References:
[Foll13]
[StSh05]
Metrics and convergence
A metric on a set \(X\) is a function \(\rho\colon X\times X\to[0,\infty)\) such that \(\rho(x,y)=0\) if and only if \(x=y\), \(\rho(x,y)=\rho(y,x)\) for all \(x,y\in X\), and \(\rho(x,z)\leq\rho(x,y)+\rho(y,z)\) for all \(x,y,z\in X\).
To prove \(x=y\) in a metric space it suffices to show \(\rho(x,y)<\varepsilon\) for every \(\varepsilon>0\), and the triangle inequality splits an estimate of \(\rho(x,z)\) through an intermediate point.
Let \(f_n, f\colon S\to\RR\). The sequence \((f_n)\) converges uniformly to \(f\) if \[ (\forall\varepsilon>0)\,(\exists n_0)\,(\forall x\in S)\,(\forall n>n_0)\quad \abs{f_n(x)-f(x)}<\varepsilon, \] and converges pointwise to \(f\) 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 fails exactly when \[ (\exists\varepsilon>0)\,(\forall n_0)\,(\exists n>n_0)\,(\exists x = x(n)\in S)\quad \abs{f_n(x)-f(x)}\geq\varepsilon, \] where the point \(x\) may depend on \(n\). Uniform convergence implies pointwise convergence, and \(x^n\) on \([0,1]\) converges pointwise but not uniformly. For a series \(\sum_{k\geq1}a_k(x)\), uniform convergence means uniform convergence of the partial sums \(f_n\coloneqq\sum_{k\leq n}a_k\).
With \(\norm{f}_\infty \coloneqq \sup_{x\in S} \abs{f(x)}\), \(f_n\to f\) uniformly if and only if there are \(M_n\to0\) with \(\norm{f_n-f}_\infty\leq M_n\) for all large \(n\).
(Uniform limit theorem)
If \(f_n\to f\) uniformly and each \(f_n\) is continuous, then \(f\) is continuous.
Given \(\varepsilon>0\) and \(x\), choose \(N\) with \(\norm{f_N-f}_\infty<\varepsilon/3\) and \(\delta>0\) with \(\abs{f_N(x)-f_N(y)}<\varepsilon/3\) for \(\abs{x-y}<\delta\). Then \(\abs{f(x)-f(y)}\leq\abs{f(x)-f_N(x)}+\abs{f_N(x)-f_N(y)}+\abs{f_N(y)-f(y)}<\varepsilon\) for \(\abs{x-y}<\delta\).
(Weierstrass $M$-test)
Let \(f_n\) be real- or complex-valued functions on a set \(A\) and \(M_n\geq0\) with \(\abs{f_n(x)}\leq M_n\) for all \(n\geq1\) and \(x\in A\), and \(\sum_{n\geq1}M_n<\infty\). Then \(\sum_{n\geq1}f_n(x)\) converges absolutely and uniformly on \(A\).
The supremum of a set of real numbers is its least upper bound and the infimum its greatest lower bound. The series \(\sum_{n\geq 1} n^{-p}\) converges if and only if \(p>1\). If \(\sum_{n\geq 1} a_n\) converges, then \(\lim_{N\to\infty}\sum_{n\geq N} a_n = 0\), so bounding \(\norm{f_n-f}_\infty\) by a tail of a convergent series proves uniform convergence. For \(\varepsilon>0\), \(\sum_{n\geq1}\varepsilon2^{-n} = \varepsilon\), which distributes an error \(\varepsilon\) over countably many terms.
Measure theory
An \(F_\sigma\) set is a countable union of closed sets, and a \(G_\delta\) set is a countable intersection of open sets. A \(\sigma\)-algebra on \(X\) is a collection of subsets of \(X\) containing \(X\) and closed under complements and countable unions, hence under countable intersections.
Let \(\mathcal M\) be a \(\sigma\)-algebra on \(X\). A measure on \(\mathcal M\) is a function \(\mu\colon\mathcal M\to[0,\infty]\) with \(\mu(\varnothing)=0\) and \(\mu\qty{\bigcup_{j\geq1}E_j} = \sum_{j\geq1}\mu(E_j)\) for every sequence of disjoint \(E_j\in\mathcal M\) (countable additivity), which implies finite additivity.
Let \((X,\mathcal M,\mu)\) be a measure space.
If \(E,F\in\mathcal M\) and \(E\subseteq F\), then \(\mu(E)\leq\mu(F)\).
If \(E_j\in\mathcal M\), then \(\mu\qty{\bigcup_jE_j}\leq\sum_j\mu(E_j)\).
If \(E_1\subseteq E_2\subseteq\cdots\) in \(\mathcal M\), then \(\mu\qty{\bigcup_jE_j} = \lim_j\mu(E_j)\).
If \(E_1\supseteq E_2\supseteq\cdots\) in \(\mathcal M\) and \(\mu(E_1)<\infty\), then \(\mu\qty{\bigcap_jE_j} = \lim_j\mu(E_j)\).
The proof of continuity of measure replaces a sequence of sets by disjoint sets with the same union: if \(A_1 \subseteq A_2 \subseteq \cdots\), set \(F_1\coloneqq A_1\) and \(F_k \coloneqq A_k \setminus A_{k-1}\) for \(k\geq 2\); then \(\bigcup_{k\geq 1} A_k = \bigsqcup_{k\geq 1} F_k\).
An outer measure on a nonempty set \(X\) is a function \(\mu^*\colon\mathcal P(X)\to[0,\infty]\) with \(\mu^*(\varnothing)=0\), \(\mu^*(A)\leq\mu^*(B)\) for \(A\subseteq B\), and \(\mu^*\qty{\bigcup_jA_j}\leq\sum_j\mu^*(A_j)\).
Let \(\mathcal E\subseteq\mathcal P(X)\) and \(\rho\colon\mathcal E\to[0,\infty]\) with \(\varnothing, X\in\mathcal E\) and \(\rho(\varnothing)=0\). For \(A\subseteq X\) let \[ \mu^*(A)\coloneqq\inf\theset{\sum_{j\geq1}\rho(E_j) \st E_j\in\mathcal E,\ A\subseteq\bigcup_{j\geq1}E_j}. \] Then \(\mu^*\) is an outer measure.
The Lebesgue outer measure of \(E\subseteq\RR^n\) is \(m_*(E)\coloneqq\inf\sum_i\abs{Q_i}\), the infimum over countable covers of \(E\) by closed cubes \(Q_i\).
A property holds almost everywhere if the set where it fails has measure zero.
For sets \(E_n\), \[ \limsup_n E_n \coloneqq \bigcap_{k\geq1}\bigcup_{n\geq k}E_n, \qquad \liminf_n E_n\coloneqq\bigcup_{k\geq1}\bigcap_{n\geq k}E_n, \] so \(\limsup_nE_n = \theset{x \st x\in E_n\text{ for infinitely many } n}\) and \(\liminf_nE_n = \theset{x \st x\in E_n\text{ for all but finitely many } n}\). De Morgan’s laws state \(\qty{\bigcup_\alpha E_\alpha}^c = \bigcap_\alpha E_\alpha^c\) and \(\qty{\bigcap_\alpha E_\alpha}^c = \bigcup_\alpha E_\alpha^c\).
Let \(\mu\) be a Lebesgue–Stieltjes measure on \(\RR\) with \(\sigma\)-algebra \(\mathcal M_\mu\). If \(E\in\mathcal M_\mu\), then \[ \mu(E) = \inf\theset{\mu(U) \st U\supseteq E,\ U\text{ open}} = \sup\theset{\mu(K) \st K\subseteq E,\ K\text{ compact}}. \] For \(E\subseteq\RR\) the following are equivalent: (a) \(E\in\mathcal M_\mu\); (b) \(E = V\setminus N_1\) with \(V\) a \(G_\delta\) set and \(\mu(N_1)=0\); (c) \(E = H\cup N_2\) with \(H\) an \(F_\sigma\) set and \(\mu(N_2)=0\).
Fubini–Tonelli
For \(f\) on \(\RR^{d_1}\times\RR^{d_2}\), \(x\in\RR^{d_1}\), and \(y\in\RR^{d_2}\), the slices of \(f\) are \(f^y(x)\coloneqq f(x,y)\) and \(f_x(y)\coloneqq f(x,y)\); for \(E\subseteq\RR^{d_1}\times\RR^{d_2}\), the slices are \(E^y\coloneqq\theset{x \st (x,y)\in E}\) and \(E_x\coloneqq\theset{y \st (x,y)\in E}\).
(Fubini's theorem [@SS05, Chapter 2, Theorem 3.1])
If \(f\) is integrable on \(\RR^{d_1}\times\RR^{d_2}\), then for almost every \(y\in\RR^{d_2}\) the slice \(f^y\) is integrable on \(\RR^{d_1}\), the function \(y\mapsto\int_{\RR^{d_1}}f^y(x)\dx\) is integrable on \(\RR^{d_2}\), and \[ \int_{\RR^{d_2}}\qty{\int_{\RR^{d_1}}f(x,y)\dx}\dy = \int_{\RR^d}f. \] By symmetry the same holds with \(x\) and \(y\) exchanged, so both iterated integrals equal \(\int_{\RR^d}f\).
(Tonelli's theorem [@SS05, Chapter 2, Theorem 3.2])
If \(f\) is a nonnegative measurable function on \(\RR^{d_1}\times\RR^{d_2}\), then for almost every \(y\in\RR^{d_2}\) the slice \(f^y\) is measurable on \(\RR^{d_1}\), the function \(y\mapsto\int_{\RR^{d_1}}f^y(x)\dx\) is measurable on \(\RR^{d_2}\), and \[ \int_{\RR^{d_2}}\qty{\int_{\RR^{d_1}}f(x,y)\dx}\dy = \int_{\RR^d}f \] in \([0,\infty]\).
To compute \(\int_{\RR^d}f\) for measurable \(f\), apply Tonelli’s theorem to \(\abs f\): if an iterated integral of \(\abs f\) is finite, then \(f\) is integrable and Fubini’s theorem applies to \(f\).
Let \(L^+\) denote the measurable functions \(X\to[0,\infty]\) on a measure space \((X,\mathcal M,\mu)\).
(Fubini--Tonelli theorem [@Fol13, Theorem 2.37])
Let \((X,\mathcal M,\mu)\) and \((Y,\mathcal N,\nu)\) be \(\sigma\)-finite measure spaces.
If \(f\in L^+(X\times Y)\), then \(g(x)\coloneqq\int f_x\,d\nu\) and \(h(y)\coloneqq\int f^y\,d\mu\) are in \(L^+(X)\) and \(L^+(Y)\), and \[ \int f\,d(\mu\times\nu) = \int\qty[\int f(x,y)\,d\nu(y)]d\mu(x) = \int\qty[\int f(x,y)\,d\mu(x)]d\nu(y). \]
If \(f\in L^1(\mu\times\nu)\), then \(f_x\in L^1(\nu)\) for almost every \(x\), \(f^y\in L^1(\mu)\) for almost every \(y\), the almost everywhere defined \(g\) and \(h\) are in \(L^1(\mu)\) and \(L^1(\nu)\), and the same equalities hold.
If \(E\subseteq\RR^{d_1}\times\RR^{d_2}\) is measurable, then for almost every \(y\in\RR^{d_2}\) the slice \(E^y\) is measurable, \(y\mapsto m(E^y)\) is measurable, and \(m(E) = \int_{\RR^{d_2}}m(E^y)\dy\).
Apply Tonelli’s theorem to \(\chi_E\).
The converse fails: a set whose slices are all measurable need not be measurable. For a non-measurable \(\mathcal N\subseteq[0,1]\), let \(E\coloneqq\mathcal N\times[0,1]\). Every slice \(E_x\) is \([0,1]\) or \(\varnothing\), hence measurable, but \(E\) is not measurable: Tonelli’s theorem applied to \(\chi_E\) would make \(E^y\) measurable for almost every \(y\), whereas \(E^y = \mathcal N\) for every \(y\in[0,1]\).
Let \(f\geq0\) on \(\RR^d\) and \(\mathcal A\coloneqq\theset{(x,y)\in\RR^d\times\RR \st 0\leq y\leq f(x)}\). Then \(f\) is measurable if and only if \(\mathcal A\) is measurable in \(\RR^{d+1}\), and in that case \(\int_{\RR^d}f(x)\dx = m(\mathcal A)\).
If \(f\) is measurable, then \(F(x,y)\coloneqq y-f(x)\) is measurable on \(\RR^{d+1}\), so \(\mathcal A = \theset{y\geq0}\cap\theset{F\leq0}\) is measurable. If \(\mathcal A\) is measurable, each slice \(\mathcal A_x = [0,f(x)]\) has \(m(\mathcal A_x)=f(x)\), so Tonelli’s theorem applied to \(\chi_{\mathcal A}\), with the roles of \(x\) and \(y\) exchanged, shows that \(f\) is measurable and \(m(\mathcal A) = \int m(\mathcal A_x)\dx = \int f(x)\dx\).
Qual problems
Problems suggested by Peter Woolfitt:
Integrability of $\sin\circ f$ and a Fubini identity for nonnegative integrable $f$ (Tag P-8RA35)
Let \(f:\mathbb{R}\to\mathbb{R}\) be a nonnegative integrable function.
Show that \(\sin\circ f\) is integrable.
Use Fubini’s theorem to show that \[ \int_{[0,\infty)} m(\{x:f(x)\ge y\})\cos y\,dy=\int_{\mathbb{R}}\sin\bigl(f(x)\bigr)\,dx. \]
Solution
<1>1. Prove that \(\sin\circ f\) is integrable.
Since \(f\ge0\) and \[ |\sin t|\le t \qquad(t\ge0), \] we have \[ |\sin(f(x))|\le f(x) \] for almost every \(x\). Because \(f\in L^1(\mathbb R)\), \[ \int_{\mathbb R}|\sin(f(x))|\,dx \le \int_{\mathbb R}f(x)\,dx<\infty. \] Hence \(\sin\circ f\in L^1(\mathbb R)\).
<1>2. Set up an absolutely integrable function on the product space.
Define \[ H(x,y):=\mathbf1_{\{0\le y\le f(x)\}}\cos y, \qquad (x,y)\in\mathbb R\times[0,\infty). \] This function is measurable. Moreover, \[ |H(x,y)|\le \mathbf1_{\{0\le y\le f(x)\}}. \] By Tonelli’s theorem, \[ \begin{aligned} \int_{\mathbb R}\int_0^\infty |H(x,y)|\,dy\,dx &\le \int_{\mathbb R}\int_0^{f(x)}1\,dy\,dx\\ &= \int_{\mathbb R}f(x)\,dx<\infty. \end{aligned} \] Thus \(H\in L^1(\mathbb R\times[0,\infty))\), so Fubini’s theorem applies.
<1>3. Integrate in the two orders.
Integrating first in \(y\) gives \[ \int_0^\infty H(x,y)\,dy = \int_0^{f(x)}\cos y\,dy = \sin(f(x)). \] Hence \[ \iint H(x,y)\,dy\,dx = \int_{\mathbb R}\sin(f(x))\,dx. \]
On the other hand, for fixed \(y\ge0\), \[ \int_{\mathbb R}H(x,y)\,dx = \cos y\,m(\{x:f(x)\ge y\}). \] Therefore Fubini yields \[ \boxed{ \int_0^\infty m(\{x:f(x)\ge y\})\cos y\,dy = \int_{\mathbb R}\sin(f(x))\,dx.} \]
Equal integrals of measurable functions: equality a.e. or a strict inequality on a subset (Tag P-OPH7A)
Let \(f,g\in L^1([a,b])\) be real-valued with \[ \int_{a}^{b} f(x) ~d x=\int_{a}^{b} g(x) ~d x. \] Show that either
- \(f(x) = g(x)\) almost everywhere, or
- There exists a measurable set \(E \subset [a, b]\) such that \[ \int _{E} f(x) \, dx > \int _{E} g(x) \, dx \]
- 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.
Solution
Let \[ h=f-g. \] Then \(h\in L^1([a,b])\) and \[ \int_a^b h\,dx=0. \] If \(h=0\) almost everywhere, then \(f=g\) almost everywhere and alternative (1) holds.
Assume instead that \(h\ne0\) on a set of positive measure. Write \[ E_+=\{h>0\}, \qquad E_-=\{h<0\}. \] If \(m(E_+)=0\), then \(h\le0\) almost everywhere and \(h<0\) on a set of positive measure. Consequently \[ \int_a^b h\,dx<0, \] contradicting \(\int h=0\). Thus \(m(E_+)>0\).
Because \(h>0\) on \(E_+\), \[ E_+=\bigcup_{k=1}^\infty\{h\ge 1/k\}. \] Hence some \(k\) satisfies \[ m(\{h\ge1/k\})>0. \] It follows that \[ \int_{E_+}h\,dx \ge \frac1k\,m(\{h\ge1/k\})>0. \] Taking \(E=E_+\) gives \[ \int_E f\,dx-\int_E g\,dx =\int_E h\,dx>0. \] Therefore alternative (2) holds.
Measurable set with dense complement and no isolated points (Tag P-4NYI7)
Does there exist a nonempty measurable set \(E \subset \mathbb{R}\) satisfying the following two properties:
given distinct \(x,y\in E\), there exists \(z\notin E\) strictly between \(x\) and \(y\);
\(E\) has no isolated points?
The distinctness requirement is necessary. If \(x=y\in E\), there is no point outside \(E\) between them, whether “between” is interpreted strictly or inclusively. Without that qualification no nonempty set can satisfy (a).
Solution
<1>1. Take the middle-thirds Cantor set.
Start with \(C_0=[0,1]\) and obtain \(C_{n+1}\) by removing the open middle third of each component interval of \(C_n\). Then \(C_n\) is a union of \(2^n\) disjoint closed intervals, each of length \(3^{-n}\), by induction. Set \(E=C=\bigcap_{n\geq0}C_n\). This is closed, hence Lebesgue measurable [@Fol13], and nonempty since \(0\in C_n\) for every \(n\).
<1>2. Verify property (a).
Let \(x<y\) belong to \(C\) and choose \(n\) with \(3^{-n}<y-x\). If \((x,y)\subset C\), it is contained in \(C_n\). An interval contained in the disjoint union defining \(C_n\) must lie in a single component interval: meeting two components would force it to contain the gap between them. But a component of length \(3^{-n}\) cannot contain \((x,y)\). This contradiction shows that some \(z\in(x,y)\) lies outside \(C\), proving (a).
<1>3. Verify property (b).
Every endpoint of a component interval of \(C_n\) survives all later stages, because only open middle thirds are removed; it therefore belongs to \(C\). For \(x\in C\) and \(\varepsilon>0\), choose \(n\) with \(3^{-n}<\varepsilon\) and let \(J\) be the component interval of \(C_n\) containing \(x\). At least one of the two endpoints of \(J\) is different from \(x\), lies in \(C\), and is at distance at most \(3^{-n}<\varepsilon\) from \(x\). Thus \(x\) is not isolated.
Therefore the answer is yes; the standard Cantor set satisfies both requested properties.
The subgraph of a nonnegative function is measurable if and only if the function is, and $m(\mathcal{A})=\int_{\RR^n}f=\int_0^\infty m(\{f\ge t\})\,dt$ (Tag P-ZCE6E)
Let \(f:\mathbb R^n\to[0,\infty]\) and define \[ \mathcal A:=\{(x,t)\in\mathbb R^n\times\mathbb R:0\le t\le f(x)\}. \]
Prove that \(f\) is Lebesgue measurable if and only if \(\mathcal A\) is Lebesgue measurable in \(\mathbb R^{n+1}\).
If \(f\) is measurable, prove \[ m_{n+1}(\mathcal A) =\int_{\mathbb R^n}f(x)\,dx =\int_0^\infty m_n(\{x:f(x)\ge t\})\,dt. \]
Solution
<1>1. If \(f\) is measurable, then \(\mathcal A\) is measurable.
Define \[ H(x,t):=f(x)-t. \] The map \((x,t)\mapsto f(x)\) is measurable, as is \((x,t)\mapsto t\), so \(H\) is measurable. Therefore \[ \mathcal A =(\mathbb R^n\times[0,\infty))\cap H^{-1}([0,\infty]) \] is Lebesgue measurable.
<1>2. If \(\mathcal A\) is measurable, then \(f\) is measurable.
For every fixed \(x\in\mathbb R^n\), the vertical section is \[ \mathcal A_x :=\{t:(x,t)\in\mathcal A\} =[0,f(x)]. \] Hence \[ f(x)=m_1(\mathcal A_x) =\int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt. \] Since \(\mathbf1_{\mathcal A}\) is a nonnegative measurable function on \(\mathbb R^{n+1}\), Tonelli’s theorem implies that the function \[ x\longmapsto \int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt \] is measurable. Thus \(f\) is measurable.
<1>3. Compute the measure of the subgraph.
Assume \(f\) is measurable. By Step 1, \(\mathcal A\) is measurable, so Tonelli gives \[ \begin{aligned} m_{n+1}(\mathcal A) &=\int_{\mathbb R^{n+1}}\mathbf1_{\mathcal A}(x,t)\,d(x,t)\\ &=\int_{\mathbb R^n}\left(\int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt\right)dx\\ &=\int_{\mathbb R^n}f(x)\,dx. \end{aligned} \]
<1>4. Derive the layer-cake formula.
For \(t\ge0\), the horizontal section is \[ \mathcal A^t :=\{x:(x,t)\in\mathcal A\} =\{x:f(x)\ge t\}. \] A second application of Tonelli gives \[ \begin{aligned} m_{n+1}(\mathcal A) &=\int_0^\infty\left(\int_{\mathbb R^n}\mathbf1_{\mathcal A}(x,t)\,dx\right)dt\\ &=\int_0^\infty m_n(\{x:f(x)\ge t\})\,dt. \end{aligned} \] Combining with Step 3 yields \[ \boxed{ m_{n+1}(\mathcal A) =\int_{\mathbb R^n}f(x)\,dx =\int_0^\infty m_n(\{f\ge t\})\,dt.} \]