0\).

Fall 2015.2 #real_analysis/qual/work

Let \(f: {\mathbf{R}}\to {\mathbf{R}}\) be Lebesgue measurable.

  • Show that there is a sequence of simple functions \(s_n(x)\) such that \(s_n(x) \to f(x)\) for all \(x\in {\mathbf{R}}\).
  • Show that there is a Borel measurable function \(g\) such that \(g = f\) almost everywhere.

Spring 2015.3 #real_analysis/qual/work

Let \(\mu\) be a finite Borel measure on \({\mathbf{R}}\) and \(E \subset {\mathbf{R}}\) Borel. Prove that the following statements are equivalent:

  • \(\forall \varepsilon > 0\) there exists \(G\) open and \(F\) closed such that \begin{align*} F \subseteq E \subseteq G \quad \text{and} \quad \mu(G\setminus F) < \varepsilon. \end{align*}
  • There exists a \(V \in G_\delta\) and \(H \in F_\sigma\) such that \begin{align*} H \subseteq E \subseteq V \quad \text{and}\quad \mu(V\setminus H) = 0 \end{align*}

Spring 2014.3 #real_analysis/qual/work

Let \(f: {\mathbf{R}}\to {\mathbf{R}}\) and suppose \begin{align*} \forall x\in {\mathbf{R}},\quad f(x) \geq \limsup _{y \rightarrow x} f(y) \end{align*} Prove that \(f\) is Borel measurable.

Spring 2014.4 #real_analysis/qual/work

Let \((X, \mathcal M, \mu)\) be a measure space and suppose \(f\) is a measurable function on \(X\). Show that \begin{align*} \lim _{n \rightarrow \infty} \int_{X} f^{n} ~d \mu = \begin{cases} \infty & \text{or} \\ \mu(f^{-1}(1)), \end{cases} \end{align*} and characterize the collection of functions of each type.

Measure Theory: Functions

Spring 2021.1 #real_analysis/qual/completed

Let \((X, \mathcal{M},\mu)\) be a measure space and let \(E_n \in \mathcal{M}\) be a measurable set for \(n\geq 1\). Let \(f_n \coloneqq\chi_{E_n}\) be the indicator function of the set \(E\) and show that

  • \(f_n \overset{n\to\infty}\to 1\) uniformly \(\iff\) there exists \(N\in {\mathbb{N}}\) such that \(E_n = X\) for all \(n\geq N\).

  • \(f_n(x) \overset{n\to\infty}\to 1\) for almost every \(x\) \(\iff\) \begin{align*} \mu \qty{ \bigcap_{n \geq 0} \bigcup_{k \geq n} (X \setminus E_k) } = 0 .\end{align*}

Part a:


  • Suppose \(\chi_{E_n}\to 1\) uniformly, we want to produce an \(N\) such that \(n\geq N \implies x\in E_n\) for all \(x\in X\).
  • Take \({\varepsilon}\coloneqq 1/2\). By uniform convergence, for \(N\) large enough, \begin{align*} & \forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} < 1/2 && \forall x\in X\\ &\iff \forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\iff \forall n\geq N \quad x\in E_n && \forall x\in X &\iff \forall n\geq N \quad E_n = X ,\end{align*} where we’ve used that \(E_n \subseteq X\) by definition and this shows \(X \subseteq E_n\). So this \(N\) suffices.


  • Let \({\varepsilon}> 0\) be arbitrary.
  • Choose \(N\) such that \(n\geq N \implies X = E_n\). Then \begin{align*} &\forall n\geq N \quad x\in E_n && \forall x\in X \\ &\forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} = 0 < {\varepsilon}&& \forall x\in X ,\end{align*} so \(\chi_{E_n} \to 1\) uniformly.

Part b:

  • Define \begin{align*} S &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\chi_{E_k}(x) \to 1}\right\}\\ &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\forall {\varepsilon},\, \exists N\, \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon},\forall k\geq N}\right\}\\ L &\coloneqq\bigcap_{n\geq 0} \bigcup_{k\geq n} \qty{X\setminus E_k} ,\end{align*} so \(S\) is the set where \(f_n\to f\) and \(X\setminus S\) is the exceptional set where \(f_n\not\to f\) doesn’t converge pointwise.

  • Claim: \(L = X\setminus S\), so if \(x\in S \iff x\in X\setminus L\).

  • Proof of claim: Suppose there exists an \(N\) such that the first line below is true. Then for a fixed \(x\), there are equivalent statements: \begin{align*} &\qquad x \in S \\ &\iff \exists N \text{ s.t. } \forall {\varepsilon}>0,\quad {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon}&& \forall k\geq N \\ &\iff \exists N \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} = 0 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } \chi_{E_k}(x) = 1 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\in E_k && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in X\setminus E_k &&\forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in \bigcup_{k\geq N} X\setminus E_k \\ &{\color{blue} \iff} x\not\in \bigcap_{n\geq 0}\bigcup_{k\geq n} X\setminus E_k \\ &\iff x\not\in L \\ &\iff x\in X\setminus L .\end{align*}

  • Proving the iff: \(f_n\to f\) almost everywhere \(\iff \mu(X\setminus S) = 0 \iff \mu(L) = 0\).

Spring 2021.3 #real_analysis/qual/work

Let \((X, \mathcal{M}, \mu)\) be a finite measure space and let \(\left\{{ f_n}\right\}_{n=1}^{\infty } \subseteq L^1(X, \mu)\). Suppose \(f\in L^1(X, \mu)\) such that \(f_n(x) \overset{n\to \infty }\to f(x)\) for almost every \(x \in X\). Prove that for every \({\varepsilon}> 0\) there exists \(M>0\) and a set \(E\subseteq X\) such that \(\mu(E) \leq {\varepsilon}\) and \({\left\lvert {f_n(x)} \right\rvert}\leq M\) for all \(x\in X\setminus E\) and all \(n\in {\mathbb{N}}\).

Fall 2020.2 #real_analysis/qual/work

  • Let \(f: {\mathbf{R}}\to {\mathbf{R}}\). Prove that \begin{align*} f(x) \leq \liminf_{y\to x} f(y)~ \text{for each}~ x\in {{\mathbf{R}}} \iff \{ x\in {{\mathbf{R}}} \mathrel{\Big|}f(x) > a \}~\text{is open for all}~ a\in {{\mathbf{R}}} \end{align*}

  • Recall that a function \(f: {{\mathbf{R}}} \to {{\mathbf{R}}}\) is called lower semi-continuous iff it satisfies either condition in part (a) above.

Prove that if \(\mathcal{F}\) is any family of lower semi-continuous functions, then \begin{align*} g(x) = \sup\{ f(x) \mathrel{\Big|}f\in \mathcal{F}\} \end{align*} is Borel measurable.

Note that \(\mathcal{F}\) need not be a countable family.

Fall 2016.2 #real_analysis/qual/completed

Let \(f, g: [a, b] \to {\mathbf{R}}\) be measurable with \begin{align*} \int_{a}^{b} f(x) ~d x=\int_{a}^{b} g(x) ~d x. \end{align*} Show that either

  • \(f(x) = g(x)\) almost everywhere, or
  • There exists a measurable set \(E \subset [a, b]\) such that \begin{align*} \int _{E} f(x) \, dx > \int _{E} g(x) \, dx \end{align*}

  • Monotonicity of the Lebesgue integral: \(f\leq g\) on \(A\) \(\implies \int_A f \leq \int_A g\)

Take the assumption and the negation of (1) and show (2). The obvious move: define the set \(A\) where they differ. The non-obvious move: split \(A\) itself up to get a strict inequality.

  • Write \(X\coloneqq[a, b]\),
  • Suppose it is not the case that \(f=g\) almost everywhere; then letting \(A\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}f(x) \neq g(x)}\right\}\), we have \(m(A) > 0\).
  • Write \begin{align*} A = A_1 {\textstyle\coprod}A_2 \coloneqq\left\{{f > g}\right\} {\textstyle\coprod}\left\{{f < g}\right\} .\end{align*}
  • Both \(A_i\) are measurable:
    • Since \(f,g\) are measurable functions, so is \(h\coloneqq f-g\).
    • We can write \begin{align*} A_1 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h > 0 }\right\} = h^{-1}((0, \infty)) \\ A_2 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h < 0 }\right\} = h^{-1}((-\infty, 0)) ,\end{align*} and pullbacks of Borel sets by measurable functions are measurable.
  • Then on \(E\), we have \(f(x)>g(x)\) pointwise. This is preserved by monotonicity of the integral, thus \begin{align*} f(x) > g(x) \text{ on } E \implies \int_{E} f(x)\,dx > \int_{E} g(x)\, dx .\end{align*}

Spring 2016.4 #real_analysis/qual/work

Let \(E \subset {\mathbf{R}}\) be measurable with \(m(E) < \infty\). Define \begin{align*} f(x)=m(E \cap(E+x)). \end{align*}

Show that

  • \(f\in L^1({\mathbf{R}})\).
  • \(f\) is uniformly continuous.
  • \(\lim _{|x| \to \infty} f(x) = 0\).

Hint: \begin{align*} \chi_{E \cap(E+x)}(y)=\chi_{E}(y) \chi_{E}(y-x) \end{align*}

#real_analysis/qual/work #real_analysis/qual/completed