Which convergence theorem? compares the hypotheses of these theorems.
Monotone convergence
For measurable \(0\leq f_1\leq f_2\leq\cdots\) with \(f_n\to f\) pointwise, \(\int f_n\to\int f\) in \([0,\infty]\). Applied to partial sums, it gives \(\int\sum_n g_n = \sum_n\int g_n\) for measurable \(g_n\geq0\).
Monotone convergence theorem (Tag T-5K3IO)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\) for \(n\geq1\) and \(f\) be functions in $L^+(X,\mcm)$ such that, for almost every \(x\in X\), \(f_n(x)\leq f_{n+1}(x)\) for all \(n\) and \(f_n(x)\to f(x)\). Then \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]
Monotone convergence theorem (Tag FT-5G4Y3)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\) for \(n\geq1\) and \(f\) belong to $L^+$. If \(f_n\le f_{n+1}\) almost everywhere for every \(n\) and \(f_n\to f\) almost everywhere, then \[ \lim_{n\to\infty} \int_X f_n \dmu = \int_X f \dmu . \]
Dominated convergence
If \(f_n\to f\) almost everywhere and \(\abs{f_n}\leq g\) with \(g\in L^1\), then \(f\in L^1\) and \(\int f_n\to\int f\); applying the theorem to \(\abs{f_n-f}\leq 2g\) gives \(\norm{f_n-f}_1\to0\).
Dominated convergence theorem (Tag T-IJQQG)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\colon X\to\CC\) for \(n\geq1\) and \(f\colon X\to\CC\) be measurable with \(f_n\to f\) almost everywhere. Suppose there is an integrable \(g\colon X\to[0,\infty)\) with \(\abs{f_n} \leq g\) almost everywhere for every \(n\). Then \(f\) is integrable and \[ \lim_{n\to\infty}\int_X \abs{f_n - f}\dmu = 0 ; \] consequently \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]
The functions \(f_n\) may be complex-valued and of arbitrary sign; the only sign condition is \(g\geq0\) on the dominating function.
Dominated convergence theorem (Tag FT-LCR5P)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\colon X\to\CC\) for \(n\geq1\) and \(f\colon X\to\CC\) be measurable. Suppose \(f_n\to f\) almost everywhere and there exists \(g \in L^1(X,\mu)\) with \(\abs{f_n} \leq g\) almost everywhere for every \(n\). Then \(f\in L^1(X,\mu)\) and \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]
Generalized dominated convergence theorem (Tag T-WYX24)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\colon X\to\CC\) and \(g_n\colon X\to[0,\infty)\) for \(n\geq1\), \(f\colon X\to\CC\), and \(g\colon X\to[0,\infty)\) be measurable. Suppose that
\(f_n\to f\) almost everywhere;
\(g_n\in L^1(X,\mu)\) and \(\abs{f_n}\leq g_n\) almost everywhere, for every \(n\);
\(g_n\to g\) almost everywhere, \(g\in L^1(X,\mu)\), and \(\lim_{n\to\infty}\int_X g_n\dmu=\int_X g\dmu\).
Then \(f\in L^1(X,\mu)\) and \[ \lim_{n\to\infty}\int_X f_n\dmu=\int_X f\dmu. \]
The dominated convergence theorem is the case \(g_n=g\) for all \(n\): the single dominating function \(g\) is replaced by dominating functions \(g_n\) that converge to \(g\) almost everywhere and in integral.
Convergence in $L^1$ implies convergence of norms (Tag PR-H4ZVI)
Let \((X,\mcm,\mu)\) be a measure space and let \(f, f_n\in L^1(\mu)\) for \(n\geq1\).
If \(\norm{f_n-f}_1\to0\), then \(\norm{f_n}_1\to\norm{f}_1\).
If \(f_n\to f\) \(\mu\)-almost everywhere and \(\norm{f_n}_1\to\norm{f}_1\), then \(\norm{f_n-f}_1\to0\) [@Fol13].
Without almost everywhere convergence, convergence of norms does not imply convergence in \(L^1\): for \(f\in L^1(\mu)\) with \(\norm{f}_1>0\) and \(f_n\coloneqq-f\), \(\norm{f_n}_1=\norm{f}_1\) for all \(n\) but \(\norm{f_n-f}_1=2\norm{f}_1\).
Fatou’s lemma
Fatou's lemma and reverse Fatou's lemma (Tag T-LDJNS)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\colon X\to[0,\infty]\) be measurable for \(n\geq1\).
(Fatou’s lemma [@Fol13]) \[ \liminf_{n\to\infty} \int_X f_n\dmu \geq \int_X \liminf_{n\to\infty} f_n\dmu . \]
(Reverse Fatou’s lemma) If moreover there is an integrable \(g\colon X\to[0,\infty)\) with \(f_n\leq g\) for all \(n\), then \[ \limsup_{n\to\infty} \int_X f_n\dmu \leq \int_X \limsup_{n\to\infty} f_n\dmu . \]
The domination hypothesis in (2) cannot be dropped. On \(\RR\) with Lebesgue measure, \(f_n\coloneqq\chi_{[n,n+1]}\) satisfies \(\int f_n=1\) for all \(n\), while \(\limsup_n f_n=0\) pointwise, so \(\limsup_n\int f_n=1>0=\int\limsup_n f_n\).
Fatou's lemma (Tag FT-P5UNP)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\) for \(n\geq1\) belong to $L^+$. Then \[ \int_X \liminf_{n\to\infty} f_n \dmu \leq \liminf_{n\to\infty} \int_X f_n \dmu . \]
The inequality can be strict. On \(\RR\) with Lebesgue measure, let \(f_n\coloneqq\one_{[n,n+1]}\). Then \(\liminf_n f_n=0\) pointwise, so the left side is \(0\), while \(\int_\RR f_n=1\) for every \(n\).
(Relations among the three theorems)
The monotone convergence theorem and Fatou’s lemma can each be proved from the other. The dominated convergence theorem follows from Fatou’s lemma applied to \(2g - \abs{f_n - f}\geq0\), which converges to \(2g\) and gives \(\limsup_n\int\abs{f_n-f}\leq 0\). For measurable \(f_n\geq0\), the inequality \(\int\liminf_n f_n\leq\liminf_n\int f_n\) of Fatou’s lemma can be strict: \(f_n\coloneqq n\chi_{(0,1/n)}\) on \(\RR\) gives \(0<1\).
Egorov and Lusin
Egorov’s theorem: if \(\mu(E)<\infty\) and measurable \(f_n\to f\) almost everywhere on \(E\) with \(f\) finite almost everywhere, then for every \(\varepsilon>0\) there is \(A\subseteq E\) with \(\mu(E\setminus A)<\varepsilon\) and \(f_n\to f\) uniformly on \(A\). Lusin’s theorem: if \(E\subseteq\RR^d\) has finite Lebesgue measure and \(f\) is measurable and finite almost everywhere on \(E\), then for every \(\varepsilon>0\) there is a closed \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon\) and \(f|_F\) continuous.
Egorov's theorem (Tag FT-OGS76)
Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq \RR^n\) be Lebesgue measurable with \(m(E) < \infty\), and let \(f_n\colon E\to\RR\) for \(n\geq1\) and \(f\colon E\to\RR\) be measurable with \(f_n \to f\) almost everywhere on \(E\). Then for every \(\varepsilon > 0\) there exists a closed set \(F\subseteq E\) such that \(m(E\setminus F)<\varepsilon\) and \(f_n\to f\) uniformly on \(F\).
Lusin's theorem (Tag T-CGFCU)
Let \(E\subseteq\RR^n\) be Lebesgue measurable with \(m(E)<\infty\), and let \(f\colon E\to\RR\) be measurable. For every \(\varepsilon>0\) there exists a closed set \(F_\varepsilon\subseteq\RR^n\) with \[ F_\varepsilon \subseteq E \quad\text{and}\quad m(E \setminus F_\varepsilon) \leq \varepsilon \] such that the restriction \(\ro{f}{F_\varepsilon}\colon F_\varepsilon\to\RR\) is continuous.
Continuity of the restriction \(\ro{f}{F_\varepsilon}\) does not make \(f\) continuous at points of \(F_\varepsilon\). Let \(E=[0,1]\) and \(f=\chi_{\QQ\cap[0,1]}\), which is discontinuous at every point of \([0,1]\). Enumerate \(\QQ\cap[0,1]=\theset{q_1,q_2,\ldots}\) and put \(F_\varepsilon\coloneqq[0,1]\setminus\bigcup_{k\geq1}(q_k-\varepsilon2^{-k-1},q_k+\varepsilon2^{-k-1})\). Then \(F_\varepsilon\) is closed, \(F_\varepsilon\subseteq E\), \(m(E\setminus F_\varepsilon)\leq\varepsilon\), and \(\ro{f}{F_\varepsilon}=0\) is continuous.
Interchanging limits, sums, and integrals
The equality \(\lim_n\int f_n=\int\lim_n f_n\) holds when \(f_n\to f\) uniformly and all \(f_n\) vanish outside one set of finite measure, when \(f_n\to f\) almost everywhere with \(\abs{f_n}\leq g\in L^1\), and when \(0\leq f_n\uparrow f\). It fails for the pointwise convergent sequences \(n\chi_{(0,1/n)}\), \(\chi_{[n,n+1]}\), and \(\frac1n\chi_{[0,n]}\) on $L^1$, each with limit \(0\) and integral \(1\). If \(\sum_n\norm{f_n}_1<\infty\), then \(\sum_n f_n\) converges absolutely almost everywhere and in \(L^1\), and \(\int\sum_n f_n = \sum_n\int f_n\).
Interchanging sums and integrals for absolutely summable series (Tag T-MN6WQ)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\colon X\to\CC\) be integrable for \(n\geq1\). The two quantities \[ \sum_{n=1}^\infty \int_X \abs{f_n}\dmu \quad\text{and}\quad \int_X\sum_{n=1}^\infty \abs{f_n}\dmu \] are equal, and if they are finite, then \(\sum_{n}f_n(x)\) converges absolutely for almost every \(x\), its sum is integrable, and \[ \int_X\sum_{n=1}^\infty f_n\dmu = \sum_{n=1}^\infty \int_X f_n\dmu . \]
Interchanging a series and an integral by Fubini--Tonelli on $X\times\NN$ (Tag FS-BM2PV)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\colon X\to\CC\) for \(n\geq1\) be measurable functions with \[ \sum_{n=1}^\infty \int_X \abs{f_n} \dmu < \infty . \] Then \(\sum_{n}f_n(x)\) converges absolutely for almost every \(x\), and \[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_X \sum_{n=1}^\infty f_n \dmu. \] The method views the sum as an integral against counting measure on \(\NN\) and applies the Fubini–Tonelli theorem on \(X\times\NN\).
<1>1. Encode the series as an integral over \(\NN\).
<2>1. Let \(\nu\) be counting measure on \(\NN\), so that \(\int_{\NN} g \, d\nu = \sum_{n=1}^\infty g(n)\) for every \(g\colon\NN\to[0,\infty]\) and every \(\nu\)-integrable \(g\colon\NN\to\CC\).
<2>2. Define \(F \colon X \times \NN \to \CC\) by \(F(x, n) \coloneqq f_n(x)\). It is measurable for the product \(\sigma\)-algebra, since \(F^{-1}(B)=\bigcup_n f_n^{-1}(B)\times\theset{n}\) for every Borel set \(B\subseteq\CC\).
<2>3. The two sides of the identity are the two iterated integrals of \(F\): \[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_{\NN} \int_X F(x, n) \, d\mu(x) \, d\nu(n), \qquad \int_X \sum_{n=1}^\infty f_n \dmu = \int_X \int_{\NN} F(x, n) \, d\nu(n) \, d\mu(x). \]
<1>2. Reduce to a \(\sigma\)-finite measure space.
<2>1. For each \(n\) and \(k\geq1\), \(\mu\qty{\theset{\abs{f_n}>1/k}}\le k\int_X\abs{f_n}\dmu<\infty\), so \(A\coloneqq\bigcup_n\theset{x\in X\suchthat f_n(x)\neq0}\) is a countable union of sets of finite measure.
<2>2. Every \(f_n\) vanishes outside \(A\), so both sides of the identity are unchanged when \(X\) is replaced by \(A\) and \(\mu\) by its restriction to \(A\), which is \(\sigma\)-finite. Counting measure on \(\NN\) is \(\sigma\)-finite as well.
<1>3. Verify the hypothesis of Fubini’s theorem.
<2>1. Since \(\abs{F}\) is nonnegative and measurable on \(A \times \NN\), Tonelli’s theorem gives \[ \int_{A \times \NN} \abs{F} \, d(\mu \times \nu) = \int_{\NN} \int_A \abs{F} \dmu \, d\nu = \sum_{n=1}^\infty \int_X \abs{f_n} \dmu < \infty. \]
<2>2. Hence \(F \in L^1(A \times \NN, \mu \times \nu)\).
<1>4. Apply Fubini’s theorem.
<2>1. Because \(F\) is integrable on the product space, Fubini’s theorem shows that \(\int_{\NN}\abs{F(x,n)}\,d\nu(n)=\sum_n\abs{f_n(x)}\) is finite for almost every \(x\) and that the iterated integrals agree: \[ \int_{\NN} \int_A F \dmu \, d\nu = \int_A \int_{\NN} F \, d\nu \dmu. \]
<2>2. Substituting the definitions of \(F\) and \(\nu\) into step <1>1 yields \[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_X \sum_{n=1}^\infty f_n \dmu. \]
The same argument, with counting measure on \(\NN\times\NN\), applies to a doubly indexed family \((f_{m,n})_{m,n\geq1}\) of measurable functions: the hypothesis \(\sum_{m,n} \int_X \abs{f_{m,n}}\dmu < \infty\) justifies interchanging the double sum with the integral.
Absolutely summable series in $L^1$ converge almost everywhere and in $L^1$ (Tag PR-EHIXY)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_k\in L^1(\mu)\) for \(k\geq1\) satisfy \(\sum_{k=1}^\infty \norm{f_k}_1 < \infty\). Then \(\sum_{k=1}^\infty f_k(x)\) converges absolutely for \(\mu\)-almost every \(x\in X\), its sum \(F\) lies in \(L^1(\mu)\), and \[ \norm{F-\sum_{k=1}^N f_k}_1 \convergesto{N\to\infty} 0, \qquad \int_X F\dmu = \sum_{k=1}^\infty \int_X f_k\dmu . \]
A series of functions with summable $L^1$ norms converges absolutely almost everywhere (Tag FR-KEWV2)
Let \((X,\mcm,\mu)\) be a measure space, let \(f_n\colon X\to\CC\) for \(n\geq1\) be measurable, and let \(S(x) \coloneqq \sum_{n\geq1} \abs{f_n(x)}\in[0,\infty]\). Then \[ \int_X S\dmu = \sum_{n\geq1} \norm{f_n}_1 . \] In particular, if \(\sum_{n\geq1} \norm{f_n}_1<\infty\), then \(S\in L^1(X,\mu)\) and \(\sum_{n\geq1} \abs{f_n(x)} < \infty\) for \(\mu\)-almost every \(x\in X\).
The partial sums \(S_N \coloneqq \sum_{n=1}^N \abs{f_n}\) are nonnegative, measurable, and increase pointwise to \(S\). By the monotone convergence theorem, \[ \int_X S\dmu = \lim_{N\to\infty} \int_X S_N\dmu = \lim_{N\to\infty} \sum_{n=1}^N \int_X \abs{f_n}\dmu = \sum_{n\geq1} \norm{f_n}_1. \] Now assume \(\sum_{n\geq1} \norm{f_n}_1<\infty\), so \(\int_X S\dmu<\infty\). Let \(A\coloneqq\theset{x\in X\suchthat S(x)=\infty}\), which is measurable. For every \(t>0\) we have \(S\geq t\,\one_A\), so \(t\,\mu(A)\le\int_X S\dmu\); letting \(t\to\infty\) gives \(\mu(A)=0\). Therefore \(\sum_{n\geq1} \abs{f_n(x)} < \infty\) for almost every \(x\).
Term-by-term integration of series of functions (Tag PR-NKZBT)
Let \((X,\mcm,\mu)\) be a measure space.
If \(f_n\in\) $L^+$ for \(n\geq1\), then \(\int_X\sum_n f_n\dmu=\sum_n\int_X f_n\dmu\) in \([0,\infty]\) [@Fol13].
If \(f_n\in L^1(\mu)\) for \(n\geq1\) and \(\sum_n\int_X\abs{f_n}\dmu<\infty\), then \(\sum_n f_n\) converges almost everywhere to a function in \(L^1(\mu)\) and \(\int_X\sum_n f_n\dmu=\sum_n\int_X f_n\dmu\) [@Fol13].
By (a) applied to \(\abs{f_n}\), the hypothesis \(\sum_n\int_X\abs{f_n}\dmu<\infty\) in (b) is the same as \(\int_X\sum_n\abs{f_n}\dmu<\infty\).
$L^1$-summable series converge almost everywhere and in $L^1$ (Tag PR-YJJSY)
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\in L^1(X,\mu)\) for \(n\geq1\) satisfy \[ \sum_{n=1}^\infty \norm{f_n}_1 < \infty. \] Then for almost every \(x\in X\) the series \(\sum_{n=1}^\infty f_n(x)\) converges absolutely, its sum \(f\), defined almost everywhere, lies in \(L^1(X,\mu)\), and \[ \norm{f-\sum_{n=1}^N f_n}_1 \longrightarrow 0 \quad\text{as } N\to\infty. \]
Continuity and differentiation under the integral sign (Tag PR-2CZUM)
Let \((X,\mcm,\mu)\) be a measure space, let \(I=[a,b]\subseteq\RR\) with \(a<b\), and let \(f\colon X\times I \to \CC\) be such that \(x\mapsto f(x,t)\) is integrable for each \(t\in I\). Put \[ F(t)\coloneqq\int_X f(x, t) \dmu(x), \qquad t\in I . \]
Suppose there exists \(g\in L^1(X,\mu)\) with \(\abs{f(x, t)} \leq g(x)\) for all \(x\in X\) and \(t\in I\). If \(t_0\in I\) and \(\lim_{t\to t_0}f(x,t)=f(x,t_0)\) for every \(x\in X\), then \(\lim_{t\to t_0}F(t)=F(t_0)\). In particular, if \(t\mapsto f(x,t)\) is continuous on \(I\) for every \(x\in X\), then \(F\) is continuous on \(I\).
Suppose the partial derivative \(\frac{\partial f}{\partial t}(x,t)\) exists for all \(x\in X\) and \(t\in I\), and there exists \(g\in L^1(X,\mu)\) with \(\abs{\frac{\partial f}{\partial t}(x, t)} \leq g(x)\) for all \(x\in X\) and \(t\in I\). Then \(F\) is differentiable on \(I\) and \[ F'(t) = \int_X \frac{\partial f}{\partial t}(x, t) \dmu(x) . \]