Limits superior
Limit superior of a sequence (Tag D-G2I5Q)
Let \((x_n)_{n\geq 1}\) be a sequence of real numbers. Its limit superior is \[ \limsup_{n\to\infty} x_{n} \coloneqq \lim_{N \to \infty} \sup_{n>N} x_{n} \in [-\infty,\infty], \] where the limit exists in \([-\infty,\infty]\) because \(\sup_{n>N}x_n\) is nonincreasing in \(N\).
(The supremum and the limit superior differ)
For \(x_n \coloneqq \frac{n+1}{n}\), \(\sup_n x_n = x_1 = 2\), while \(x_n\to1\), so \(\limsup_n x_n = 1\).
Convergence theorems
Bounded convergence theorem (Tag T-TRW6N)
Let \(a<b\), let \(m\) be Lebesgue measure on \(\RR\), and let \(f_n\colon\RR\to\RR\) for \(n\geq1\) be Lebesgue measurable functions that vanish outside \([a,b]\). Suppose there is \(M\geq0\) with \(\abs{f_n(x)}\leq M\) for all \(n\) and all \(x\), and that \(f_n\to f\) pointwise for a function \(f\colon\RR\to\RR\). Then \[ \lim_{n\to\infty}\int f_n\,dm=\int f\,dm. \]
As a pointwise limit of measurable functions, \(f\) is measurable; it vanishes outside \([a,b]\) and satisfies \(\abs{f}\leq M\), so \(f_n\) and \(f\) are integrable. Let \(\varepsilon>0\). By Egorov's theorem applied on \([a,b]\), there is a closed set \(B\subseteq[a,b]\) with \(m(A)<\varepsilon\) for \(A\coloneqq[a,b]\setminus B\), such that \(f_n\to f\) uniformly on \(B\). Then \[ \abs{\int_B (f_n-f)\,dm}\leq(b-a)\sup_{x\in B}\abs{f_n(x)-f(x)}\longrightarrow0, \] and, since \(\abs{f_n-f}\leq 2M\), \[ \abs{\int_A (f_n-f)\,dm}\leq 2M\,m(A)<2M\varepsilon. \] Since \(f_n-f\) vanishes outside \(A\cup B\), \[ \limsup_{n\to\infty}\abs{\int f_n\,dm-\int f\,dm}\leq 0+2M\varepsilon. \] As \(\varepsilon>0\) was arbitrary, \(\int f_n\,dm\to\int f\,dm\).
The theorem is a special case of the dominated convergence theorem, with dominating function \(M\chi_{[a,b]}\).
(Fatou's lemma)
If \(f_n \geq 0\) are Lebesgue measurable functions on \(\RR^d\) and \(f_n\to f\) pointwise, then \[ \int f \leq \liminf_{n\to\infty} \int f_n. \]
Let \(g\) be a bounded measurable function supported on a set of finite measure with \(0\leq g\leq f\), and let \(g_n \coloneqq \min(g, f_n)\). Then \(g_n\to\min(g,f) = g\) pointwise, and the \(g_n\) are uniformly bounded and supported in the support of \(g\), so the bounded convergence theorem gives \(\int g_n\to\int g\). Since \(\int g_n\leq\int f_n\), this gives \(\int g\leq\liminf_n\int f_n\). Taking the supremum over all such \(g\) gives \(\int f\leq\liminf_n\int f_n\).
(Monotone convergence theorem)
If \(0 \leq f_1 \leq f_2 \leq \cdots\) are Lebesgue measurable functions and \(f(x) \coloneqq \lim_n f_n(x)\), then \[ \int f = \lim_{n\to\infty} \int f_n. \]
Since \(f_n\leq f\), \(\limsup_n\int f_n\leq\int f\), and Fatou’s lemma gives \(\int f\leq\liminf_n\int f_n\).
Differentiation
Let \(f\colon[a,b]\to\RR\).
If \(f\) is monotone, or more generally of bounded variation, then \(f'(x)\) exists for almost every \(x\) and \(\int_a^b\abs{f'} < \infty\).
The Cantor function has bounded variation and \(f'=0\) almost everywhere, but it is not constant, so \(f(x) - f(a) = \int_a^x f'\) can fail for \(f\) of bounded variation.
If \(f\) is absolutely continuous, then \(f(x) = f(a) + \int_a^x f'\) for all \(x\in[a,b]\).
If \(f\in L^1([a,b])\), then \(F(x) \coloneqq\int_a^x f\) is absolutely continuous and \(F' = f\) almost everywhere.
(Lebesgue density theorem)
If \(E\subseteq\RR\) is a Lebesgue measurable set of finite measure and \(m\) is Lebesgue measure, then for almost every \(x\in E\), \[ \lim_{r\to 0}\frac{m(E\cap B(x,r))}{m(B(x,r))} = 1. \]
Let \(F(x)\coloneqq m(E\cap(-\infty,x])\), an indefinite integral of \(\chi_E\in L^1(\RR)\). By the Lebesgue differentiation theorem, \(F'=\chi_E\) almost everywhere, and at each point \(x\) where \(F\) is differentiable, \[ \frac{m(E\cap B(x,r))}{m(B(x,r))} = \frac{F(x+r)-F(x-r)}{2r}\to F'(x)\quad (r\to0). \]
(A continuous nowhere differentiable function)
There exists a continuous function \(f\colon\RR\to\RR\) that is differentiable at no point.
Weierstrass’s function \(\sum_{n\geq0} a^n\cos(b^n\pi x)\) with \(0<a<1\), \(b\) an odd positive integer, and \(ab>1+3\pi/2\) is such a function; see Undergraduate counterexamples. Another such function is \(f(x)\coloneqq\sum_{n\geq1} a_n\sin(b_nx)\) with \(\sum a_n<\infty\) and \(a_nb_n\to\infty\) rapidly, for example \(a_n = 10^{-n}\) and \(b_n = 10^{6n}\). For each \(n\) and \(x\), an increment \(\Delta x\) of size comparable to \(1/b_n\) makes the \(n\)th term change by an amount comparable to \(a_n\), while the terms with \(k<n\) change by at most \(\sum_{k<n}a_kb_k/b_n\), which is small compared with \(a_n\), and the terms with \(k>n\) change by at most \(\sum_{k>n}2a_k\), also small compared with \(a_n\). So \(\abs{\Delta f/\Delta x}\) is comparable to \(a_nb_n\to\infty\), and \(f'(x)\) does not exist.
Baire category
Nowhere dense sets (Tag D-2MJRE)
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 \(S\).
Meager sets (Tag D-5NODS)
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 \(X\).
A subset of a topological space is residual if its complement is meager.
Every countable subset of \(\RR\), such as \(\QQ\), is meager, since each point is closed with empty interior.
In a topological space \(X\):
a subset of a meager set is meager;
a countable union of meager sets is meager;
a countable intersection of residual sets is residual;
a countable intersection of dense open sets is residual.
If \(X\) is a complete metric space, the Baire category theorem states that every residual subset of \(X\) is dense, and in particular nonempty.
No function continuous exactly on $\QQ$ (Tag P-HCA6D)
Can there be a function \(f:I\to \RR\) that is continuous on \(\QQ\) and discontinuous otherwise?
Solution
No. Let \(I \subseteq \RR\) be an open interval and \(f\colon I \to \RR\) any function.
<1>1. The set of points of continuity of \(f\) is a countable intersection of open sets.
Let \(U_n\) be the set of \(x \in I\) having an open neighborhood \(U \subseteq I\) with \(\operatorname{diam} f(U) < 1/n\). If \(U\) witnesses \(x \in U_n\), it witnesses \(y \in U_n\) for every \(y \in U\), so \(U_n\) is open. The function \(f\) is continuous at \(x\) if and only if \(x \in U_n\) for every \(n\).
<1>2. \(\QQ \cap I\) is not a countable intersection of open subsets of \(I\).
Suppose \(\QQ \cap I = \bigcap_n V_n\) with \(V_n\) open. Each \(V_n\) contains \(\QQ \cap I\), so it is dense in \(I\), and each \(I \setminus \theset q\), \(q \in \QQ \cap I\), is open and dense in \(I\). This countable family of open dense subsets of \(I\) has empty intersection, contradicting the Baire category theorem, which applies to \(I\) because \(I\) is homeomorphic to \(\RR\).
<1>3. Q.E.D.
Steps <1>1 and <1>2. See also E-QSNCL.