The Lebesgue Integral and the Convergence Theorems

For any extended-real-valued $h$, the identities $\{h<a\}={\bigcup }_{q\in \mathbb{Q},q<a}(X\setminus \{h>q\})$ and $\{h>a\}={\bigcup }_{q\in \mathbb{Q},q>a}(X\setminus \{h<q\})$ follow from rational density, including when $h(x)$ is infinite. Thus [L2] and [L3] show that measurability of all strict sublevels is equivalent to measurability of all strict superlevels.

Put $h=f{\chi }_E$ as defined in clause 4. For $a<0$, $\{h>a\}=X$, since $h\ge 0$. For $a\ge 0$, $\{h>a\}=E\cap \{f>a\}$. Both sets are measurable, including at $a=0$, so clause 4 follows.

For a pointwise-defined sum, $\{f+g>a\}={\bigcup }_{q\in \mathbb{Q}}(\{f>q\}\cap \{g>a-q\})$. Indeed, if both summands are finite and their sum exceeds $a$, choose a rational strictly between $a-g(x)$ and $f(x)$. If one summand is $+\mathrm{\infty }$, the other is not $-\mathrm{\infty }$, and a rational meeting the two inequalities still exists; if a summand is $-\mathrm{\infty }$, the defined sum cannot exceed $a$. The reverse inclusion follows by adding the inequalities. The union is countable and measurable, proving clause 1.

For $c>0$, $\{cf>a\}=\{f>a\mathrm{/}c\}$; for $c<0$, $\{cf>a\}=\{f<a\mathrm{/}c\}$. If $c=0$, each superlevel is either $X$ or $\mathrm{\varnothing }$. Step 1.1 and [L1] therefore prove clause 2.

Put $u={\inf }_n{f}_n$ and $v={\sup }_n{f}_n$. The defining order properties of infimum and supremum give $\{u<a\}={\bigcup }_n\{f_n<a\}$ and $\{v>a\}={\bigcup }_n\{f_n>a\}$, including infinite values. By step 1.1 and [L2] these sets are measurable, so $u$ and $v$ are measurable by [L1]. If $f_n\uparrow f$, then $f=v$ pointwise. This proves clause 5.

The function $-f$ is measurable by step 2.1. For any real-valued measurable $h$, the superlevel of $\max (h,0)$ is $X$ when $a<0$ and $\{h>a\}$ when $a\ge 0$. Apply this to $h=f$ and $h=-f$ to obtain measurable $f^{+}$ and $f^{-}$. They are finite-valued, so step 1.3 makes $\mathrm{\mid }f\mathrm{\mid }=f^{+}+f^{-}$ measurable, proving clause 3.

Clauses 1–5 follow respectively from steps 1.3, 2.1, 3.1, 1.2, and 2.2.

Set $s_0:=0$. For $n\ge 1$ and $0\le k<n{2}^n$, put $E_{n,k}:=\{x:k{2}^{-n}\le f(x)<(k+1)2^{-n}\}$ and set $s_n:={\sum }_{k=0}^{n{2}^n-1}k{2}^{-n}{\chi }_{E_{n,k}}+n{\chi }_{\{f\ge n\}}\mathrm{.}$ Each $E_{n,k}$ and $\{f\ge n\}$ is measurable by [L1], the range of $s_n$ is finite, and therefore each $s_n$ is simple by [L2]; $s_0$ is also simple.

For each $x$, one has $0\le s_n(x)\le f(x)$. If $f(x)<+\mathrm{\infty }$ and $n>f(x)$, then $f(x)-2^{-n}<s_n(x)\le f(x)$; if $f(x)=+\mathrm{\infty }$, then $s_n(x)=n$. Hence $s_n(x)\to f(x)$.

The functions are increasing. Indeed, $s_n(x)$ is the largest multiple of $2^{-n}$ at most $f(x)\wedge n$, hence is also a multiple of $2^{-(n+1)}$ at most $f(x)\wedge (n+1)$. The maximality of the latter dyadic truncation gives $s_n(x)\le s_{n+1}(x)$. Together with step 2.1, this proves $s_n\uparrow f$.

Complete both representations to partitions of $X$. [given, L1] Put $E_0=X\setminus {\bigcup }_{i=1}^m{E}_i$ and $F_0=X\setminus {\bigcup }_{j=1}^n{F}_j$, with coefficients $c_0=d_0=0$. Both are measurable. Adding these zero terms leaves the represented function and each coefficient sum unchanged, including when a complement has infinite measure, by the convention in [L1]. The augmented families $(E_i{)}_{i=0}^m$ and $(F_j{)}_{j=0}^n$ are finite measurable partitions of $X$.

Refine the two partitions by their intersections. [step 1.1] For $0\le i\le m$ and $0\le j\le n$ set $G_{ij}=E_i\cap F_j$. These sets are measurable and pairwise disjoint, and $E_i={⨆}_{j=0}^n{G}_{ij},F_j={⨆}_{i=0}^m{G}_{ij}\mathrm{.}$ On every nonempty $G_{ij}$ the two formulas give the same value of $s$, so $c_i=d_j$.

Apply finite additivity and the nonnegative extended-real finite-sum rules. [L1, L2, step 2.1] They give ${\sum }_{i=0}^m{c}_i\mu (E_i)={\sum }_{i=0}^m{\sum }_{j=0}^n{c}_i\mu (G_{ij})={\sum }_{i=0}^m{\sum }_{j=0}^n{d}_j\mu (G_{ij})={\sum }_{j=0}^n{d}_j\mu (F_j)\mathrm{.}$ For a zero coefficient, every product with an infinite measure is $0$ by [L1]; for a positive coefficient the usual extended-real distributivity applies. Thus no subtraction of infinities occurs.

Removing the added zero terms from step 3.1 proves equality of the original coefficient sums.

Complete the representations of $s$ and $t$ with their zero-valued complements. Take their finite measurable common refinement $(E_r)$. On each cell write $s=a_r$ and $t=b_r$. If $s\le t$, then $a_r\le b_r$.

The zero-scalar case is separate. [L2] When $c=0$, the function $cs$ is zero. Representing it by $0{\chi }_X$ gives $\int 0sd\mu =0$, even if $\mu (X)=+\mathrm{\infty }$, by the definition's local zero-times-infinity convention.

Monotonicity follows cell by cell. [step 1.1, L2] On the common partition, $\int sd\mu ={\sum }_r{a}_r\mu (E_r),\int td\mu ={\sum }_r{b}_r\mu (E_r)\mathrm{.}$ For finite or infinite $\mu (E_r)$, the local simple-integral convention makes $a_r\mu (E_r)\le b_r\mu (E_r)$ whenever $0\le a_r\le b_r$. Summing these nonnegative extended-real inequalities proves clause 1.

Additivity follows on the same partition. [step 1.1, L2] The coefficient of $s+t$ on $E_r$ is $a_r+b_r$, and $(a_r+b_r)\mu (E_r)=a_r\mu (E_r)+b_r\mu (E_r)$ under the local zero-times-infinity convention. Finite sums in $[0,+\mathrm{\infty }]$ can be regrouped without subtraction, so clause 3 follows.

For $c>0$, scalar multiplication holds cell by cell. [step 1.1, L2] The identity $(c{a}_r)\mu (E_r)=c(a_r\mu (E_r))$ is valid in $[0,+\mathrm{\infty }]$ for positive $c$, and finite summation gives $\int csd\mu =c\int sd\mu$. Together with the preceding cases, this proves all three clauses.

The function $s$ is one of its own admissible simple minorants, so [L1] gives $\int sd\mu \ge {\int }_{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}}sd\mu \mathrm{.}$

If $u$ is any admissible simple minorant of $s$, then $u\le s$, so [L2] gives ${\int }_{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}}ud\mu \le {\int }_{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}}sd\mu \mathrm{.}$ Taking the supremum over all such $u$ in [L1] yields the reverse inequality.

The two inequalities from steps 1.1 and 1.2 give equality of the nonnegative and simple integrals on $s$.

By [L4], choose a finite measurable partition $X={⨆}_{j=0}^m{E}_j$ on which $s=c_j\ge 0$, including its zero-valued complement. For every measurable $A$, the sets $A\cap E_j$ partition $A$, so [L1] and [L2] give ${\nu }_s(A)={\sum }_{j=0}^m{c}_j\mu (A\cap E_j)$. A term with $c_j=0$ is defined to be zero even when $\mu (A\cap E_j)=+\mathrm{\infty }$.

Step 1.1 gives ${\nu }_s(\mathrm{\varnothing })=0$. If $(A_n)$ is pairwise disjoint, then for each fixed $j$, the sets $(A_n\cap E_j)$ are pairwise disjoint. Countable additivity of $\mu$ and interchange of one finite sum with a nonnegative series give ${\nu }_s({\bigcup }_n{A}_n)={\sum }_{j=0}^m{c}_j{\sum }_n\mu (A_n\cap E_j)={\sum }_n{\sum }_{j=0}^m{c}_j\mu (A_n\cap E_j)={\sum }_n{\nu }_s(A_n)$. Zero-coefficient terms remain zero by the simple-integral convention.

Therefore ${\nu }_s$ satisfies the two conditions in [L3], so it is a measure.

If $f\le g$, every simple minorant of $f$ is also a simple minorant of $g$. Taking suprema in [L1] gives $\int fd\mu \le \int gd\mu$.

The zero function has just one nonnegative simple minorant: itself. [L1, L2, L3] Its simple integral is $0$ by [L3], so [L1] gives integral $0$ for the zero function, even on a space of infinite measure.

For $c>0$, multiplication by $c$ bijects simple minorants of $f$ with those of $cf$. The inverse divides by $c$ and preserves nonnegativity and simplicity. By [L2] and [L3], the corresponding simple integrals differ by the factor $c$. Multiplication by a positive finite real commutes with the supremum in $[0,+\mathrm{\infty }]$, including when that supremum is infinite. Hence $\int cfd\mu =c\int fd\mu$. Together with steps 1.1 and 1.2, this proves both clauses.

By [L1], the integrals $\int f_{n}d\mu$ increase and are bounded above by $\int fd\mu$. Write $L={\sup }_n\int f_{n}d\mu$, so $L\le \int fd\mu$ in $[0,+\mathrm{\infty }]$.

Fix a finite-valued nonnegative simple function $s\le f$ and $0<c<1$. Set $A_n=\{f_n\ge cs\}$. The sets $A_n$ increase to $X$: where $s=0$ membership is automatic, and where $s>0$, the limit $f\ge s>cs$ eventually forces $f_n\ge cs$. Since $A\mapsto {\int }_{A}sd\mu$ is a measure [L2], continuity from below [L3] gives ${\int }_{A_n}sd\mu \uparrow \int sd\mu$.

On $A_n$, $cs\le f_n$, hence $cs{\chi }_{A_n}\le f_n$ everywhere. By monotonicity [L1] and simple-integral agreement and homogeneity [L4], $c{\int }_{A_n}sd\mu \le \int f_{n}d\mu \le L$. Letting $n\to \mathrm{\infty }$ in step 1.2 yields $c\int sd\mu \le L$. Letting a fixed sequence $c_m\uparrow 1$ shows $\int sd\mu \le L$, also when the simple integral is infinite.

The inequality from step 2.1 holds for every admissible simple minorant $s\le f$. Taking their supremum, as in The nonnegative Lebesgue integral, gives $\int fd\mu \le L$. Combine this with step 1.1 to obtain $\int f_{n}d\mu \uparrow \int fd\mu$.

Choose simple functions $s_n\uparrow f$ and $t_n\uparrow g$ by [L1]. Then $s_n+t_n$ is simple for each $n$, and $s_n+t_n\uparrow f+g$ by [L2].

By [L3] and [L4], $\int (f+g)d\mu ={\lim }_n\int (s_n+t_n)d\mu ={\lim }_n(\int s_{n}d\mu +\int t_{n}d\mu )=\int fd\mu +\int gd\mu \mathrm{.}$ The last limit equality also holds when either limiting integral is infinite because addition is continuous for increasing sequences in $[0,\mathrm{\infty }]$.

Each partial sum $S_n$ is measurable by [L1], the sequence $(S_n)$ is [L1, given] increasing, and $S_n\uparrow S$ pointwise.

By [L2], $\int S_{n}d\mu ={\sum }_{k<n}\int f_{k}d\mu$ for every $n$. Applying [step 1.1, L2, L3, algebra] ∎ [L3] to step 1.1 gives $\int Sd\mu ={\lim }_n\int S_{n}d\mu ={\lim }_n{\sum }_{k<n}\int f_{k}d\mu ,$ which is exactly the displayed series identity.

One has ${\nu }_f(\mathrm{\varnothing })=0$. If $(A_n)$ is a pairwise disjoint sequence, put $B_n:={\bigcup }_{k<n}{A}_k$. Then ${\chi }_{B_n}\uparrow {\chi }_{{\bigcup }_k{A}_k}$, so $f{\chi }_{B_n}\uparrow f{\chi }_{{\bigcup }_k{A}_k}$, including where $f=+\mathrm{\infty }$ under the convention $0\cdot \mathrm{\infty }=0$. By [L2], ${\nu }_f({\bigcup }_k{A}_k)={\lim }_n{\nu }_f(B_n)$.

Because the sets $A_k$ are disjoint, $f{\chi }_{B_n}={\sum }_{k<n}f{\chi }_{A_k}$. Repeated use of [L3] gives ${\nu }_f(B_n)={\sum }_{k<n}{\nu }_f(A_k)$. Taking the increasing limit in step 1.1 proves countable additivity.

Steps 1.1 and 2.1 verify the two conditions in [L4], so ${\nu }_f$ is a measure.

Suppose first that $g={\sum }_{j=1}^m{c}_j{\chi }_{E_j}$ is simple with pairwise disjoint measurable $E_j$. By [L5] for the measure $fd\mu$ and then [L1], $\int gd(fd\mu )={\sum }_j{c}_j(fd\mu )(E_j)={\sum }_j{c}_j{\int }_{E_j}fd\mu$. Also $gf={\sum }_j{c}_{j}f{\chi }_{E_j}$ has pairwise disjoint summand supports, so [L2] and [L5] give $\int gfd\mu ={\sum }_j{c}_j{\int }_{E_j}fd\mu$. Hence $\int gd(fd\mu )=\int gfd\mu$.

For general measurable $g\ge 0$, choose simple $g_n\uparrow g$ by [L3]. Then $g_{n}f\uparrow gf$ pointwise (using $0\cdot \mathrm{\infty }=0$). Applying [L4] to both measures and step 1.1 to each $g_n$ yields $\int gd(fd\mu )={\lim }_n\int g_{n}d(fd\mu )={\lim }_n\int g_{n}fd\mu =\int gfd\mu$.

For each $n$, define $g_n:={\inf }_{k\ge n}{f}_k$. Then each $g_n$ is measurable by [L1], one has $g_n\le g_{n+1}$ and $g_n\uparrow {\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$ pointwise. Also $g_n\le f_n$ for every $n$.

By [L2] and step 1.1, $\int {\mathrm{lim\hspace{0.17em}inf}}_n{f}_{n}d\mu ={\lim }_n\int g_{n}d\mu$. For every fixed $n$ and all $k\ge n$, one has $g_n\le g_k\le f_k$. By [L3], $\int g_{n}d\mu \le {\inf }_{k\ge n}\int f_{k}d\mu$. Taking the supremum over $n$ and using monotone convergence on the left gives $\int {\mathrm{lim\hspace{0.17em}inf}}_n{f}_{n}d\mu \le {\mathrm{lim\hspace{0.17em}inf}}_n\int f_{n}d\mu$, including infinite values.

For each $m\ge 1$, put $g_m:=g\wedge m$ and $u_{n,m}:=f_n\wedge m$. Then $g_m$ and $u_{n,m}$ are measurable, $0\le u_{n,m}\le g_m\le m$, and $g_m-u_{n,m}$ is a nonnegative measurable function.

Apply [L1] to the sequence $g_m-u_{n,m}$. Since $g_m$ is finite-valued, ${\mathrm{lim\hspace{0.17em}inf}}_n(g_m-u_{n,m})=g_m-{\mathrm{lim\hspace{0.17em}sup}}_n{u}_{n,m},$ $\int (g_m-u_{n,m})d\mu =\int g_{m}d\mu -\int u_{n,m}d\mu ,$ and $\int (g_m-{\mathrm{lim\hspace{0.17em}sup}}_n{u}_{n,m})d\mu =\int g_{m}d\mu -\int {\mathrm{lim\hspace{0.17em}sup}}_n{u}_{n,m}d\mu \mathrm{.}$ Rearranging Fatou's inequality therefore gives ${\mathrm{lim\hspace{0.17em}sup}}_n\int u_{n,m}d\mu \le \int {\mathrm{lim\hspace{0.17em}sup}}_n{u}_{n,m}d\mu \le \int {\mathrm{lim\hspace{0.17em}sup}}_n{f}_{n}d\mu \mathrm{.}$

Since $0\le f_n-u_{n,m}\le g-g_m$, [L2] and [L3] give $\int f_{n}d\mu =\int u_{n,m}d\mu +\int (f_n-u_{n,m})d\mu \le \int u_{n,m}d\mu +\int (g-g_m)d\mu \mathrm{.}$ Taking ${\mathrm{lim\hspace{0.17em}sup}}_n$ and using step 2.1 yields ${\mathrm{lim\hspace{0.17em}sup}}_n\int f_{n}d\mu \le \int {\mathrm{lim\hspace{0.17em}sup}}_n{f}_{n}d\mu +\int (g-g_m)d\mu \mathrm{.}$

Because $g_m\uparrow g$, [L5] gives $\int g_{m}d\mu \uparrow \int gd\mu$. Applying [L2] to $g=(g-g_m)+g_m$ shows $\int (g-g_m)d\mu =\int gd\mu -\int g_{m}d\mu ⟶0.$ Letting $m\to \mathrm{\infty }$ in step 3.1 proves ${\mathrm{lim\hspace{0.17em}sup}}_n\int f_{n}d\mu \le \int {\mathrm{lim\hspace{0.17em}sup}}_n{f}_{n}d\mu \mathrm{.}$

Assume $\int fd\mu =0$. For $n\ge 1$ let $E_n:=\{f\ge 1\mathrm{/}n\}$. Then $(1\mathrm{/}n){\chi }_{E_n}\le f$, so [L1] and [L4] give $\frac{1}{n}\mu (E_n)=\int (1\mathrm{/}n){\chi }_{E_n}d\mu \le \int fd\mu =0$. Hence $\mu (E_n)=0$ for every $n$. Since $\{f>0\}={\bigcup }_n{E}_n$, the exceptional set where $f\mathrm{\ne }0$ is null, so $f=0$ almost everywhere by [L2].

Assume $f=0$ almost everywhere, and let $N$ be a measurable null set containing $\{f>0\}$. For any disjoint simple representation $s={\sum }_j{c}_j{\chi }_{E_j}\le f$, every $E_j$ with $c_j>0$ lies inside $N$ and has measure zero. Every zero-coefficient term contributes zero, including an omitted complement cell of infinite measure, by the simple integral's $0\cdot (+\mathrm{\infty })=0$ convention. Thus $\int sd\mu =0$. The now-complete finite refinement proof of The simple integral is independent of the chosen representation makes this value independent of representation. Taking the supremum over all simple minorants in [L3] gives $\int fd\mu =0$.

Step 1.1 proves the forward implication and step 1.2 proves the reverse implication.

Let $F:=\{f=+\mathrm{\infty }\}$, which is measurable. For every positive integer $n$, $n{\chi }_F\le f$. By [L1] and [L2], $n\mu (F)=\int n{\chi }_{F}d\mu \le \int fd\mu <+\mathrm{\infty }$.

If $\mu (F)>0$, the inequalities in step 1.1 fail for sufficiently large $n$; if $\mu (F)=+\mathrm{\infty }$, they fail already for $n=1$. Thus $\mu (F)=0$, so the exceptional set where $f$ is infinite is null. Equivalently, $f<+\mathrm{\infty }$ almost everywhere.