Proof techniques
Limits via \(\limsup\) and \(\liminf\). For a real sequence \((a_n)\), \(\limsup_n a_n\) and \(\liminf_n a_n\) exist in \([-\infty,\infty]\), and \(\lim_n a_n = c\) if and only if \(\limsup_n a_n\leq c\leq\liminf_n a_n\). The limit fails to exist exactly when \(\liminf_n a_n<\limsup_n a_n\).
Equality from two inequalities. \(a=b\) if and only if \(a\leq b\) and \(b\leq a\).
An \(\varepsilon\) of room. If \(a<b+\varepsilon\) for every \(\varepsilon>0\), then \(a\leq b\); if \(\norm a<\varepsilon\) for every \(\varepsilon>0\), then \(a=0\).
Approximate infima. If \(s = \inf S\) for a nonempty \(S\subseteq\RR\) bounded below, then for every \(\varepsilon>0\) there is \(x\in S\) with \(s\leq x<s+\varepsilon\).
Local to global. A property of a function that is local, such as continuity or differentiability at each point, holds on \(\RR^n\) if it holds on every ball \(B_R(0)\).
Good and bad sets. An integral \(\int_X\abs f\) is bounded by splitting \(X = G\sqcup B\), with \(f\) controlled pointwise on \(G\) and \(\mu(B)\) small; for two functions, \(X = \theset{f>g}\sqcup\theset{f=g}\sqcup\theset{f<g}\).
Local singularities and tails. \(\int_{\RR^n}\abs f = \int_{\abs x\leq1}\abs f + \int_{\abs x>1}\abs f\), and on \((1,\infty)\), \(\int_1^\infty\abs f = \sum_{k\geq0}\int_{2^k}^{2^{k+1}}\abs f\).
Add and subtract. For example, \(\norm{T_nx_n - Tx}\leq\norm{T_nx_n - Tx_n} + \norm{Tx_n - Tx}\leq\norm{T_n-T}\norm{x_n}+\norm T\norm{x_n-x}\).
Reduction to nice sets and functions. For Lebesgue measurable \(E\subseteq\RR^n\), \(m(E) = \lim_n m(E\cap B_n(0))\) by continuity from below; if \(m(E)<\infty\), then for every \(\varepsilon>0\) there are a compact \(K\subseteq E\) with \(m(E\setminus K)<\varepsilon\) and a finite union of rectangles \(R\) with \(m(E\,\triangle\,R)<\varepsilon\). For \(1\leq p<\infty\), two continuous maps on \(L^p\) that agree on simple functions, or on \(C_c(\RR^n)\), agree on \(L^p\).
Subsequences. A sequence converging in \(L^p\) has an almost everywhere convergent subsequence, and a bounded sequence in \(\RR^n\) has a convergent subsequence.
Sequential criteria. \(\lim_{\varepsilon\to0}g(\varepsilon) = L\) if and only if \(g(\varepsilon_n)\to L\) for every sequence \(\varepsilon_n\to0\) with \(\varepsilon_n\neq0\).
Standard facts
If \(\sup_x\abs{f_n(x)} = M_n\to0\), then \(f_n\to0\) uniformly; \(M_n\) can often be computed from \(f_n'\) by the first derivative test.
If \(\sum_n f_n\) converges uniformly on a neighborhood of \(x\) and each \(f_n\) is continuous at \(x\), then \(\sum_n f_n\) is continuous at \(x\).
Uniform convergence of \(\sum_n f_n\) on \(A\) follows from the Weierstrass \(M\)-test: \(\sup_{A}\abs{f_n}\leq M_n\) with \(\sum_n M_n<\infty\).
A continuous function with compact support on \(\RR^n\) is bounded and uniformly continuous.
\(\ell^1(\ZZ)\subsetneq\ell^2(\ZZ)\): for example \((1/(\abs k+1))_{k\in\ZZ}\in\ell^2\setminus\ell^1\).
If \(\mu(X)<\infty\) and \(1\leq p<q\leq\infty\), then \(L^q(\mu)\subseteq L^p(\mu)\), by Hölder’s inequality. For counting measure on \(\ZZ\), \(\ell^p\subseteq\ell^q\). For Lebesgue measure on \(\RR\), neither inclusion holds.
For \(a>0\) and \(1\leq p<\infty\), \(\abs x^{-a}\chi_{\theset{\abs x\leq1}}\in L^p(\RR^n)\) if and only if \(ap<n\), and \(\abs x^{-a}\chi_{\theset{\abs x>1}}\in L^p(\RR^n)\) if and only if \(ap>n\); so \(\abs x^{-n/p}\) fails to be in \(L^p\) both near \(0\) and near \(\infty\).
Every Lebesgue measurable set \(E\subseteq\RR^n\) is \(H\sqcup N\) with \(H\) an \(F_\sigma\) set and \(N\) null.
An absolutely continuous function on \([a,b]\) has bounded variation, and \(f(x) = f(a)+\int_a^x f'\).
For \(f,g\) in conjugate spaces \(L^p\) and \(L^q\), \(\abs{\int fg}\leq\norm f_p\norm g_q\).
\(\mu(X) = \norm{1}_{L^1(\mu)} = \int_X 1\,d\mu\).
Littlewood’s principles, on a set of finite measure: measurable sets are nearly finite unions of intervals, measurable functions are nearly continuous, and almost everywhere convergent sequences are nearly uniformly convergent; see Littlewood's three principles.
Standard results
Convergent series have terms and tails tending to zero (Tag PR-IGVTV)
Let \((a_n)_{n\geq1}\) be a sequence in \(\CC\) such that \(\sum_{n=1}^\infty a_n\) converges. Then \(a_n\to0\) as \(n\to\infty\), and \[ \sum_{n=N}^\infty a_n \convergesto{N\to\infty} 0 . \]
Uniform limit theorem (Tag T-ERNLN)
Let \(X\) be a topological space, and let \(f_n\colon X\to\CC\) for \(n\geq1\) and \(f\colon X\to\CC\) be functions. If each \(f_n\) is continuous and \(f_n\to f\) uniformly on \(X\), then \(f\) is continuous.
Let \(\varepsilon>0\) and \(x\) in the domain. Choose \(N\) with \(\sup_y\abs{F_N(y)-F(y)}<\varepsilon/3\), and \(\delta>0\) with \(\abs{F_N(x)-F_N(y)}<\varepsilon/3\) for \(\abs{x-y}<\delta\), by continuity of \(F_N\). Then for \(\abs{x-y}<\delta\), \[ \abs{F(x) - F(y)} \leq \abs{F(x) - F_N(x)} + \abs{F_N(x) - F_N(y)} + \abs{F_N(y) - F(y)} < \varepsilon. \]
Uniform limits commute with integrals on finite measure spaces (Tag PR-L7LNZ)
Let \((X,\mcm,\mu)\) be a measure space with \(\mu(X)<\infty\). If \(f_n,f\in L^1(X,\mu)\) and \(f_n\to f\) uniformly on \(X\), then \[ \abs{\int_X(f_n-f)\dmu} \leq \mu(X)\sup_{x\in X}\abs{f_n(x)-f(x)} \convergesto{n\to\infty} 0, \] so \(\int_X f_n\dmu\to\int_X f\dmu\).
\[ \abs{\int_X(f_n-f)\dmu} \leq \int_X\abs{f_n-f}\dmu \leq \mu(X)\sup_{x\in X}\abs{f_n(x)-f(x)}, \] and the right-hand side tends to \(0\) by uniform convergence.
Weak $M$-test (Tag PR-6WMSR)
Let \(S\) be a set and let \(f_n\colon S\to\CC\) for \(n\geq1\). Suppose that for each \(x\in S\) there are constants \(M_n(x)\geq0\) with \(\abs{f_n(x)} \leq M_n(x)\) for all \(n\) and \(\sum_{n} M_n(x) < \infty\). Then the series \(f(x) \coloneqq \sum_{n} f_n(x)\) converges absolutely at every \(x\in S\).
The constants \(M_n(x)\) may depend on \(x\), and the conclusion is only pointwise convergence. When one sequence \(M_n\) serves every \(x\in S\), the Weierstrass $M$-test gives uniform convergence on \(S\).
Weierstrass $M$-test for uniform convergence (Tag FF-IAUQG)
Let \(A\) be a set and let \(f_n\colon A\to\CC\) for \(n\geq 0\). Put \(\norm{f_n}_{\infty, A}\coloneqq\sup_{x\in A}\abs{f_n(x)}\). If \(\sum_{n\geq 0}\norm{f_n}_{\infty, A} < \infty\), then for every \(x\in A\) the series \(\sum_{n\geq 0} f_n(x)\) converges absolutely, and its partial sums converge uniformly on \(A\) to \(\sum_{n\geq 0} f_n\).
Equivalently, the conclusion holds whenever there are real numbers \(M_n\) with \(\abs{f_n(x)} \leq M_n\) for all \(x\in A\) and \(n\geq 0\) and \(\sum_{n\geq 0} M_n < \infty\).
For each \(x\), \(\sum_n\abs{f_n(x)}\leq\sum_n M_n<\infty\), so \(S(x)\coloneqq\sum_{n\geq 0}f_n(x)\) exists. For \(x\in A\) and \(N\geq 0\), \(\abs{S(x) - \sum_{n=0}^{N}f_n(x)}\leq\sum_{n>N}M_n\), which tends to \(0\) as \(N\to\infty\) independently of \(x\). The first form is the case \(M_n = \norm{f_n}_{\infty,A}\), and conversely \(\norm{f_n}_{\infty,A}\leq M_n\).
Conversely, if \(\sum_n f_n\) converges uniformly on \(A\), then \(\sup_{x\in A}\abs{f_n(x)}\to0\); this necessary condition does not imply \(\sum_n\sup_{x\in A}\abs{f_n(x)}<\infty\), as the constant functions \(f_n\coloneqq(-1)^n/n\) show.
Lebesgue measurable sets are $F_\sigma$ sets up to a null set (Tag PR-PIVFR)
Let \(E\subseteq\RR^n\) be Lebesgue measurable. Then \(E = H \cup N\) with \(H\cap N=\emptyset\), where \(H\) is an $F_\sigma$ set and \(N\) has Lebesgue measure \(0\).
(Borel characterization of measurable sets)
For each \(n\geq1\) there is a closed \(K_n\subseteq E\) with \(m(E\setminus K_n) \leq \frac 1 n\). Let \(H\coloneqq\bigcup_n K_n\), an \(F_\sigma\) set contained in \(E\). Then \(N\coloneqq E\setminus H\subseteq E\setminus K_n\) for every \(n\), so \(m(N)\leq\frac1n\) for every \(n\) and \(m(N)=0\).
Approximation of measurable sets by open, closed, and compact sets (Tag T-IIKSW)
Let \(E\subseteq\RR^n\) be Lebesgue measurable and let \(\varepsilon>0\).
There exists an open set \(O\supseteq E\) with \(m(O\setminus E) < \varepsilon\).
There exists a closed set \(F\subseteq E\) with \(m(E\setminus F) < \varepsilon\).
If \(m(E)<\infty\), there exists a compact set \(K\subseteq E\) with \(m(E\setminus K) < \varepsilon\) [@Fol13].
The finiteness hypothesis in (3) cannot be dropped. For \(E=\RR^n\), every compact \(K\subseteq\RR^n\) is bounded, hence has finite measure, so \(m(\RR^n\setminus K)=\infty\).
(Approximation of measurable sets)
Suppose first \(m(E)<\infty\). By the definition of outer measure there are closed cubes \(Q_i\) with \(E\subseteq\bigcup_iQ_i\) and \(\sum_i\abs{Q_i}<m(E)+\varepsilon/2\); enlarging each \(Q_i\) to an open cube \(Q_i'\) with \(\abs{Q_i'}<\abs{Q_i}+\varepsilon2^{-i-1}\), the open set \(O\coloneqq\bigcup_iQ_i'\) satisfies \(m(O)<m(E)+\varepsilon\), so \(m(O\setminus E)<\varepsilon\). In general write \(E=\bigcup_kE_k\) with \(m(E_k)<\infty\), choose open \(O_k\supseteq E_k\) with \(m(O_k\setminus E_k)<\varepsilon2^{-k}\), and let \(O\coloneqq\bigcup_kO_k\).
Apply (1) to \(E^c\) to get an open \(O\supseteq E^c\) with \(m(O\setminus E^c)<\varepsilon\), and let \(F\coloneqq O^c\). Then \(F\) is closed, \(F\subseteq E\), and \(E\setminus F = O\setminus E^c\), so \(m(E\setminus F)<\varepsilon\).
Assume \(m(E)<\infty\); for \(E=\RR^n\) no compact \(K\) has \(m(E\setminus K)<\infty\). By (2) choose a closed \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon/2\), and let \(K_n\coloneqq F\cap\overline{B_n(0)}\), which is compact. The sets \(E\setminus K_n\) decrease to \(E\setminus F\) and \(m(E)<\infty\), so \(m(E\setminus K_n)\to m(E\setminus F)\), and \(m(E\setminus K_n)<\varepsilon\) for \(n\) large.
Exercises
Translation invariance of Lebesgue integral, $L^{1}$ continuity, and regularity of measurable sets (Tag E-OMK54)
Prove the Lebesgue integral is translation/dilation invariant.
Prove continuity in \(L_1\): \(\norm{\tau_hf - f}\converges{h\to 0}\too 0\).
Prove that \(E\) is measurable \(\iff\) \(E = F \disjoint Z\) with \(F\in F_\sigma\) and \(Z\) null \(\iff\) \(E = G\sm Z\) with \(G\in G_\delta\) and \(Z\) null.
Show that \(m(E) = \sup_{K \subseteq E}m(K) \iff\) there exists \(K = K(\eps)\) with \(m(K) \in [m(E) - \eps, m(E)]\).
- What’s most useful here is the proof technique, not so much the result itself.
Apply Fubini and Tonelli to literally anything.
Prove that \(\norm{f}_p\to \norm{f}_\infty\) over a finite measure space.
Apply Cauchy-Schwarz to literally anything, in the form of \(\norm{fg}_1 \leq \norm{f}_2 \norm{g}_2\).
Solution
Let \(m\) be Lebesgue measure on \(\RR^n\), \(\tau_h f(x) = f(x + h)\), and \(f_\delta(x) = \delta^{-n}f(x/\delta)\) for \(\delta > 0\).
<1>1. For measurable \(f \geq 0\) or \(f \in L^1\), \(\int \tau_h f = \int f\) and \(\int f_\delta = \int f\).
<2>1. The identities hold for \(f = \chi_E\) with \(E\) measurable.
\(\tau_h\chi_E = \chi_{E-h}\) and \((\chi_E)_\delta = \delta^{-n}\chi_{\delta E}\), and Lebesgue measure satisfies \(m(E - h) = m(E)\) and \(m(\delta E) = \delta^n m(E)\).
<2>2. Q.E.D.
By linearity, step <2>1 gives the identities for nonnegative simple functions. For measurable \(f \geq 0\) choose simple \(0 \leq s_k \nearrow f\); then \(\tau_h s_k \nearrow \tau_h f\) and \((s_k)_\delta \nearrow f_\delta\), and the monotone convergence theorem passes the identities to \(f\). For \(f \in L^1\) apply this to \(f^+\) and \(f^-\).
<1>2. For \(f \in L^1\), \(\|\tau_h f - f\|_1 \to 0\) as \(h \to 0\).
<2>1. The claim holds for \(f = \chi_I\) with \(I\) a bounded box.
\(\|\tau_h\chi_I - \chi_I\|_1 = m(I \triangle (I - h))\), and \(I \triangle (I-h)\) is contained in the set of points within distance \(|h|\) of \(\partial I\), whose measure tends to \(0\) with \(h\).
<2>2. Step functions, finite linear combinations of indicators of bounded boxes, are dense in \(L^1\).
Simple functions are dense in \(L^1\). For \(m(E) < \infty\) and \(\eps > 0\), outer regularity gives an open \(U \supseteq E\) with \(m(U \setminus E) < \eps\). \(U\) is a countable union of almost disjoint closed cubes, so a finite union \(A\) of them has \(m(U \setminus A) < \eps\), and \(\|\chi_E - \chi_A\|_1 < 2\eps\).
<2>3. Q.E.D.
Given \(\eps > 0\), step <2>2 gives a step function \(s\) with \(\|f - s\|_1 < \eps/3\), and step <2>1 with the triangle inequality gives \(\|\tau_h s - s\|_1 < \eps/3\) for \(|h|\) small. By step <1>1, \(\|\tau_h f - \tau_h s\|_1 = \|f - s\|_1\), so \(\|\tau_h f - f\|_1 < \eps\) for \(|h|\) small.
<1>3. \(E\) is measurable if and only if \(E = F \sqcup Z\) with \(F\) an \(F_\sigma\) set and \(Z\) null, if and only if \(E = G \setminus Z'\) with \(G\) a \(G_\delta\) set and \(Z'\) null.
<2>1. If \(E\) is measurable, then \(E = F \sqcup Z\) as stated.
Outer regularity applied to \(E^c\) gives closed \(F_k \subseteq E\) with \(m(E \setminus F_k) < 1/k\). Put \(F = \bigcup_k F_k\) and \(Z = E \setminus F\). Then \(F\) is \(F_\sigma\), \(F \cap Z = \emptyset\), and \(m(Z) \leq m(E \setminus F_k) < 1/k\) for every \(k\).
<2>2. If \(E\) is measurable, then \(E = G \setminus Z'\) as stated.
Outer regularity gives open \(G_k \supseteq E\) with \(m(G_k \setminus E) < 1/k\). Put \(G = \bigcap_k G_k\) and \(Z' = G \setminus E\). Then \(G\) is \(G_\delta\), \(E = G \setminus Z'\), and \(m(Z') \leq m(G_k \setminus E) < 1/k\) for every \(k\).
<2>3. Q.E.D.
Steps <2>1 and <2>2 are the forward implications. Conversely, \(F_\sigma\) and \(G_\delta\) sets are Borel, null sets are Lebesgue measurable, and unions and differences of measurable sets are measurable.
<1>4. \(m(E) = \sup\theset{m(K) : K \subseteq E \text{ compact}}\) if and only if for every \(\eps > 0\) there is a compact \(K \subseteq E\) with \(m(E) - \eps \leq m(K) \leq m(E)\).
Every compact \(K \subseteq E\) has \(m(K) \leq m(E)\), so \(m(E)\) is an upper bound of the set \(S = \theset{m(K)}\). The supremum equals the upper bound \(m(E)\) exactly when no \(m(E) - \eps\) with \(\eps > 0\) is an upper bound of \(S\), which is the stated condition.
<1>5. If \(\mu(X) < \infty\) and \(f \in L^\infty(\mu)\), then \(\|f\|_p \to \|f\|_\infty\) as \(p \to \infty\).
If \(\mu(X) = 0\) both sides vanish; assume \(\mu(X) > 0\). Upper bound: \(\|f\|_p \le \|f\|_\infty\mu(X)^{1/p} \to \|f\|_\infty\). Lower bound: for \(0 < M < \|f\|_\infty\) the set \(\theset{|f| > M}\) has positive finite measure, and \(\|f\|_p \ge M\mu\theset{|f| > M}^{1/p} \to M\). So \(M \le \liminf_p \|f\|_p \le \limsup_p \|f\|_p \le \|f\|_\infty\) for every such \(M\).
Measurable slices (Tag PR-6NDTF)
Let \(n=n_1+n_2\), let \(m\) denote Lebesgue measure on each \(\RR^k\), and let \(E\subseteq\RR^{n}=\RR^{n_1}\times\RR^{n_2}\) be Lebesgue measurable. For \(x\in\RR^{n_1}\), let \(E_x \coloneqq \theset{y \in \RR^{n_2} \suchthat (x,y) \in E}\). Then:
For almost every \(x\in \RR^{n_1}\), the slice \(E_x\) is measurable in \(\RR^{n_2}\).
The function \[ F\colon \RR^{n_1} \to [0,\infty], \qquad x \mapsto m(E_x) = \int_{\RR^{n_2}} \chi_{E_x}(y) \,dy, \] defined for almost every \(x\), is measurable, and \[ m(E) = \int_{\RR^{n_1}} m(E_x) \,dx = \int_{\RR^{n_1}} \int_{\RR^{n_2}} \chi_{E_x}(y) \,dy \,dx . \]
(The region under a graph)
Let \(f\colon\RR^n\to[0,\infty)\) and \(A\coloneqq\theset{(x,y)\in\RR^n\times\RR \st 0\leq y\leq f(x)}\). Then \(f\) is Lebesgue measurable if and only if \(A\) is Lebesgue measurable in \(\RR^{n+1}\), and in that case \(m(A) = \int_{\RR^n}f(x)\dx\).
Suppose \(f\) is measurable. The functions \(F(x,y)\coloneqq f(x)\) and \(G(x,y)\coloneqq y\) are measurable on \(\RR^{n+1}\), since \(\theset{F>a} = \theset{f>a}\times\RR\) and \(\theset{G>a} = \RR^n\times(a,\infty)\). Hence \(A = \theset{G\leq F}\cap\theset{G\geq0}\) is measurable.
Suppose \(A\) is measurable. For every \(x\), the slice \(A_x = [0,f(x)]\) has \(m(A_x) = f(x)\). By Measurable slices, \(x\mapsto m(A_x) = f(x)\) is measurable and \(m(A) = \int_{\RR^n}m(A_x)\dx = \int_{\RR^n}f(x)\dx\).