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    Definitions

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    Convergence and continuity

    Limit superior and limit inferior of a sequence

    Let \((a_n)_{n\geq 0}\) be a sequence of real numbers. Its limit superior and limit inferior are \[ \begin{aligned} \limsup_{n\to\infty} a_n &\coloneqq \lim_{n\to \infty} \sup_{j\geq n} a_j = \inf_{n\geq 0} \sup_{j\geq n} a_j, \\ \liminf_{n\to\infty} a_n &\coloneqq \lim_{n\to \infty} \inf_{j\geq n} a_j = \sup_{n\geq 0} \inf_{j\geq n} a_j, \end{aligned} \] with values in \([-\infty,\infty]\); the limits exist and equal the infimum and supremum because \(\sup_{j\geq n}a_j\) is nonincreasing and \(\inf_{j\geq n}a_j\) is nondecreasing in \(n\).

    Continuity and uniform continuity

    Let \(X\subseteq\RR\) and let \(f\colon X\to \RR\).

    • The function \(f\) is continuous on \(X\) if \[ \forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon, x_0)>0 \quad \forall y\in X: \quad \abs{x_0 - y} < \delta \implies \abs{f(x_0) - f(y)} < \varepsilon. \]

    • The function \(f\) is uniformly continuous on \(X\) if \[ \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon)>0 \quad \forall x, y \in X: \quad \abs{x - y} < \delta \implies \abs{f(x) - f(y)} < \varepsilon. \]

    Substituting \(y = x_0 + h\), the two conditions read \[ \begin{aligned} &\forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta(\varepsilon, x_0)>0 \quad \forall h \text{ with } x_0+h\in X: && \abs{h} < \delta \implies \abs{f(x_0) - f(x_0 + h)} < \varepsilon, \\ &\forall \varepsilon>0 \quad \exists \delta(\varepsilon)>0 \quad \forall x\in X \quad \forall h \text{ with } x+h\in X: && \abs{h} < \delta \implies \abs{f(x) - f(x + h)} < \varepsilon. \end{aligned} \] In the first, \(\delta\) may depend on the point \(x_0\); in the second, one \(\delta\) serves every point of \(X\).

    Uniform continuity

    Let \((X, d_1)\) and \((Y, d_2)\) be metric spaces. A function \(f\colon X \to Y\) is uniformly continuous if for every \(\varepsilon > 0\) there exists \(\delta=\delta(\varepsilon) > 0\) such that for all \(x,y\in X\), \[ x\in B_\delta(y) \implies f(x) \in B_\varepsilon(f(y)). \]

    For continuity of \(f\) at every point, the same condition is required with \(\delta=\delta(\varepsilon,y)\) allowed to depend on the point \(y\); uniform continuity requires one \(\delta\) for all \(y\in X\).

    What does it mean for $f: (X, d_1) \to (Y, d_2)$ to be uniformly continuous?

    In the definition of continuity on a set, \(\delta\) may depend on both \(\varepsilon\) and the point \(x_0\); in the definition of uniform continuity, \(\delta\) depends only on \(\varepsilon\), and the condition holds for all pairs \(x,y\).

    Topological notions in a metric space

    Let \((X,d)\) be a metric space, let \(A\subseteq X\), and let \(p\in X\).

    • A neighborhood of \(p\) is an open set \(U_p\subseteq X\) containing \(p\).

    • An \(\varepsilon\)-neighborhood of \(p\) is an open ball \(B_\varepsilon(p) \coloneqq \theset{q\in X \suchthat d(p, q) < \varepsilon}\) for some \(\varepsilon>0\).

    • The point \(p\) is an accumulation point or a limit point of \(A\) if every punctured neighborhood \(U_p\setminus\theset{p}\) contains a point \(q\in A\). Write \(A'\) for the set of limit points of \(A\) and \(\overline{A} \coloneqq A\cup A'\) for the closure of \(A\).

    • If \(p\in A\) and \(p\) is not a limit point of \(A\), then \(p\) is an isolated point of \(A\).

    • \(A\) is closed if \(A' \subseteq A\), so that \(A\) contains all of its limit points.

    • A point \(p\in A\) is an interior point of \(A\) if there is a neighborhood \(U_p\) of \(p\) with \(U_p \subseteq A\).

    • \(A\) is open if every point of \(A\) is an interior point of \(A\).

    • \(A\) is perfect if \(A\) is closed and \(A\subseteq A'\), so that every point of \(A\) is a limit point of \(A\).

    • \(A\) is bounded if there are a real number \(M\) and a point \(q\in X\) such that \(d(p, q) < M\) for all \(p\in A\).

    • \(A\) is dense in \(X\) if every point of \(X\) lies in \(A\) or is a limit point of \(A\), that is, \(X = A\cup A' = \overline{A}\).

    Diameter of a subset of a metric space

    Let \((X,d)\) be a metric space and let \(A\subseteq X\) be nonempty. The diameter of \(A\) is \[ \diam(A) \coloneqq \sup_{x, y\in A} d(x, y) \in [0,\infty]. \]

    What is the diameter of a set $A$ in a metric space?

    Diameter of a set

    The diameter is the supremum of the distances between pairs of points of the set.Suggest a slogan

    Let \((X, d)\) be a metric space and let \(A\subseteq X\) be nonempty. The diameter of \(A\) is \[ \diam(A) = \sup_{x, y\in A} d(x, y)\in[0,\infty]. \]

    What is the diameter of set?

    Pointwise convergence

    Let \(S\) be a set, and let \(f_n\colon S\to\CC\) for \(n\geq 1\) and \(f\colon S\to\CC\) be functions. The sequence \((f_n)\) converges pointwise to \(f\) on \(S\) if \[ (\forall \varepsilon>0)\,(\forall x \in S)\,(\exists n_0 = n_0(x, \varepsilon))\,(\forall n>n_0)\quad \abs{f_n(x)-f(x)}<\varepsilon. \]

    Uniform convergence

    Let \(S\) be a set, and let \(f_n\colon S\to\CC\) for \(n\geq 1\) and \(f\colon S\to\CC\) be functions. The sequence \((f_n)\) converges uniformly to \(f\) on \(S\) if \[ (\forall \varepsilon>0)\,(\exists n_0 = n_0(\varepsilon))\,(\forall x \in S)\,(\forall n>n_0)\quad \abs{f_n(x)-f(x)}<\varepsilon. \]

    Negating the quantifiers, \((f_n)\) does not converge uniformly to \(f\) on \(S\) if and only if \[ (\exists \varepsilon>0)\,(\forall n_0)\,(\exists x = x(n_0) \in S)\,(\exists n>n_0)\quad \abs{f_n(x)-f(x)} \geq \varepsilon. \] The point \(x\) at which the estimate fails may depend on \(n_0\).

    Uniform convergence of a sequence of functions

    Let \(E\) be a set, and let \(f_n\colon E\to\CC\) for \(n\geq 1\) and \(f\colon E\to\CC\) be functions. The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if for every \(\varepsilon>0\) there exists \(N\in\NN\) such that \(\abs{f_n(x) - f(x)} < \varepsilon\) for all \(n\geq N\) and all \(x\in E\).

    Let \(E\), \((f_n)\), and \(f\) be as in the definition, and put \(\norm{f_n - f}_\infty \coloneqq \sup_{x\in E}\abs{f_n(x) - f(x)}\in[0,\infty]\). Then \((f_n)\) converges uniformly to \(f\) on \(E\) if and only if \(\norm{f_n - f}_\infty\to 0\) as \(n\to\infty\).

    If \((f_n)\) converges uniformly to \(f\), then for \(\varepsilon>0\) and \(N\) as in the definition, \(\norm{f_n - f}_\infty \leq \varepsilon\) for all \(n\geq N\), so \(\norm{f_n - f}_\infty\to 0\). Conversely, if \(\norm{f_n - f}_\infty\to 0\), then for \(\varepsilon>0\) there exists \(N\) with \(\norm{f_n - f}_\infty < \varepsilon\) for all \(n\geq N\), and then \(\abs{f_n(x) - f(x)}\leq\norm{f_n - f}_\infty<\varepsilon\) for all \(n\geq N\) and \(x\in E\).

    What does uniform convergence of $\theset{f_n}$ to $f$ on $E$ require of $N$?

    Uniform convergence of a sequence of functions

    Let \(E\) be a set, and let \(f_n\colon E\to\RR\) for \(n\geq 1\) and \(f\colon E\to\RR\) be functions. The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if for every \(\varepsilon>0\) there exists \(N=N(\varepsilon)\) such that \(\abs{f_n(x) - f(x)} < \varepsilon\) for all \(n\geq N\) and all \(x\in E\).

    With \(\norm{g}_\infty \coloneqq \sup_{x\in E}\abs{g(x)}\) for \(g\colon E\to\RR\), the sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if and only if \(\norm{f_n - f}_\infty\to 0\) as \(n\to\infty\).

    What does it mean for $f_n$ to converge uniformly to $f$ on $E$?

    Function spaces

    Completeness of a metric space

    A metric space \((X,d)\) is complete if every Cauchy sequence in \(X\) converges to a point of \(X\).

    Completeness of a normed space via absolutely convergent series

    Let \((V,\norm{\cdot})\) be a normed vector space. The following are equivalent:

    • \(V\) is complete: every Cauchy sequence in \(V\) converges.

    • Every absolutely convergent series in \(V\) converges: if \(x_n\in V\) and \(\sum_{n=1}^\infty\norm{x_n}<\infty\), then \(\sum_{n=1}^\infty x_n\) converges in \(V\).

    Assume \(V\) is complete and \(\sum_n\norm{x_n}<\infty\). For \(M>N\) the partial sums \(s_N\coloneqq\sum_{n=1}^N x_n\) satisfy \(\norm{s_M-s_N}\leq\sum_{n=N+1}^M\norm{x_n}\), so \((s_N)\) is Cauchy and converges.

    Conversely, assume every absolutely convergent series converges, and let \((y_m)\) be a Cauchy sequence. Choose \(m_1<m_2<\cdots\) with \(\norm{y_p-y_q}<2^{-k}\) for all \(p,q\geq m_k\). The series \(y_{m_1}+\sum_{k\geq 1}(y_{m_{k+1}}-y_{m_k})\) is absolutely convergent, so it converges; its partial sums are \(y_{m_{k}}\), so the subsequence \((y_{m_k})\) converges to some \(y\in V\). A Cauchy sequence with a convergent subsequence converges to the same limit, so \(y_m\to y\).

    What two conditions each characterise completeness?

    Completeness of a normed space via absolutely convergent series

    A normed space is complete exactly when every absolutely convergent series converges in the space.Suggest a slogan

    Let \((V,\norm{\cdot})\) be a normed vector space. Then \(V\) is complete if and only if every absolutely convergent series in \(V\) converges: whenever \(x_n\in V\) and \(\sum_{n=1}^\infty\norm{x_n}<\infty\), the series \(\sum_{n=1}^\infty x_n\) converges in \(V\).

    This is the equivalence proved in FD-WN55Z.

    Characterize completeness of a normed space in terms of absolutely convergent series.

    If \(X\) is a Banach space and \(\sum_n\norm{x_n}<\infty\), then \(\sum_n x_n\) converges in \(X\).

    Dense subsets

    Let \(X\) be a topological space. A subset \(A\subseteq X\) is dense in \(X\) if its closure is \(X\): \(\cl_X(A) = X\).

    When is a subset $A \subseteq X$ dense in $X$?

    Dense subset

    A dense subset meets every nonempty open set.Suggest a slogan

    Let \(X\) be a topological space. A subset \(A\subseteq X\) is dense in \(X\) if and only if \(\cl_X(A) = X\), if and only if every nonempty open subset of \(X\) meets \(A\).

    What is a dense subset?

    Separable spaces

    A topological space \(X\) is separable if \(X\) has a countable dense subset.

    Which countability condition makes a space separable?

    Nowhere dense sets

    Let \(X\) be a topological space. A subset \(S\subseteq X\) is nowhere dense in \(X\) if its closure \(\overline{S}\) has empty interior, that is, no nonempty open subset of \(X\) is contained in \(\overline{S}\).

    Equivalently, every nonempty open set \(U\subseteq X\) contains a nonempty open set \(V\) with \(V\cap S=\emptyset\), so \(S\) is not dense in any nonempty open subset of \(X\). Indeed, if \(\overline{S}\) has empty interior then \(V\coloneqq U\setminus\overline{S}\) is nonempty and open; conversely, a nonempty open \(W\subseteq\overline{S}\) contains no nonempty open set disjoint from \(S\). For \(X=\RR\), \(S\) is nowhere dense if and only if every open interval contains an open subinterval disjoint from \(S\).

    Nowhere dense sets

    Let \(X\) be a topological space. A set \(A\subseteq X\) is nowhere dense in \(X\) if its closure has empty interior: \(\qty{\overline{A}}^\circ=\emptyset\).

    Equivalently, \(A\) is not dense in any nonempty open subset of \(X\). For \(X=\RR\), a set \(A\) is nowhere dense if and only if every open interval \(I\) contains an open subinterval \(S\subseteq I\) with \(S\cap A = \emptyset\), that is, \(\overline{A}\) contains no open interval.

    In \(\RR\), the sets \(\theset{1/n \suchthat n\geq 1}\) and \(\ZZ\) are nowhere dense: their closures \(\theset{0}\cup\theset{1/n\suchthat n\geq 1}\) and \(\ZZ\) contain no open interval. The sets \(\QQ\) and \(\ZZ\cup\qty{(a, b)\cap \QQ}\) with \(a<b\) are not nowhere dense: their closures contain \(\RR\) and \([a,b]\) respectively.

    What does it mean for a set to be nowhere dense?

    Nowhere dense sets

    Nowhere dense means every nonempty open set contains a nonempty open subset disjoint from the closure.Suggest a slogan

    Let \(X\) be a topological space and \(A\subseteq X\). The following are equivalent:

    • \(A\) is nowhere dense in \(X\), that is, \(\qty{\overline{A}}^\circ = \emptyset\);

    • every nonempty open set \(U\subseteq X\) contains a nonempty open set \(V\) with \(V\cap A = \emptyset\);

    • \(A\) is not dense in any nonempty open subset of \(X\).

    For \(X = \RR\), \(A\) is nowhere dense if and only if every open interval \(I\) contains an open interval \(S\subseteq I\) with \(S\cap A = \emptyset\), that is, \(\overline{A}\) contains no open interval.

    What does it mean for a set to be nowhere dense?

    $\{1/n\}$ and $\ZZ$ are nowhere dense; $\QQ$ is not

    The closures of $\{1/n\}$ and $\ZZ$ contain no interval, while $\QQ$ is dense in $\RR$.Suggest a slogan

    In \(\RR\):

    • \(\theset{1/n\suchthat n\geq 1}\) and \(\ZZ\) are nowhere dense, since their closures \(\theset{0}\cup\theset{1/n\suchthat n\geq 1}\) and \(\ZZ\) contain no open interval;

    • \(\QQ\) and \(\ZZ\cup\qty{(a, b)\cap\QQ}\) for \(a<b\) are not nowhere dense, since their closures \(\RR\) and \(\ZZ\cup[a,b]\) contain the open interval \((a,b)\).

    Give an example of a set that is not nowhere dense.

    Meager sets

    Let \(X\) be a topological space. A subset \(S\subseteq X\) is meager in \(X\) if it is a countable union of nowhere dense subsets of \(X\).

    Meagre sets

    Let \(X\) be a topological space. A subset \(A\subseteq X\) is meagre in \(X\) if it is a countable union of nowhere dense subsets of \(X\).

    What is a meagre set?

    Meagre sets

    Meagre means a countable union of nowhere dense sets.Suggest a slogan

    Let \(X\) be a topological space and \(A\subseteq X\). Then \(A\) is meagre in \(X\) if and only if \(A\) is a countable union of nowhere dense subsets of \(X\).

    What is a meagre set?

    First and second category

    Let \(X\) be a topological space and let \(A\subseteq X\).

    • \(A\) is of first category in \(X\) if it is a countable union of nowhere dense subsets of \(X\).

    • \(A\) is of second category in \(X\) if it is not of first category in \(X\).

    What does it mean for a subset to be of first category, and of second category?

    First and second category sets

    First category means a countable union of nowhere dense sets; second category means not first category.Suggest a slogan

    Let \(X\) be a topological space and \(A\subseteq X\). Then \(A\) is of first category in \(X\) if and only if \(A\) is a countable union of nowhere dense subsets of \(X\), and of second category in \(X\) otherwise.

    What is a first category set? A second category?

    Baire space

    A topological space \(X\) is a Baire space if every countable intersection of open dense subsets of \(X\) is dense in \(X\).

    Baire space

    A topological space \(X\) is a Baire space if whenever \((U_n)_{n\geq 1}\) is a countable family of open dense subsets of \(X\), the intersection \(\bigcap_{n\geq 1} U_n\) is dense in \(X\).

    What does it mean for $X$ to be a Baire space?

    Baire spaces

    In a Baire space, every countable intersection of open dense sets is dense.Suggest a slogan

    A topological space \(X\) is a Baire space if and only if for every countable family \((U_n)_{n\geq 1}\) of open dense subsets of \(X\), the intersection \(\bigcap_{n\geq 1} U_n\) is dense in \(X\).

    What is a Baire space?

    Measure theory

    Limit superior and limit inferior of a sequence of sets

    Let \((E_n)_{n\geq 1}\) be a sequence of subsets of a set \(X\). Its limit inferior and limit superior are \[ \begin{aligned} \liminf_{n\to\infty} E_{n} &\coloneqq \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_{n} = \theset{x\in X \suchthat x\in E_{n} \text{ for all but finitely many } n}, \\ \limsup_{n\to\infty} E_{n} &\coloneqq \bigcap_{N=1}^\infty \bigcup_{n=N}^{\infty} E_{n} = \theset{x\in X \suchthat x\in E_{n} \text{ for infinitely many } n}. \end{aligned} \]

    Limit superior and limit inferior of a sequence of sets

    Let \((A_n)_{n\geq 1}\) be a sequence of subsets of a set \(X\). Its limit superior and limit inferior are \[ \begin{aligned} \limsup_{n\to\infty} A_n &\coloneqq \bigcap_{n\geq 1} \bigcup_{j\geq n} A_j = \theset{x\in X \suchthat x\in A_n \text{ for infinitely many } n}, \\ \liminf_{n\to\infty} A_n &\coloneqq \bigcup_{n\geq 1} \bigcap_{j\geq n} A_j = \theset{x\in X \suchthat x\in A_n \text{ for all but finitely many } n}. \end{aligned} \]

    What are $\limsup_n A_n$ and $\liminf_n A_n$ for a sequence of sets?

    $\limsup$ and $\liminf$ of a sequence of sets

    $\liminf$ means eventually always; $\limsup$ means infinitely often.Suggest a slogan

    Let \((A_n)_{n\geq 1}\) be a sequence of subsets of a set \(X\). Its limit inferior and limit superior are \[ \liminf _{n \to \infty} A_n=\bigcup_{n \geq 1} \bigcap_{j \geq n} A_j, \qquad \limsup _{n \to \infty} A_n=\bigcap_{n \geq 1} \bigcup_{j \geq n} A_j. \] A point \(x\in X\) lies in \(\liminf_n A_n\) if and only if \(x\in A_j\) for all but finitely many \(j\), and in \(\limsup_n A_n\) if and only if \(x\in A_j\) for infinitely many \(j\). In particular, \(\liminf _{n \to \infty} A_n \subseteq \limsup _{n \to \infty} A_n\).

    A point lies in \(\bigcap_{j\geq n}A_j\) for some \(n\) exactly when it lies in every \(A_j\) with \(j\geq n\) for some \(n\), that is, in all but finitely many \(A_j\). A point lies in \(\bigcup_{j\geq n}A_j\) for every \(n\) exactly when for every \(n\) it lies in some \(A_j\) with \(j\geq n\), that is, in infinitely many \(A_j\). A point in all but finitely many \(A_j\) lies in infinitely many of them.

    Define $\limsup, \liminf$ for sequences of sets. What are their containments?

    For sets \(E_n\subseteq X\), \[ \liminf_n E_n = \bigcup_{N\geq1}\bigcap_{n\geq N}E_n, \qquad \limsup_n E_n = \bigcap_{N\geq1}\bigcup_{n\geq N}E_n, \] so \(x\in\liminf_nE_n\) if and only if \(x\in E_n\) for all but finitely many \(n\), and \(x\in\limsup_nE_n\) if and only if \(x\in E_n\) for infinitely many \(n\). These parallel \(\liminf_n x_n = \sup_N\inf_{n\geq N}x_n\) and \(\limsup_n x_n = \inf_N\sup_{n\geq N}x_n\), with suprema corresponding to unions and infima to intersections. In particular \(\liminf_n E_n\subseteq\limsup_n E_n\).

    For a measure \(\mu\) and measurable \(E_n\), \[ \mu\qty{\liminf_n E_n }\leq \liminf_n \mu(E_n), \] and if \(\mu\qty{\bigcup_nE_n}<\infty\), then also \(\limsup_n\mu(E_n)\leq\mu\qty{\limsup_n E_n}\). If \(\liminf_nE_n = \limsup_nE_n\eqqcolon E\) and \(\mu\qty{\bigcup_nE_n}<\infty\), then \(\mu(E) = \lim_n\mu(E_n)\). For \(E_n\nearrow E=\bigcup_nE_n\) the equality \(\mu(E)=\lim_n\mu(E_n)\) holds for every measure \(\mu\), and for \(E_n\searrow E=\bigcap_nE_n\) it holds when \(\mu(E_1)<\infty\).

    $F_\sigma$ and $G_\delta$ sets

    Let \(X\) be a topological space. A subset of \(X\) is an \(F_\sigma\) set if it is a countable union of closed sets, and a \(G_\delta\) set if it is a countable intersection of open sets.

    In \(F_\sigma\), “F” stands for fermé, French for “closed”, and \(\sigma\) stands for a sum, that is, a union.

    $F_\sigma$ sets

    Let \(X\) be a topological space. A subset \(A\subseteq X\) is an \(F_\sigma\) set if \(A\) is a countable union of closed subsets of \(X\).

    What is an $F_\sigma$ set?

    $G_\delta$ sets

    Let \(X\) be a topological space. A subset \(A\subseteq X\) is a \(G_\delta\) set if \(A\) is a countable intersection of open subsets of \(X\).

    What is a $G_\delta$ set?

    Borel sets

    Let \(X\) be a topological space. The Borel \(\sigma\)-algebra of \(X\) is the smallest \(\sigma\)-algebra of subsets of \(X\) that contains every open subset of \(X\), namely the intersection of all such \(\sigma\)-algebras. Its elements are the Borel sets of \(X\).

    Since a \(\sigma\)-algebra is closed under complements, countable unions, and countable intersections, every closed set, every $F_\sigma$ set, and every $G_\delta$ set is a Borel set.

    What is a Borel set?

    Lebesgue outer measure

    A closed cube in \(\RR^n\) is a product \(Q=\prod_{i=1}^n[a_i,a_i+\ell]\) with \(\ell\geq 0\), and its volume is \(\abs{Q}\coloneqq\ell^n\). The outer measure of a set \(E\subseteq\RR^n\) is \[ m_*(E) \coloneqq \inf \theset{\sum_{i=1}^\infty \abs{Q_i} \suchthat Q_1, Q_2, \ldots \text{ closed cubes with } E\subseteq\bigcup_{i=1}^\infty Q_i} \in [0,\infty]. \]

    Lebesgue outer measure

    For a closed cube \(Q=\prod_{i=1}^n[a_i,a_i+\ell]\subseteq\RR^n\), write \(\abs{Q}\coloneqq\ell^n\) for its volume. The outer measure of \(E\subseteq\RR^n\) is \[ m_*(E) \coloneqq \inf \theset{ \sum_{i=1}^\infty \abs{Q_i} \suchthat Q_1,Q_2,\ldots \text{ closed cubes with } E\subseteq\bigcup_{i=1}^\infty Q_i}. \]

    How is the outer measure $m_*(E)$ defined?

    Basic properties of Lebesgue outer measure

    Let \(m_*\) be the outer measure on \(\RR^n\).

    1. Monotonicity. If \(E_1\subseteq E_2\subseteq\RR^n\), then \(m_*(E_1)\leq m_*(E_2)\).

    2. Countable subadditivity. If \(E=\bigcup_{j=1}^\infty E_j\) with \(E_j\subseteq\RR^n\), then \(m_*(E)\leq\sum_{j=1}^\infty m_*(E_j)\).

    3. Approximation from above by open sets. For every \(E\subseteq\RR^n\), \(m_*(E)=\inf\theset{m_*(O)\suchthat O\supseteq E \text{ open}}\).

    4. Additivity for almost disjoint cubes. If \(E=\bigcup_{j=1}^\infty Q_j\) is a countable union of closed cubes that are pairwise almost disjoint, then \(m_*(E)=\sum_{j=1}^\infty\abs{Q_j}\).

    What four properties does outer measure have?

    Lebesgue measurable sets

    Let \(m_*\) be the outer measure on \(\RR^n\). A subset \(E\subseteq \RR^n\) is Lebesgue measurable if for every \(\varepsilon> 0\) there exists an open set \(O \supseteq E\) such that \(m_*(O\setminus E) < \varepsilon\). For a Lebesgue measurable set \(E\), its Lebesgue measure is \(m(E) \coloneqq m_*(E)\).

    Lebesgue measurability of a set

    Let \(m_*\) be the outer measure on \(\RR^n\). A set \(E\subseteq \RR^n\) is measurable if for every \(\varepsilon>0\) there exists an open set \(G \supseteq E\) with \(m_*(G\setminus E)<\varepsilon\).

    When is a set $E \subseteq \RR^n$ measurable?

    Equivalent characterizations of measurability of a set

    Lebesgue measurability is equivalent to open, closed, $F_\sigma$, or $G_\delta$ approximation up to null sets.Suggest a slogan

    Let \(m_*\) be the outer measure on \(\RR^n\) and let \(E\subseteq \RR^n\). The following are equivalent:

    1. \(E\) is Lebesgue measurable: for every \(\varepsilon>0\) there exists an open set \(G\supseteq E\) with \(m_*(G\setminus E)<\varepsilon\).

    2. For every \(\varepsilon>0\) there exists a closed set \(F\subseteq E\) with \(m_*(E\setminus F) < \varepsilon\).

    3. \(E = H \cup Z\) with \(H\) an $F_\sigma$ set and \(Z\) a null set.

    4. \(E = V\setminus Z\) with \(V\) a $G_\delta$ set and \(Z\) a null set.

    If \(m_*(E)<\infty\), these are also equivalent to:

    1. For every \(\varepsilon>0\) there exists a compact set \(K\subseteq E\) with \(m_*(E\setminus K) < \varepsilon\).

    The finiteness hypothesis in (5) cannot be dropped: \(E=\RR^n\) is measurable, but \(m_*(\RR^n\setminus K)=\infty\) for every compact \(K\subseteq\RR^n\).

    What conditions each characterise measurability of a set $E \subset \RR^n$?

    Classes of Lebesgue measurable sets

    Open, closed, and null sets are measurable, and measurable sets remain measurable after taking cylinders.Suggest a slogan

    The following subsets of \(\RR^n\) are Lebesgue measurable:

    • open sets;

    • closed sets;

    • sets \(E\) with outer measure \(m_*(E) = 0\).

    If \(M\subseteq\RR^n\) is Lebesgue measurable, then the cylinder \(M\times\RR\subseteq\RR^{n+1}\) is Lebesgue measurable.

    Which sets are known to be measurable?

    Null sets in $\RR^n$

    Let \(m\) be Lebesgue measure on \(\RR^n\). A set \(A\subseteq\RR^n\) is null if for every \(\varepsilon>0\) there exist open sets \(U_1,U_2,\ldots\subseteq\RR^n\) with \(A\subseteq\bigcup_{j} U_j\) and \(\sum_j m(U_j) < \varepsilon\).

    A set \(A\) is null if and only if its outer measure \(m_*(A)\) is \(0\). Indeed, \(m_*(A)\leq\sum_j m(U_j)\) for every such cover; conversely, if \(m_*(A)=0\), a cover of \(A\) by closed cubes of total volume less than \(\varepsilon/2^{n+1}\) can be replaced by the open cubes with the same centers and twice the side lengths, of total volume less than \(\varepsilon\).

    What does it mean for a set to be null?

    Almost disjoint sets

    Let \(X\) be a topological space. Subsets \(A,B\subseteq X\) are almost disjoint if their interiors are disjoint: \(A^\circ \cap B^\circ = \emptyset\).

    When are two sets almost disjoint?

    Almost disjoint sets

    Almost disjoint means the interiors do not meet.Suggest a slogan

    Let \(X\) be a topological space. Subsets \(A, B\subseteq X\) are almost disjoint if and only if \(A^\circ \cap B^\circ = \emptyset\).

    What are almost disjoint sets?

    Complete measure

    Let \((X,\mcm,\mu)\) be a measure space. The measure \(\mu\) is complete if every subset of a null set is measurable: whenever \(N\in\mcm\), \(\mu(N)=0\), and \(A\subseteq N\), then \(A\in\mcm\).

    What is a complete measure?

    Convergence in measure

    Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq\RR^n\) be Lebesgue measurable, and let \(f_k\colon E\to\RR\) for \(k\geq 1\) and \(f\colon E\to\RR\) be measurable. The sequence \((f_k)\) converges in measure to \(f\) on \(E\) if for every \(\alpha>0\), \[ \lim_{k \to \infty} m\qty{\theset{x \in E \suchthat \abs{f_{k}(x)-f(x)}>\alpha}}=0. \]

    What does it mean for $f_k$ to converge to $f$ in measure?

    The class $L^+$ of nonnegative measurable functions

    Let \((X,\mcm)\) be a measurable space. The class \(L^+\), or \(L^+(X,\mcm)\), is the set of measurable functions \(f\colon X\to[0,\infty]\).

    Integrals and \(L^p\) spaces

    Integrable functions

    Let \((X,\mcm,\mu)\) be a measure space. A measurable function \(f\colon X\to\CC\) is integrable if \[ \norm{f}_1 \coloneqq \int_X \abs{f} \dmu < \infty. \]

    Essential supremum norm and essentially bounded functions

    Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to\CC\) be measurable. The essential supremum of \(\abs{f}\) is \[ \norm{f}_\infty \coloneqq \inf \theset{\alpha \geq 0 \suchthat \mu\qty{\theset{x\in X \suchthat \abs{f(x)} \geq \alpha}} = 0} \in [0,\infty], \] with \(\inf\emptyset=\infty\). The function \(f\) is essentially bounded if there exists a real number \(c\) such that \(\mu\qty{\theset{x\in X \suchthat \abs{f(x)} > c}} = 0\), equivalently \(\norm{f}_\infty < \infty\).

    When \(\norm{f}_\infty<\infty\), the set \(\theset{\abs{f}>\norm{f}_\infty}=\bigcup_{k\geq 1}\theset{\abs{f}\geq\norm{f}_\infty+1/k}\) is a countable union of null sets, so \(\abs{f(x)} \leq \norm{f}_\infty\) for almost every \(x\), and \(\norm{f}_\infty\) is the smallest constant with this property.

    The $L^\infty$ norm

    Let \((X,\mcm,m)\) be a measure space and let \(f\colon X\to\CC\) be measurable. The \(L^\infty\) norm of \(f\) is \[ \norm{f}_\infty \coloneqq \inf \theset{\alpha \geq 0 \suchthat m\qty{\theset{x\in X\suchthat \abs{f(x)} \geq \alpha}} = 0} \in[0,\infty]. \]

    When \(\norm{f}_\infty<\infty\), it is the least real number \(c\) with \(\abs{f}\leq c\) almost everywhere.

    What is the infinity norm $\norm{f}_\infty$?

    $L^\infty$

    Let \((X,\mcm,\mu)\) be a measure space. The space \(L^\infty(X)\) is \[ L^\infty(X) \coloneqq \theset{f\colon X\to \CC \suchthat f \text{ is measurable and essentially bounded}} = \theset{f\colon X\to \CC \suchthat f \text{ is measurable and } \norm{f}_\infty < \infty}, \] where \(\norm{f}_\infty\) is the essential supremum of \(\abs{f}\).

    Convolution of functions on $\RR^n$

    Let \(f,g\colon\RR^n\to\CC\) be measurable. The convolution of \(f\) and \(g\) is the function \[ (f * g)(x) \coloneqq \int_{\RR^n} f(x-y)\, g(y) \, dy, \] defined at every \(x\in\RR^n\) for which the integral converges absolutely.

    Convolution $(f\ast g)(\xi) = \int f(\xi - y)g(y) \, dy$

    Convolution averages one function against translated copies of the other.Suggest a slogan

    Let \(f, g\colon\RR^n\to\CC\) be measurable. At every \(\xi\in\RR^n\) for which the integral converges absolutely, the convolution of \(f\) and \(g\) is \[ (f\ast g)(\xi) = \int_{\RR^n} f(\xi - y)g(y) \, dy. \]

    How is the convolution $(f \ast g)(\xi)$ defined?

    Fourier transform

    For \(f\in L^1(\RR^n)\), the Fourier transform of \(f\) is the function \(\widehat f\colon\RR^n\to\CC\) given by \[ \widehat f(\xi) \coloneqq \int_{\RR^n} f(x)e^{-2\pi i x\cdot\xi}\,dx, \qquad \xi\in\RR^n. \]

    Dilation

    Let \(\phi\colon\RR^n\to\CC\) be a function and let \(t>0\). The dilation of \(\phi\) by \(t\) is the function \(\phi_t\colon\RR^n\to\CC\) given by \[ \phi_{t}(x) \coloneqq t^{-n} \phi\qty{t^{-1} x}. \]

    Approximations to the identity

    Let \(\phi\in L^1(\RR^n)\) with \(\int_{\RR^n} \phi = 1\), and for \(t>0\) let \(\phi_t\) be its dilation \(\phi_t(x)=t^{-n}\phi(t^{-1}x)\). Then \(\phi\), or the family \((\phi_t)_{t>0}\), is called an approximate identity.

    For every \(\phi\in L^1(\RR^n)\) and \(t>0\), the substitution \(x=ty\) gives \(\int \phi_{t} = \int \phi\), so every dilation of an approximate identity also has integral \(1\).

    Approximations to the identity

    Let \(\phi\in L^1(\RR^n)\) with \(\int_{\RR^n}\phi = 1\), and for \(t>0\) let \(\phi_t(x)\coloneqq t^{-n}\phi(t^{-1}x)\) be its dilations. Then \(\phi\), or the family \((\phi_t)_{t>0}\), is called an approximate identity.

    Let \(\phi\in L^1(\RR^n)\) with \(\int\phi=1\). Then:

    • \(\int \phi_t = \int\phi = 1\) and \(\norm{\phi_t}_1 = \norm{\phi}_1\) for every \(t>0\); in particular \(\norm{\phi_t}_1=1\) when \(\phi\geq 0\).

    • \(\sup_{t>0} \norm{\phi_t}_1 < \infty\).

    • For every \(h>0\), \(\lim_{t\to 0^+} \int_{\abs{x} \geq h} \abs{\phi_t(x)}\dx = 0\).

    The substitution \(x=ty\), \(dx=t^n\,dy\), gives \(\int\phi_t=\int\phi\) and \(\norm{\phi_t}_1=\norm{\phi}_1\), which proves the first two items. The same substitution gives \[ \int_{\abs{x}\geq h}\abs{\phi_t(x)}\dx=\int_{\abs{y}\geq h/t}\abs{\phi(y)}\,dy, \] and the right side tends to \(0\) as \(t\to 0^+\) by dominated convergence, with dominating function \(\abs{\phi}\in L^1\).

    Functional analysis

    Dual norm

    Let \(X\neq\theset{0}\) be a normed vector space over \(\CC\), and let \(\dualof{X}\) be the space of continuous linear functionals \(X\to\CC\). For \(L \in \dualof{X}\), the dual norm or operator norm of \(L\) is \[ \norm{L}_{\dualof{X}} \coloneqq \sup_{ \substack{x\in X \\ \norm{x} = 1} } \abs{L(x)} = \sup_{ \substack{x\in X \\ \norm{x} \leq 1} } \abs{L(x)}. \]

    Orthonormal sequence

    Let \(H\) be an inner product space with inner product \(\inner{\cdot}{\cdot}\) and norm \(\norm{u}\coloneqq\sqrt{\inner{u}{u}}\), and let \(I\) be a countable index set. A family \((u_i)_{i\in I}\) of elements of \(H\) is orthonormal if

    1. \(\inner{u_i}{u_j} = 0\) for all \(i,j\in I\) with \(i \neq j\), and

    2. \(\norm{u_j} = 1\) for all \(j\in I\).

    Basis of a Hilbert space

    Let \(\mch\) be a Hilbert space. An orthonormal sequence \((u_n)_{n\geq 1}\) in \(\mch\) is a basis of \(\mch\) if the linear span of \(\theset{u_n \suchthat n\geq 1}\), the set of finite linear combinations of the \(u_n\), is dense in \(\mch\).

    Complete families in a Hilbert space

    Let \(H\) be a Hilbert space. A family \((u_{n})_{n\geq 1}\) of vectors in \(H\) is complete if the only \(x\in H\) with \(\inner{x}{u_{n}} = 0\) for all \(n\) is \(x = 0\).

    Dual of a Hilbert space

    Let \(H\) be a Hilbert space over \(\CC\). The dual of \(H\) is the set \(\dualof{H}\) of linear functionals \(L\colon H\to \CC\) that are continuous with respect to the norm of \(H\).

    Linear functionals

    Let \(X\) be a vector space over \(\CC\). A map \(L\colon X \to \CC\) is a linear functional if \[ L(\alpha x + y) = \alpha L(x) + L(y) \quad \text{for all } x,y\in X \text{ and } \alpha\in\CC. \]

    Banach space

    A Banach space is a normed vector space \((V,\norm{\cdot})\) over \(\RR\) or \(\CC\) that is complete as a metric space with the metric \(d(x,y)\coloneqq\norm{x-y}\).

    Hilbert space

    A Hilbert space is an inner product space \((H,\inner{\cdot}{\cdot})\) over \(\RR\) or \(\CC\) that is a Banach space under the norm \(\norm{x}\coloneqq\sqrt{\inner{x}{x}}\).

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