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    Constructing the integral

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    Let \((X,\mathcal M,\mu)\) be a measure space. The integral is defined first for simple functions, then for nonnegative measurable functions as a supremum over simple functions below them, and then for measurable \(f\) with \(\int\abs f<\infty\) as \(\int f^+-\int f^-\). Linearity and monotonicity of the integral are proved in the same order: first for simple functions, then for nonnegative measurable functions, where additivity uses the monotone convergence theorem, then for integrable functions.

    Measurable and simple functions

    A function \(f\colon X\to\RR\) is measurable if \(f^{-1}(B)\in\mathcal M\) for every Borel set \(B\), equivalently if \(\theset{f>a}\in\mathcal M\) for every \(a\in\RR\). Sums, products, countable suprema and infima, and pointwise limits of sequences of measurable functions are measurable. A simple function is a measurable function with finitely many values, and \(\int\sum_i a_i\chi_{E_i} = \sum_i a_i\mu(E_i)\) for \(a_i\geq0\).

    Measurable function

    Let \((X, \mcm_X)\) and \((Y, \mcm_Y)\) be measurable spaces. A function \(f\colon X \to Y\) is \((\mcm_X, \mcm_Y)\)-measurable if \(\inverseof{f}(B)\in\mcm_X\) for every \(B\in\mcm_Y\). Let \(\mcb_\RR\) and \(\mcb_{\RR^d}\) denote the Borel \(\sigma\)-algebras and \(\mcl_{\RR^d}\) the \(\sigma\)-algebra of Lebesgue measurable subsets of \(\RR^d\).

    • A function \(f\colon X\to \RR\) on a measurable space \((X, \mcm_X)\) is measurable if it is \((\mcm_X, \mcb_\RR)\)-measurable.

    • A function \(f\colon \RR^d\to \RR\) is Borel measurable if it is \((\mcb_{\RR^d}, \mcb_\RR)\)-measurable.

    • A function \(f\colon \RR^d\to \RR\) is Lebesgue measurable if it is \((\mcl_{\RR^d}, \mcb_\RR)\)-measurable.

    Let \((X, \mcm_X)\) and \((Y, \mcm_Y)\) be measurable spaces, and let \(\mce\subseteq\mcm_Y\) be a family of sets that generates \(\mcm_Y\) as a \(\sigma\)-algebra. A function \(f\colon X\to Y\) is \((\mcm_X,\mcm_Y)\)-measurable if and only if \(\inverseof{f}(E)\in\mcm_X\) for every \(E\in\mce\).

    In particular, since \(\mcb_\RR\) is generated by each of the families of rays below, a function \(f\colon X\to\RR\) is measurable if and only if any one of the following holds for every \(\alpha \in \RR\):

    • \(\inverseof{f}((\alpha, \infty))\in \mcm_X\);

    • \(\inverseof{f}([\alpha, \infty))\in \mcm_X\);

    • \(\inverseof{f}((-\infty, \alpha))\in \mcm_X\);

    • \(\inverseof{f}((-\infty, \alpha])\in \mcm_X\).

    The forward implication holds because \(\mce\subseteq\mcm_Y\). Conversely, the family \(\theset{B\subseteq Y \suchthat \inverseof{f}(B)\in\mcm_X}\) is a \(\sigma\)-algebra, because preimages commute with complements and countable unions, and it contains \(\mce\); hence it contains the \(\sigma\)-algebra \(\mcm_Y\) generated by \(\mce\).

    Lebesgue and Borel measurable functions

    Let \(\mathcal{M}_L\) and \(\mathcal{M}_B\) be the \(\sigma\)-algebras of Lebesgue measurable and Borel subsets of \(\RR\), let \(E\in\mathcal{M}_L\), and let \(f\colon E \to [-\infty,\infty]\). The function \(f\) is Lebesgue measurable if \[ \theset{x \in E \suchthat f(x)>a}=f^{-1}((a, \infty]) \in \mathcal{M}_L \quad\text{for every } a\in\RR. \] When \(E\in\mathcal{M}_B\), the function \(f\) is Borel measurable if \(f^{-1}((a, \infty]) \in \mathcal{M}_B\) for every \(a\in\RR\).

    Which preimages must be measurable for $f$ to be a measurable function?

    Lebesgue measurable functions

    Let \(X\subseteq\RR^n\) be Lebesgue measurable and let \(f\colon X\to[-\infty,\infty]\). The function \(f\) is measurable if for every \(\alpha \in \RR\) the set \[ S_\alpha \coloneqq \theset{ x\in X \suchthat f(x) > \alpha} \] is Lebesgue measurable.

    Which sets must be Lebesgue measurable for $f$ to be a measurable function?

    Closure of measurable functions under algebraic operations and limits

    Measurability is preserved by sums, products, $\max$, $\min$, countable $\sup$, $\inf$, $\limsup$, $\liminf$, and pointwise limits.Suggest a slogan

    Let \((X,\mcm)\) be a measurable space [@Fol13].

    1. If \(f,g\colon X\to\RR\) are measurable, then \(f+g\), \(fg\), \(\max(f,g)\), and \(\min(f,g)\) are measurable.

    2. If \(f_n\colon X\to[-\infty,\infty]\) are measurable for \(n\geq1\), then \(\sup_n f_n\), \(\inf_n f_n\), \(\limsup_n f_n\), and \(\liminf_n f_n\) are measurable; if \(f(x)\coloneqq\lim_n f_n(x)\) exists for every \(x\in X\), then \(f\) is measurable.

    3. For \(E\subseteq X\), the characteristic function \(\chi_E\) is measurable if and only if \(E\in\mcm\).

    Simple function

    Let \((X,\mcm)\) be a measurable space. A simple function on \(X\) is a function \(s\colon X\to\CC\) of the form \[ s(x) = \sum_{j=1}^n c_j \chi_{E_j}(x), \] where \(n\geq 1\), \(c_1,\ldots,c_n\in\CC\), \(E_1,\ldots,E_n\in\mcm\), and \(\chi_E\) is the indicator function of \(E\).

    Approximation of measurable functions by simple functions

    A measurable function is the pointwise limit of simple functions with increasing moduli, and the convergence is uniform on every set where the function is bounded.Suggest a slogan

    Let \((X,\mcm)\) be a measurable space and let \(f\colon X\to\CC\) be measurable. There is a sequence \((\phi_n)\) of simple functions such that \(0\leq\abs{\phi_1}\leq\abs{\phi_2}\leq\cdots\leq\abs{f}\), \(\phi_n\to f\) pointwise on \(X\), and \(\phi_n\to f\) uniformly on every set on which \(f\) is bounded [@Fol13].

    The Lebesgue integral

    For measurable \(f\geq0\), \(\int f\coloneqq\sup\theset{\int\phi \st 0\leq\phi\leq f,\ \phi\text{ simple}}\). A measurable \(f\) is integrable if \(\int\abs f<\infty\), and then \(\int f\coloneqq\int f^+-\int f^-\) with \(f^\pm\coloneqq\max(\pm f,0)\).

    Lebesgue integral of a nonnegative function

    Let \((X,\mcm,\mu)\) be a measure space. For a nonnegative simple function \(s = \sum_{j=1}^n c_j \chi_{E_j}\) with \(c_j\in[0,\infty)\) and \(E_j\in\mcm\), put \[ \int_X s \dmu \coloneqq \sum_{j=1}^n c_j \mu(E_j), \] with the convention \(0\cdot\infty=0\). For \(f\in\) $L^+$, the Lebesgue integral of \(f\) is \[ \int_X f \dmu \coloneqq \sup \theset{ \int_X s \dmu \suchthat s \text{ simple and } 0\leq s \leq f }. \]

    Lebesgue integral of a nonnegative measurable function

    Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to[0,\infty]\) be measurable. The Lebesgue integral of \(f\) is \[ \int f \coloneqq \sup \theset{\int \phi \suchthat \phi \text{ a simple function with } 0 \leq \phi \leq f}, \] where a nonnegative simple function \(\phi=\sum_{j=1}^n c_j\chi_{E_j}\) has integral \(\int\phi\coloneqq\sum_{j=1}^n c_j\,\mu(E_j)\).

    How is the Lebesgue integral of $f$ defined from simple functions?

    Essential supremum and infimum, essentially bounded functions

    Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to\RR\) be measurable.

    • An essential lower bound of \(f\) is a real number \(b\) such that \(S_{b} \coloneqq \theset{x\in X\suchthat f(x) < b } = \inverseof{f}((-\infty, b))\) has measure zero. The essential infimum of \(f\) is the supremum of its essential lower bounds: \[ \ess\inf f \coloneqq \sup \theset{b\in\RR\suchthat \mu (S_b) = 0}. \]

    • An essential upper bound of \(f\) is a real number \(c\) such that \(T_c \coloneqq \theset{x\in X\suchthat f(x) > c} = \inverseof{f}((c, \infty))\) has measure zero. The essential supremum of \(f\) is the infimum of its essential upper bounds: \[ \ess\sup f \coloneqq \inf \theset{c\in\RR\suchthat \mu (T_c) = 0}. \]

    • A measurable function \(g\colon X\to\CC\) is essentially bounded if \(\norm{g}_\infty \coloneqq \ess\sup \abs{g} < \infty\).

    The essential infimum is the largest \(b\) with \(f\geq b\) almost everywhere, and the essential supremum is the smallest \(c\) with \(f\leq c\) almost everywhere, when these are finite. An essentially bounded function is bounded outside a set of measure zero.

    Essentially bounded functions agree almost everywhere with bounded functions

    An essentially bounded $f$ agrees almost everywhere with a bounded measurable $g$ with $\sup\abs{g}\le\norm{f}_\infty$.Suggest a slogan

    Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to\CC\) be measurable with \(\norm{f}_\infty<\infty\), that is, \(f\) is essentially bounded. Then there is a bounded measurable function \(g\colon X\to\CC\) with \(\sup_{x\in X}\abs{g(x)}\leq\norm{f}_\infty\) and \(f=g\) \(\mu\)-almost everywhere.

    $p$-tests for series and integrals

    $\sum k^{-p}$ and $\int_\varepsilon^\infty x^{-p}\,dx$ converge exactly for $p>1$ and $\int_0^1 x^{-p}\,dx$ exactly for $p<1$; in $\RR^n$ the threshold for $\abs{x}^{-p}$ is $p=n$.Suggest a slogan

    Let \(p\in\RR\), let \(\varepsilon>0\), let \(n\geq1\), and let \(B \coloneqq \theset{x\in \RR^n \suchthat \abs{x} \leq 1}\) be the closed unit ball for the Euclidean norm. Then \[ \begin{aligned} \sum_{k=1}^\infty \frac 1 {k^p} < \infty &\iff p>1, \\ \int_\varepsilon^\infty \frac {dx} {x^p} < \infty &\iff p>1, \\ \int_0^1 \frac {dx} {x^p} < \infty &\iff p<1, \\ \int_B \frac{dx}{\abs{x}^p} < \infty &\iff p < n, \\ \int_{\RR^n\setminus B} \frac{dx}{\abs{x}^p} < \infty &\iff p > n . \end{aligned} \]

    The $p$-test for $\int_0^1 x^{-p}\,dx$ and $\int_1^\infty x^{-p}\,dx$

    The threshold is $p=1$: below it $x^{-p}$ is integrable at $0$, above it at $\infty$.Suggest a slogan

    For \(p\in\RR\), \[ \begin{aligned} \int_0^1 \frac{dx}{x^p} < \infty &\iff p < 1, \\ \int_1^\infty \frac{dx}{x^p} < \infty &\iff p > 1. \end{aligned} \]

    For \(0<a<b\), \(\int_a^b x^{-p}\dx\) equals \(\frac{b^{1-p} - a^{1-p}}{1-p}\) if \(p\neq 1\) and \(\log b - \log a\) if \(p = 1\). Letting \(a\to 0^+\) with \(b = 1\) gives a finite limit exactly when \(1-p>0\); letting \(b\to\infty\) with \(a = 1\) gives a finite limit exactly when \(1-p<0\).

    For which $p$ do $\int_0^1 x^{-p}$ and $\int_1^\infty x^{-p}$ converge?

    Integrability of $1/x^p$ on $(0,1)$ and $(1,\infty)$

    $x^{-p}$ is integrable near $0$ exactly for $p<1$, and near $\infty$ exactly for $p>1$.Suggest a slogan

    Let \(p\in\RR\) and let \(f(x)\coloneqq x^{-p}\) for \(x>0\).

    • \(f\in L^1((0, 1))\) if and only if \(p < 1\).

    • \(f\in L^1((1, \infty))\) if and only if \(p > 1\).

    For \(0<a<b\), \(\int_a^b x^{-p}\dx\) equals \(\frac{b^{1-p} - a^{1-p}}{1-p}\) if \(p\neq 1\) and \(\log b - \log a\) if \(p = 1\). By monotone convergence, \(\int_{(0,1)}x^{-p}\dx = \lim_{a\to 0^+}\int_a^1 x^{-p}\dx\), which is finite exactly when \(1-p>0\), and \(\int_{(1,\infty)}x^{-p}\dx = \lim_{b\to\infty}\int_1^b x^{-p}\dx\), which is finite exactly when \(1-p<0\).

    Where is ${1\over x^p}$ integrable in $\RR$? (Depending on $p$)

    Integrability of $x^p$ on $(0,1)$ and $(1,\infty)$

    $x^p$ is integrable near $0$ exactly for $p>-1$, and near $\infty$ exactly for $p<-1$.Suggest a slogan

    Let \(p\in\RR\) and let \(f(x)\coloneqq x^{p}\) for \(x>0\).

    • \(f\in L^1((0, 1))\) if and only if \(p > -1\).

    • \(f\in L^1((1, \infty))\) if and only if \(p < -1\).

    For \(0<a<b\), \(\int_a^b x^{p}\dx\) equals \(\frac{b^{p+1} - a^{p+1}}{p+1}\) if \(p\neq -1\) and \(\log b - \log a\) if \(p = -1\). By monotone convergence, \(\int_{(0,1)}x^{p}\dx = \lim_{a\to 0^+}\int_a^1 x^{p}\dx\), which is finite exactly when \(p+1>0\), and \(\int_{(1,\infty)}x^{p}\dx = \lim_{b\to\infty}\int_1^b x^{p}\dx\), which is finite exactly when \(p+1<0\).

    Where is $x^p$ integrable in $\RR$? (Depending on $p$)

    Changing \(f\) on a null set changes neither \(\int f\) nor its essential supremum, so elements of \(L^p\) are equivalence classes of functions equal almost everywhere. The integrability of \(x^a\) on \((0,1)\) exactly when \(a>-1\) and on \((1,\infty)\) exactly when \(a<-1\) is the comparison used for singularities at a point and for decay at infinity.

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