Proof of Integrals over Measurable Subsets

Throughout, for $A\in\mathcal{F}$, $\mathcal{F}_{A}=\{B\in\mathcal{F}:B\subseteq A\}$ and $\mu_{A}(B)=\mu(B)$ for $B\in\mathcal{F}_{A}$ (Integrals of Functions over a Measurable Subset (i)); for a function $f$ on $X$, $f|_{A}$ is its restriction to $A$; $\mathcal{S}^{+}_{A}=\mathcal{S}^{+}(A,\mathcal{F}_{A})$ and $I_{\mu_{A}}$ are the nonnegative simple functions and the simple integral of $(A,\mathcal{F}_{A},\mu_{A})$; and for $g\in\mathcal{M}^{+}$, $\Sigma_{g}$ is the set of Nonnegative Measurable Maps and Their Lebesgue Integral (ii), so that $\int_{X}g\,d\mu=\sup\Sigma_{g}$, and $\Sigma^{A}_{k}$ is the same set for $k\in\mathcal{M}^{+}(A,\mathcal{F}_{A})$ and $\mu_{A}$. Every element of $[0,\infty]$ is $\ge0$, and $\le$ is a total order on $[-\infty,\infty]$ (Order and Suprema in the Extended Real Line (viii), Order and Suprema in the Extended Real Line (i)).

Step 1: the functions $f\mathbf{1}_{A}$. Let $A\in\mathcal{F}$. Then $\mathbf{1}_{A}\in\mathcal{S}^{+}$ (Properties of Simple Functions and of the Simple Integral (i)), so $\mathbf{1}_{A}$ is measurable as a function into $\mathbb{R}$ and into $[0,\infty]$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (ii), Basic Properties of the Integral of Nonnegative Measurable Maps (i)). For a function $f$ on $X$ into $[0,\infty]$ or into $\mathbb{R}$,

$$(f\mathbf{1}_{A})(x)=f(x)\quad(x\in A),\qquad(f\mathbf{1}_{A})(x)=0\quad(x\in X\setminus A),$$

as $y\cdot1=y$ and $y\cdot0=0$ in both (Indicator Function of a Subset (i), Sums, Multiples and Products of Functions into [0,∞] (iii), Pointwise Operations on Real-Valued Maps (i), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii), Addition and Multiplication on [0,∞] (ii)). If $f\in\mathcal{M}^{+}$, then $f\mathbf{1}_{A}\in\mathcal{M}^{+}$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (iii)); if $f:X\to\mathbb{R}$ is measurable, so is $f\mathbf{1}_{A}$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)).

Step 2: measurability of zero extensions. Let $A\in\mathcal{F}$, let $Y$ be $\mathbb{R}$ or $[0,\infty]$ and $\mathcal{G}$ accordingly $\mathcal{B}(\mathbb{R})$ or $\mathcal{B}([0,\infty])$, so that measurability as a function into $Y$ is $\mathcal{G}$-measurability (Measurable Map, and the Set of Measurable Maps (iii), Measurable Map, and the Set of Measurable Maps (vi)), and let $g:A\to Y$ and $\tilde g:X\to Y$ be as in (iv). Then $\tilde g|_{A}=g$, as both have the same values (Domains and Restrictions of Maps (iii)). If $\tilde g$ is $\mathcal{F}/\mathcal{G}$-measurable, then $g=\tilde g|_{A}$ is $\mathcal{F}_{A}/\mathcal{G}$-measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (viii)). Conversely, let $g$ be $\mathcal{F}_{A}/\mathcal{G}$-measurable and $C\in\mathcal{G}$. Then

$$\tilde g^{-1}(C)=g^{-1}(C)\quad(0\notin C),\qquad\tilde g^{-1}(C)=g^{-1}(C)\cup(X\setminus A)\quad(0\in C),$$

with $g^{-1}(C)\in\mathcal{F}_{A}\subseteq\mathcal{F}$ and $X\setminus A\in\mathcal{F}$, so $\tilde g^{-1}(C)\in\mathcal{F}$ (Measurable Map, and the Set of Measurable Maps (i), σ-Algebra and Measurable Space (i), Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii)), and $\tilde g$ is $\mathcal{F}/\mathcal{G}$-measurable.

Step 3: simple functions vanishing off $A$. Let $A\in\mathcal{F}$.

(a) Let $s\in\mathcal{S}^{+}$ vanish on $X\setminus A$. Then $s|_{A}\in\mathcal{S}^{+}_{A}$ and $I_{\mu}(s)=I_{\mu_{A}}(s|_{A})$. Indeed, $s|_{A}$ has values $\ge0$, is measurable for $\mathcal{F}_{A}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (viii)), and its image $E=s|_{A}[A]$ is a subset of the finite set $C=s[X]$, hence finite (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv)); so $s|_{A}\in\mathcal{S}^{+}_{A}$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (ii), Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iii)). By Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv),

$$I_{\mu}(s)=\sum_{c\in C}c\cdot\mu\bigl(s^{-1}(\{c\})\bigr),\qquad I_{\mu_{A}}(s|_{A})=\sum_{c\in E}c\cdot\mu_{A}\bigl((s|_{A})^{-1}(\{c\})\bigr).$$

Each $c\in C\setminus E$ is a value of $s$ off $A$, so $c=0$ and its summand is $0$ (Addition and Multiplication on [0,∞] (ii)); so the first sum may be taken over $E$ (Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (i)). For $c\in E$, the two summands are both $0$ if $c=0$; if $c\neq0$, then $s^{-1}(\{c\})\subseteq A$, so $s^{-1}(\{c\})=(s|_{A})^{-1}(\{c\})\in\mathcal{F}_{A}$, on which $\mu_{A}$ and $\mu$ agree. So the two sums agree term by term, and $I_{\mu}(s)=I_{\mu_{A}}(s|_{A})$.

(b) Let $t\in\mathcal{S}^{+}_{A}$, and let $\tilde t:X\to\mathbb{R}$ be the function equal to $t$ on $A$ and to $0$ on $X\setminus A$ (Maps Defined by Cases (i)). Its values are $\ge0$, and its image lies in the finite set $t[A]\cup\{0\}$, so it is finite (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv)); by Step 2, $\tilde t$ is measurable and $\tilde t|_{A}=t$. So $\tilde t\in\mathcal{S}^{+}$, and $\tilde t$ vanishes off $A$.

Proof of (i). By Step 1, $f\mathbf{1}_{A}\in\mathcal{M}^{+}$; and $f|_{A}\in\mathcal{M}^{+}(A,\mathcal{F}_{A})$ (Integrals of Functions over a Measurable Subset (ii)). We show $\Sigma_{f\mathbf{1}_{A}}=\Sigma^{A}_{f|_{A}}$. Let $v=I_{\mu}(s)\in\Sigma_{f\mathbf{1}_{A}}$, with $s\in\mathcal{S}^{+}$ and $s\le f\mathbf{1}_{A}$. Off $A$, $0\le s(x)\le0$, so $s(x)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)); by Step 3(a), $v=I_{\mu_{A}}(s|_{A})$ with $s|_{A}\in\mathcal{S}^{+}_{A}$ and $s|_{A}\le f|_{A}$, so $v\in\Sigma^{A}_{f|_{A}}$. Conversely, let $v=I_{\mu_{A}}(t)\in\Sigma^{A}_{f|_{A}}$, with $t\in\mathcal{S}^{+}_{A}$ and $t\le f|_{A}$. With $\tilde t$ as in Step 3(b), $I_{\mu}(\tilde t)=I_{\mu_{A}}(t)=v$ by Step 3(a), and $\tilde t\le f\mathbf{1}_{A}$ (equal to $t\le f$ on $A$, and $0$ off $A$), so $v\in\Sigma_{f\mathbf{1}_{A}}$. By Nonnegative Measurable Maps and Their Lebesgue Integral (ii) and Integrals of Functions over a Measurable Subset (ii),

$$\int_{X}f\mathbf{1}_{A}\,d\mu=\sup\Sigma_{f\mathbf{1}_{A}}=\sup\Sigma^{A}_{f|_{A}}=\int f|_{A}\,d\mu_{A}=\int_{A}f\,d\mu .$$

Proof of (ii). Let $h=f\mathbf{1}_{A}$, measurable by Step 1, and note $f|_{A}$ is measurable for $\mathcal{F}_{A}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (viii)). The functions $|f|$, $f^{\pm}$, $|h|$, $h^{\pm}$ lie in $\mathcal{M}^{+}$, and $|f|_{A}|$, $(f|_{A})^{\pm}$ in $\mathcal{M}^{+}(A,\mathcal{F}_{A})$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)). By Step 1 and Positive Part, Negative Part and Absolute Value of a Real-Valued Function (iii), Positive Part, Negative Part and Absolute Value of a Real-Valued Function (i) and Positive Part, Negative Part and Absolute Value of a Real-Valued Function (ii), the values give

$$|h|=|f|\mathbf{1}_{A},\quad h^{\pm}=f^{\pm}\mathbf{1}_{A},\qquad|f|_{A}|=|f|\,|_{A},\quad(f|_{A})^{\pm}=f^{\pm}|_{A},$$

the products being those of functions into $[0,\infty]$. So by Integrals of Functions over a Measurable Subset (ii) and (i),

$$\int|f|_{A}|\,d\mu_{A}=\int_{A}|f|\,d\mu=\int_{X}|h|\,d\mu,\qquad\int(f|_{A})^{\pm}\,d\mu_{A}=\int_{A}f^{\pm}\,d\mu=\int_{X}h^{\pm}\,d\mu .\qquad(\ast)$$

Now $f$ is integrable on $A$ if and only if $f|_{A}\in\mathcal{L}^{1}(A,\mathcal{F}_{A},\mu_{A})$ (Integrals of Functions over a Measurable Subset (iii)), that is, $\int|f|_{A}|\,d\mu_{A}<\infty$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv) for $(A,\mathcal{F}_{A},\mu_{A})$); by $(\ast)$ this is $\int_{X}|h|\,d\mu<\infty$, that is, $h\in\mathcal{L}^{1}(\mu)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv)). In that case, by Integrable Functions and Their Lebesgue Integral (iii) for $\mu$ and $\mu_{A}$, $(\ast)$ and Integrals of Functions over a Measurable Subset (iv),

$$\int_{X}h\,d\mu=\int_{X}h^{+}\,d\mu-\int_{X}h^{-}\,d\mu=\int(f|_{A})^{+}\,d\mu_{A}-\int(f|_{A})^{-}\,d\mu_{A}=\int f|_{A}\,d\mu_{A}=\int_{A}f\,d\mu .$$

Proof of (iii). As $f$ is measurable into $\mathbb{R}$ with values $\ge0$, $f\in\mathcal{M}^{+}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii), Nonnegative Measurable Maps and Their Lebesgue Integral (i)). The function $f|_{A}$ is measurable for $\mathcal{F}_{A}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (viii)) with values $\ge0$. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi) for $(A,\mathcal{F}_{A},\mu_{A})$, $f|_{A}\in\mathcal{L}^{1}(A,\mathcal{F}_{A},\mu_{A})$ if and only if its integral as an element of $\mathcal{M}^{+}(A,\mathcal{F}_{A})$ is finite, and then that integral is a real number equal to its integral as an element of $\mathcal{L}^{1}(A,\mathcal{F}_{A},\mu_{A})$. By Integrals of Functions over a Measurable Subset (ii), Integrals of Functions over a Measurable Subset (iii) and Integrals of Functions over a Measurable Subset (iv), these give the remaining three assertions of (iii).

Proof of (iv). By Step 2, $g$ is measurable for $\mathcal{F}_{A}$ if and only if $\tilde g$ is measurable for $\mathcal{F}$, and $\tilde g|_{A}=g$. Let both be measurable. As $\tilde g$ vanishes off $A$, Step 1 gives $\tilde g\mathbf{1}_{A}=\tilde g$. If $g$ is $[0,\infty]$-valued, then $\tilde g\in\mathcal{M}^{+}$ and, by (i) and Integrals of Functions over a Measurable Subset (ii),

$$\int_{X}\tilde g\,d\mu=\int_{X}\tilde g\mathbf{1}_{A}\,d\mu=\int_{A}\tilde g\,d\mu=\int\tilde g|_{A}\,d\mu_{A}=\int g\,d\mu_{A}.$$

If $g$ is real-valued, then by (ii) for $\tilde g$, $\tilde g=\tilde g\mathbf{1}_{A}\in\mathcal{L}^{1}(\mu)$ if and only if $\tilde g$ is integrable on $A$, that is, $g=\tilde g|_{A}\in\mathcal{L}^{1}(A,\mathcal{F}_{A},\mu_{A})$ (Integrals of Functions over a Measurable Subset (iii)); and then the same chain of equalities holds by (ii) and Integrals of Functions over a Measurable Subset (iv).

Proof of (v). Every member of $\mathcal{F}$ is a subset of $X$ (σ-Algebra and Measurable Space (i)), so $\mathcal{F}_{X}=\mathcal{F}$, $\mu_{X}=\mu$ and $f|_{X}=f$ for every function $f$ on $X$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iii)); that is, $(X,\mathcal{F}_{X},\mu_{X})=(X,\mathcal{F},\mu)$. The three assertions are then Integrals of Functions over a Measurable Subset (ii), Integrals of Functions over a Measurable Subset (iii) and Integrals of Functions over a Measurable Subset (iv) with $A=X$.

Proof of (vi). Let $f\in\mathcal{L}^{1}(\mu)$ and $A\in\mathcal{F}$; $f$ is measurable (Integrable Functions and Their Lebesgue Integral (ii)). By Step 1, $|f\mathbf{1}_{A}|\le|f|$ pointwise, so by Basic Properties of the Integral of Nonnegative Measurable Maps (iii) and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv),

$$\int_{X}|f\mathbf{1}_{A}|\,d\mu\le\int_{X}|f|\,d\mu<\infty ,$$

so $f\mathbf{1}_{A}\in\mathcal{L}^{1}(\mu)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv)), and $f$ is integrable on $A$ by (ii).

Proof of (vii). First, $A\cup B\in\mathcal{F}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii)). As $A\cap B=\emptyset$, $\mathbf{1}_{A\cup B}=\mathbf{1}_{A}+\mathbf{1}_{B}$, checked at points of $A$, of $B$ and of $X\setminus(A\cup B)$ (Indicator Function of a Subset (i), Addition and Multiplication on [0,∞] (i)). Hence $f\mathbf{1}_{A\cup B}=f\mathbf{1}_{A}+f\mathbf{1}_{B}$, for $f\in\mathcal{M}^{+}$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iii) and Sums, Multiples and Products of Functions into [0,∞] (i), and for real-valued $f$ by Pointwise Operations on Real-Valued Maps (i). For $f\in\mathcal{M}^{+}$, by (i) and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i),

$$\int_{A\cup B}f\,d\mu=\int_{X}f\mathbf{1}_{A\cup B}\,d\mu=\int_{X}f\mathbf{1}_{A}\,d\mu+\int_{X}f\mathbf{1}_{B}\,d\mu=\int_{A}f\,d\mu+\int_{B}f\,d\mu .$$

For $f\in\mathcal{L}^{1}(\mu)$, $f$ is integrable on $A$, $B$ and $A\cup B$ ((vi)), so $f\mathbf{1}_{A}$, $f\mathbf{1}_{B}$, $f\mathbf{1}_{A\cup B}\in\mathcal{L}^{1}(\mu)$ with integrals $\int_{A}f\,d\mu$, $\int_{B}f\,d\mu$, $\int_{A\cup B}f\,d\mu$ ((ii)); and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii) with $a=b=1$ gives the same chain of equalities.

Proof of (viii). As $A\subseteq B$, $\mathbf{1}_{A}\le\mathbf{1}_{B}$ (Indicator Function of a Subset (i)), so $f\mathbf{1}_{A}\le f\mathbf{1}_{B}$ (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iv), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii)). By (i) and Basic Properties of the Integral of Nonnegative Measurable Maps (iii),

$$\int_{A}f\,d\mu=\int_{X}f\mathbf{1}_{A}\,d\mu\le\int_{X}f\mathbf{1}_{B}\,d\mu=\int_{B}f\,d\mu .$$

Proof of (ix). For $B\in\mathcal{F}_{A}$, $\mu_{A}(B)=\mu(B)\le\mu(A)=0$ (Basic Properties of a Measure (iii)), so $\mu_{A}(B)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)). Let $f\in\mathcal{M}^{+}$ and $v=I_{\mu_{A}}(t)\in\Sigma^{A}_{f|_{A}}$ with $t\in\mathcal{S}^{+}_{A}$. Every summand of

$$v=\sum_{c\in t[A]}c\cdot\mu_{A}\bigl(t^{-1}(\{c\})\bigr)$$

is $c\cdot0=0$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv), Addition and Multiplication on [0,∞] (ii)), so $v=0$ (Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (ii)). So $0$ is an upper bound of $\Sigma^{A}_{f|_{A}}$, and $\int_{A}f\,d\mu=\sup\Sigma^{A}_{f|_{A}}\le0$ (Integrals of Functions over a Measurable Subset (ii), Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), whence $\int_{A}f\,d\mu=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)).

Let $f:X\to\mathbb{R}$ be measurable. As in the proof of (ii), $|f|,f^{\pm}\in\mathcal{M}^{+}$, $|f|_{A}|=|f|\,|_{A}$ and $(f|_{A})^{\pm}=f^{\pm}|_{A}$. By the first part, $\int|f|_{A}|\,d\mu_{A}=\int_{A}|f|\,d\mu=0<\infty$, so $f|_{A}\in\mathcal{L}^{1}(A,\mathcal{F}_{A},\mu_{A})$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv)), that is, $f$ is integrable on $A$ (Integrals of Functions over a Measurable Subset (iii)). By Integrals of Functions over a Measurable Subset (iv), Integrable Functions and Their Lebesgue Integral (iii) and the first part,

$$\int_{A}f\,d\mu=\int(f|_{A})^{+}\,d\mu_{A}-\int(f|_{A})^{-}\,d\mu_{A}=\int_{A}f^{+}\,d\mu-\int_{A}f^{-}\,d\mu=0-0=0 .$$