Proof of The Lebesgue Integral and Null Sets

Throughout, an integral of an element of $\mathcal{M}^{+}$ is its integral as an element of $\mathcal{M}^{+}$, unless an element of $\mathcal{L}^{1}(\mu)$ is named. Every element of $[0,\infty]$ is $\ge0$, and $\le$ is a total order on $[-\infty,\infty]$ agreeing with that of $\mathbb{R}$ (Order and Suprema in the Extended Real Line (viii), Order and Suprema in the Extended Real Line (i)); these two facts are used without further mention.

Proof of (i). Choose $B\in\mathcal{F}$ with $N\subseteq B$ and $\mu(B)=0$ (Null Set of a Measure (i)). Let $s\in\mathcal{S}^{+}$ with $s(x)\le f(x)$ for every $x\in X$, and let $c\in s[X]$. If $c\neq0$, then every $x\in s^{-1}(\{c\})$ has $0<c\le f(x)$, so $f(x)\neq0$ and $x\in N\subseteq B$; hence, as $s^{-1}(\{c\})\in\mathcal{F}$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv)), $\mu(s^{-1}(\{c\}))\le\mu(B)=0$ (Basic Properties of a Measure (iii)), so $\mu(s^{-1}(\{c\}))=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)). In either case $c\cdot\mu(s^{-1}(\{c\}))=0$ (Addition and Multiplication on [0,∞] (ii)), and by the formula for $I_{\mu}$ in Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv) and Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (ii)

$$I_{\mu}(s)=\sum_{c\in s[X]}c\cdot\mu\bigl(s^{-1}(\{c\})\bigr)=0 .$$

So $0$ is an upper bound of the set $\Sigma_{f}$ of Nonnegative Measurable Maps and Their Lebesgue Integral (ii), whence $\int_{X}f\,d\mu=\sup\Sigma_{f}\le0$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)) and $\int_{X}f\,d\mu=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)).

Proof of (ii). Let $u(x)\le v(x)$ for $\mu$-almost every $x$. The set $X\setminus\{x\in X:u(x)\le v(x)\}$ is $\mu$-null (A Property Holding Almost Everywhere (i)); choose $B\in\mathcal{F}$ containing it with $\mu(B)=0$ (Null Set of a Measure (i)). As $\mathbf{1}_{B},\mathbf{1}_{X\setminus B}\in\mathcal{M}^{+}$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii)), the functions $u'=u\mathbf{1}_{X\setminus B}$ and $u''=u\mathbf{1}_{B}$ belong to $\mathcal{M}^{+}$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (iii)). At each $x$ one of $\mathbf{1}_{B}(x)$, $\mathbf{1}_{X\setminus B}(x)$ is $1$ and the other $0$, so

$$u=u'+u'',\qquad u'(x)=0\ (x\in B),\qquad u'(x)=u(x)\le v(x)\ (x\in X\setminus B)$$

(Indicator Function of a Subset (i), Sums, Multiples and Products of Functions into [0,∞] (iii), Sums, Multiples and Products of Functions into [0,∞] (i), Addition and Multiplication on [0,∞] (ii), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i)). As $u''$ vanishes off $B$ and $B$ is $\mu$-null (Null Sets and Properties Holding Almost Everywhere (i)), $\int_{X}u''\,d\mu=0$ by (i). Hence, by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i) and Basic Properties of the Integral of Nonnegative Measurable Maps (iii) (as $u'\le v$ everywhere),

$$\int_{X}u\,d\mu=\int_{X}u'\,d\mu+\int_{X}u''\,d\mu=\int_{X}u'\,d\mu\le\int_{X}v\,d\mu .$$

If $u(x)=v(x)$ for $\mu$-almost every $x$, then $u(x)\le v(x)$ and $v(x)\le u(x)$ for $\mu$-almost every $x$ (Null Sets and Properties Holding Almost Everywhere (vi)), and the two inequalities just proved give $\int_{X}u\,d\mu=\int_{X}v\,d\mu$.

Proof of (iii). The functions $f^{\pm}$ and $g^{\pm}$ belong to $\mathcal{M}^{+}$ (Integrable Functions and Their Lebesgue Integral (i)). Wherever $f(x)=g(x)$, also $f^{+}(x)=g^{+}(x)$ and $f^{-}(x)=g^{-}(x)$ (Positive Part, Negative Part and Absolute Value of a Real-Valued Function (i), Positive Part, Negative Part and Absolute Value of a Real-Valued Function (ii)), so $f^{\pm}=g^{\pm}$ holds $\mu$-almost everywhere (Null Sets and Properties Holding Almost Everywhere (vi)), and $\int_{X}f^{\pm}\,d\mu=\int_{X}g^{\pm}\,d\mu$ by (ii). These integrals are finite as $f\in\mathcal{L}^{1}(\mu)$, so the measurable function $g$ lies in $\mathcal{L}^{1}(\mu)$ (Integrable Functions and Their Lebesgue Integral (i), Integrable Functions and Their Lebesgue Integral (ii)), and by Integrable Functions and Their Lebesgue Integral (iii)

$$\int_{X}g\,d\mu=\int_{X}g^{+}\,d\mu-\int_{X}g^{-}\,d\mu=\int_{X}f^{+}\,d\mu-\int_{X}f^{-}\,d\mu=\int_{X}f\,d\mu .$$

Proof of (vi). The function $f$ is $\mathcal{F}/\mathcal{B}([0,\infty])$-measurable (Nonnegative Measurable Maps and Their Lebesgue Integral (i), Measurable Map, and the Set of Measurable Maps (vi)), and $A_{t}=f^{-1}(\{y\in[0,\infty]:t\le y\})$ with $\{y\in[0,\infty]:t\le y\}\in\mathcal{B}([0,\infty])$ (Borel Subsets of [0,∞]: Rays, the Point ∞, and Traces on ℝ (i)), so $A_{t}\in\mathcal{F}$ (Measurable Map, and the Set of Measurable Maps (i)). As $t$ is a real number $\ge0$, it lies in $[0,\infty]$ (The Extended Real Line, Its Order, and [0,∞] (ii)), and $t\mathbf{1}_{A_{t}}\in\mathcal{M}^{+}$ with $\int_{X}t\mathbf{1}_{A_{t}}\,d\mu=t\,\mu(A_{t})$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv)). Its value is $t\cdot1=t\le f(x)$ at $x\in A_{t}$ and $t\cdot0=0\le f(x)$ at $x\in X\setminus A_{t}$ (Sums, Multiples and Products of Functions into [0,∞] (ii), Indicator Function of a Subset (i), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii), Addition and Multiplication on [0,∞] (ii)). By Basic Properties of the Integral of Nonnegative Measurable Maps (iii),

$$t\,\mu(A_{t})=\int_{X}t\mathbf{1}_{A_{t}}\,d\mu\le\int_{X}f\,d\mu .$$

Proof of (iv): the set $P$. Let $P=\{x\in X:f(x)>0\}$. Then $P\in\mathcal{F}$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vi) with $a=0$, and $P=X\setminus\{x\in X:f(x)=0\}$, as $y>0$ if and only if $y\neq0$ for $y\in[0,\infty]$. So $f(x)=0$ for $\mu$-almost every $x$ if and only if $P$ is $\mu$-null (A Property Holding Almost Everywhere (i)), that is, if and only if $\mu(P)=0$ (Null Sets and Properties Holding Almost Everywhere (i)). If $\mu(P)=0$, then $P$ is $\mu$-null and $f$ vanishes off $P$, so $\int_{X}f\,d\mu=0$ by (i).

Proof of (iv): $\int_{X}f\,d\mu=0$ implies $\mu(P)=0$. Let $\int_{X}f\,d\mu=0$. For $n\in\mathbb{N}$, $n\in\mathbb{N}_{0}$, so $n$ is a real number (Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (i); The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i) with Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (v)); $1\le n$ in $\mathbb{R}$ and $0<1$, so $0<n$ (Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii), Rules of Arithmetic and Order in an Ordered Field (vii)); and $1/n=n^{-1}$ is a real number (Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (ii)) with $0<1/n$ (Rules of Arithmetic and Order in an Ordered Field (viii)). Let $(A_{n})_{n\in\mathbb{N}}$ be the sequence in $\mathcal{F}$ with $A_{n}=\{x\in X:1/n\le f(x)\}$ for every $n\in\mathbb{N}$, given by The Sets Where a Function Is at Least 1/n Form a Sequence of Measurable Sets (i). By (vi) with $t=1/n$, $A_{n}\in\mathcal{F}$ and

$$\tfrac{1}{n}\,\mu(A_{n})\le\int_{X}f\,d\mu=0,$$

so $\frac{1}{n}\,\mu(A_{n})=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)). As $1/n\neq0$, $\mu(A_{n})\neq\infty$, and the real product $\frac{1}{n}\,\mu(A_{n})$ vanishes, so $\mu(A_{n})=0$ (Addition and Multiplication on [0,∞] (ii); The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i) with Rules of Arithmetic and Order in an Ordered Field (i)). By Null Sets and Properties Holding Almost Everywhere (iv), $A=\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{F}$ and $\mu(A)=0$. Moreover $P\subseteq A$: let $x\in P$; if $f(x)=\infty$, then $1/1\le f(x)$ (Order and Suprema in the Extended Real Line (ii)) and $x\in A_{1}$; if $f(x)$ is real, then $0<f(x)$ and The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii) gives $n\in\mathbb{N}$ with $1/n<f(x)$, so $x\in A_{n}$ (Indexed Families of Sets and Their Union, Intersection and Product (ii)). Hence $\mu(P)\le\mu(A)=0$ (Basic Properties of a Measure (iii)), and $\mu(P)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)).

Proof of (v). The set $E=\{x\in X:f(x)=\infty\}=f^{-1}(\{\infty\})$ belongs to $\mathcal{F}$, as $\{\infty\}\in\mathcal{B}([0,\infty])$ and $f$ is $\mathcal{F}/\mathcal{B}([0,\infty])$-measurable (Borel Subsets of [0,∞]: Rays, the Point ∞, and Traces on ℝ (ii), Nonnegative Measurable Maps and Their Lebesgue Integral (i), Measurable Map, and the Set of Measurable Maps (vi), Measurable Map, and the Set of Measurable Maps (i)). The number $c=\int_{X}f\,d\mu$ is real (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)). Let $n\in\mathbb{N}$; then $n\in\mathbb{N}_{0}$, so $n$ is a real number (Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (i); The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i) with Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (v)), with $1\le n$, so $n\in[0,\infty]$ (Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii), The Extended Real Line, Its Order, and [0,∞] (ii)). The function $n\mathbf{1}_{E}$ lies in $\mathcal{M}^{+}$ with integral $n\,\mu(E)$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv)), and its value is $n\le\infty=f(x)$ at $x\in E$ and $0\le f(x)$ at $x\in X\setminus E$ (Sums, Multiples and Products of Functions into [0,∞] (ii), Indicator Function of a Subset (i), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii), Addition and Multiplication on [0,∞] (ii), Order and Suprema in the Extended Real Line (ii)). By Basic Properties of the Integral of Nonnegative Measurable Maps (iii),

$$n\,\mu(E)=\int_{X}n\mathbf{1}_{E}\,d\mu\le\int_{X}f\,d\mu=c\qquad(n\in\mathbb{N}).$$

Suppose $\mu(E)\neq0$. If $\mu(E)=\infty$, then $n=1$ gives $\infty=1\cdot\mu(E)\le c$ (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii)), so $c=\infty$ (Order and Suprema in the Extended Real Line (ii)), which is false. If $\mu(E)$ is real, then $\mu(E)>0$, and The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (ii) gives $n\in\mathbb{N}$ with $c<n\,\mu(E)$, the real product, which is that of $[0,\infty]$ (Addition and Multiplication on [0,∞] (ii)); this contradicts the display. Hence $\mu(E)=0$. Finally, $y<\infty$ fails for $y\in[0,\infty]$ exactly when $y=\infty$ (Order and Suprema in the Extended Real Line (ii)), so $X\setminus\{x\in X:f(x)<\infty\}=E$, which is $\mu$-null (Null Sets and Properties Holding Almost Everywhere (i)); that is, $f(x)<\infty$ for $\mu$-almost every $x$.

Proof of (vii), (viii) and (ix): an exceptional set. The set of $x$ at which $f_{m}(x)\to f(x)$ fails is $\mu$-null (A Property Holding Almost Everywhere (i)); choose $B_{0}\in\mathcal{F}$ containing it with $\mu(B_{0})=0$ (Null Set of a Measure (i)). For $m\in\mathbb{N}$, the function $g-|f_{m}|$ is measurable, as $g$ is (Integrable Functions and Their Lebesgue Integral (ii), Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)), so

$$E_{m}=\{x\in X:g(x)-|f_{m}(x)|<0\}=X\setminus\{x\in X:|f_{m}(x)|\le g(x)\}$$

belongs to $\mathcal{F}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v)) and is $\mu$-null by hypothesis, so $\mu(E_{m})=0$ (Null Sets and Properties Holding Almost Everywhere (i)). (The sequence $m\mapsto E_{m}$ is formed by Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (v) and Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (vi).) By Null Sets and Properties Holding Almost Everywhere (iv), $E=\bigcup_{m\in\mathbb{N}}E_{m}\in\mathcal{F}$ and $\mu(E)=0$; so $B=B_{0}\cup E\in\mathcal{F}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii)) is $\mu$-null and $\mu(B)=0$ (Null Sets and Properties Holding Almost Everywhere (iii), Null Sets and Properties Holding Almost Everywhere (i)). For every $x\in X\setminus B$, $f_{m}(x)\to f(x)$ and $|f_{m}(x)|\le g(x)$ for every $m$ (Indexed Families of Sets and Their Union, Intersection and Product (ii)).

Modified functions. Let $\chi=\mathbf{1}_{X\setminus B}$, a measurable function into $\mathbb{R}$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)), and let $\tilde f=\chi f$ and $\tilde f_{m}=\chi f_{m}$ for $m\in\mathbb{N}$, products of real-valued functions; they are measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)). (The sequence $(\tilde f_{m})_{m}$ is formed by Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iv) and Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii), from $\chi$, the sequence $(f_{m})$, the projections of Projection, Evaluation and Currying Maps (i) and the evaluation of Projection, Evaluation and Currying Maps (ii).) On $B$ all these functions vanish, and on $X\setminus B$ they agree with $f$ and $f_{m}$ (Indicator Function of a Subset (i), Pointwise Operations on Real-Valued Maps (i)). Hence $\tilde f_{m}(x)\to\tilde f(x)$ for every $x\in X$, the constant sequence $0$ converging to $0$ on $B$ (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i)). Moreover $|g|\in\mathcal{L}^{1}(\mu)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x)), and

$$|\tilde f_{m}(x)|\le|g(x)|\qquad(m\in\mathbb{N},\ x\in X),$$

the left side being $0\le|g(x)|$ for $x\in B$ and $|f_{m}(x)|\le g(x)\le|g(x)|$ for $x\in X\setminus B$ (Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ix); The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i) with Rules of Arithmetic and Order in an Ordered Field (x)).

Dominated convergence for the modified functions. By Dominated Convergence Theorem (i), Dominated Convergence Theorem (ii) and Dominated Convergence Theorem (iii), applied to $(\tilde f_{m})$, $\tilde f$ and $|g|$: $\tilde f$, every $\tilde f_{m}$ and every $|\tilde f_{m}-\tilde f|$ lie in $\mathcal{L}^{1}(\mu)$, and

$$\lim_{m\to\infty}\int_{X}|\tilde f_{m}-\tilde f|\,d\mu=0,\qquad\lim_{m\to\infty}\int_{X}\tilde f_{m}\,d\mu=\int_{X}\tilde f\,d\mu .$$

Return to the given functions. The functions $|f_{m}-f|$ are measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)). On $X\setminus B$ we have $\tilde f=f$, $\tilde f_{m}=f_{m}$ and $|\tilde f_{m}-\tilde f|=|f_{m}-f|$, so each of these equalities fails only on a subset of the $\mu$-null set $B$, which is $\mu$-null (Null Sets and Properties Holding Almost Everywhere (ii)). By (iii), $f$, every $f_{m}$ and every $|f_{m}-f|$ lie in $\mathcal{L}^{1}(\mu)$, with

$$\int_{X}f\,d\mu=\int_{X}\tilde f\,d\mu,\qquad\int_{X}f_{m}\,d\mu=\int_{X}\tilde f_{m}\,d\mu,\qquad\int_{X}|f_{m}-f|\,d\mu=\int_{X}|\tilde f_{m}-\tilde f|\,d\mu .$$

So the sequences $\bigl(\int_{X}f_{m}\,d\mu\bigr)_{m}$ and $\bigl(\int_{X}|f_{m}-f|\,d\mu\bigr)_{m}$ are those of the previous step, and converge to $\int_{X}f\,d\mu$ and to $0$. This proves (vii), (viii) and (ix).