The Lebesgue Integral and the Convergence Theorems

If $f+g$ is defined pointwise, then for every real $a$, [L1, algebra] $\{f+g>a\}={\bigcup }_{q\in \mathbb{Q}}(\{f>q\}\cap \{g>a-q\}),$ so clause 1 follows from [L1].

For $c\mathrm{\ne }0$ one has $\{cf>a\}=\{f>a\mathrm{/}c\}$ when $c>0$ and [L1, algebra] $\{cf>a\}=\{f<a\mathrm{/}c\}$ when $c<0$; for $c=0$ the function is constant. Applying [L1] proves clause 2. The formulas $f^{+}=\max \{f,0\}$ and $f^{-}=\max \{-f,0\}$ therefore give clause 3.

If $E\in \mathcal{A}$ and $f\ge 0$, then [L1, algebra] ∎ $\{f{\chi }_E>a\}=E\cap \{f>a\}(a>0),$ so clause 4 follows from [L1]. Also $\{{\inf }_n{f}_n>a\}={\bigcap }_n\{f_n>a\},\{{\sup }_n{f}_n>a\}={\bigcup }_n\{f_n>a\},$ so clause 5 follows from [L1] as well, including the monotone-limit case $f={\sup }_n{f}_n$.

Set $s_0:=0$. For $n\ge 1$ and $0\le k<n{2}^n$, put [L1, L2, construct] $E_{n,k}:=\{x:k{2}^{-n}\le f(x)<(k+1)2^{-n}\}\cap \{f<n\},$ 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 [step 1.1, algebra] $n>f(x)$, 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 a dyadic multiple of [step 2.1, L3, algebra] ∎ $2^{-n}$ below $f(x)\wedge n$, hence also a dyadic multiple of $2^{-(n+1)}$ below $f(x)\wedge (n+1)$; so the defining maximality of the $(n+1)$-st dyadic truncation gives $s_n(x)\le s_{n+1}(x)$. Therefore $s_n\uparrow f$, in accord with [L3].

For each pair $(i,j)$ put $G_{ij}:=E_i\cap F_j$. The family $(G_{ij})$ is [given, L2] measurable and pairwise disjoint, and $E_i={⨆}_j{G}_{ij},F_j={⨆}_i{G}_{ij}\mathrm{.}$ Whenever $G_{ij}\mathrm{\ne }\mathrm{\varnothing }$, the two simple formulas for $s$ agree on $G_{ij}$, so $c_i=d_j$.

Finite additivity over the partitions in step 1.1 gives[step 1.1, L1, L2, algebra] ${\sum }_i{c}_i\mu (E_i)={\sum }_{i,j}{c}_i\mu (G_{ij})={\sum }_{i,j}{d}_j\mu (G_{ij})={\sum }_j{d}_j\mu (F_j)\mathrm{.}$ If some coefficient is $0$ on a cell of infinite measure, the convention in [L1] forces both corresponding terms to be $0$, so no ambiguity occurs there either.

Therefore the simple integral does not depend on the chosen representation, [step 2.1, L1] ∎ and the definition in [L1] is well defined.

Choose one finite measurable partition $(E_r)$ on which both $s$ and $t$ [L1, construct] are constant, say $s={\sum }_r{a}_r{\chi }_{E_r}$ and $t={\sum }_r{b}_r{\chi }_{E_r}$. Then $a_r\le b_r$ for every $r$ because $s\le t$.

Using the common partition from step 1.1 and [L2], [step 1.1, L2, algebra] $\int sd\mu ={\sum }_r{a}_r\mu (E_r),\int td\mu ={\sum }_r{b}_r\mu (E_r),\int csd\mu ={\sum }_{r}c{a}_r\mu (E_r),$ and $\int (s+t)d\mu ={\sum }_r(a_r+b_r)\mu (E_r)\mathrm{.}$ Termwise comparison gives monotonicity, while ordinary finite-sum algebra gives homogeneity and additivity.

Therefore all three properties hold for the simple integral.

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 are equalities, so the two [step 1.1, step 1.2] ∎ integrals agree on simple functions.

Write $s={\sum }_{j=1}^m{c}_j{\chi }_{E_j}$. Then for every measurable $A$,[L1, L2, given, algebra] ${\nu }_s(A)={\int }_{A}sd\mu ={\sum }_{j=1}^m{c}_j\mu (A\cap E_j)\mathrm{.}$

Step 1.1 gives ${\nu }_s(\mathrm{\varnothing })=0$. If $(A_n)$ is a pairwise disjoint[step 1.1, L2, L3, algebra] measurable sequence, then each $(A_n\cap E_j)$ is pairwise disjoint, so ${\nu }_s\text{\hspace{0.17em}⁣}\left({\bigcup }_n{A}_n\right)={\sum }_{j=1}^m{c}_j\mu \text{\hspace{0.17em}⁣}({\bigcup }_n(A_n\cap E_j))={\sum }_{j=1}^m{c}_j{\sum }_n\mu (A_n\cap E_j)={\sum }_n{\nu }_s(A_n)\mathrm{.}$

Therefore ${\nu }_s$ satisfies the two conditions in [L3], so it is a [step 2.1, L3] ∎ measure.

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

If $c=0$, both sides are $0$. Assume $c>0$. Multiplication by $c$ carries [L1, L2, given, algebra] simple minorants of $f$ bijectively onto simple minorants of $cf$, and [L2] scales their integrals by the same factor. Taking suprema in [L1] yields $\int cfd\mu =c\int fd\mu$.

Steps 1.1 and 1.2 prove the monotonicity and nonnegative homogeneity rules.

By [L1], the numbers $\int f_{n}d\mu$ increase and satisfy[given, L1] $\int f_{n}d\mu \le \int fd\mu$ for every $n$. So their supremum $L$ exists in $[0,+\mathrm{\infty }]$ and $L\le \int fd\mu$.

Fix a nonnegative simple function $s\le f$ and a real $c$ with $0<c<1$.[given, L2, L3] Put $A_n:=\{f_n\ge cs\}$. Then $A_n\uparrow X$: if $s(x)=0$ then $x\in A_n$ for all $n$, while if $s(x)>0$ then $f_n(x)\uparrow f(x)\ge s(x)>cs(x)$, so eventually $x\in A_n$. Since $A\mapsto {\int }_{A}sd\mu$ is a measure by [L2], [L3] gives ${\int }_{A_n}sd\mu \uparrow {\int }_{X}sd\mu \mathrm{.}$

On $A_n$ one has $cs\le f_n$, hence $cs{\chi }_{A_n}\le f_n$. By [L1] and [L4],[step 1.2, L1, L4, algebra] $c{\int }_{A_n}sd\mu =\int cs{\chi }_{A_n}d\mu \le \int f_{n}d\mu \le L\mathrm{.}$ Letting $n\to \mathrm{\infty }$ in step 1.2 yields $c\int sd\mu \le L\mathrm{.}$ Now choose $c_m=1-2^{-m}$ and let $m\to \mathrm{\infty }$; then $\int sd\mu \le L$.

Step 2.1 holds for every simple minorant $s\le f$, so taking the supremum [step 1.1, step 2.1, given] ∎ over such $s$ gives $\int fd\mu \le L$ by the definition of the nonnegative integral. Together with step 1.1, this proves $L=\int fd\mu$, so $\int f_{n}d\mu \uparrow \int fd\mu$.

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

By [L3] and [L4],[step 1.1, L3, L4, algebra] ∎ $\int (f+g)d\mu ={\lim }_n\int (s_n+t_n)d\mu ={\lim }_n(\int s_{n}d\mu +\int t_{n}d\mu )=\int fd\mu +\int gd\mu \mathrm{.}$

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,[L1, L2, given] put $B_n:={\bigcup }_{k<n}{A}_k$. Then ${\chi }_{B_n}\uparrow {\chi }_{{\bigcup }_k{A}_k}$, so $f{\chi }_{B_n}\uparrow f{\chi }_{{\bigcup }_k{A}_k}$ and [L2] gives ${\nu }_f\text{\hspace{0.17em}⁣}\left({\bigcup }_k{A}_k\right)={\lim }_n{\nu }_f(B_n)\mathrm{.}$

Because the sets $A_k$ are disjoint, repeated use of [L3] gives [step 1.1, L3, algebra] ${\nu }_f(B_n)={\sum }_{k<n}{\nu }_f(A_k)\mathrm{.}$ Substituting this into step 1.1 proves countable additivity.

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

Suppose first that $g={\sum }_{j=1}^m{c}_j{\chi }_{E_j}$ is a simple representation, so the sets $E_j$ are pairwise disjoint. Applying [L5] on the measure space $(X,\mathcal{A},fd\mu )$ and then using [L1], one gets.

For general measurable $g\ge 0$, choose simple $g_n\uparrow g$ by [L3]. Then $g_{n}f\uparrow gf$ pointwise. Applying [L4] twice and step 1.1 to each $g_n$ yields.

For each $n$, define [L1, given, construct] $g_n:={\inf }_{k\ge n}{f}_k\mathrm{.}$ 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], [step 1.1, L2, L3] ∎ $\int {\mathrm{lim\hspace{0.17em}inf}}_n{f}_{n}d\mu ={\lim }_n\int g_{n}d\mu \mathrm{.}$ Since $g_n\le f_n$, [L3] gives $\int g_{n}d\mu \le \int f_{n}d\mu$ for every $n$. Taking the limit in $n$ yields the claimed inequality.

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[L1, L2, given, algebra] $(1\mathrm{/}n){\chi }_{E_n}\le f$, so [L1] gives $\frac{1}{n}\mu (E_n)=\int (1\mathrm{/}n){\chi }_{E_n}d\mu \le \int fd\mu =0.$ 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[L2, L3, given] containing $\{f>0\}$. If $s={\sum }_j{c}_j{\chi }_{E_j}$ is a simple minorant of $f$, then every set $E_j$ with $c_j>0$ lies inside $\{f>0\}\subseteq N$, so $\mu (E_j)=0$; the remaining coefficients are $0$. Therefore $\int sd\mu =0$. 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 [step 1.1, step 1.2] ∎ implication.

Let $F:=\{f=+\mathrm{\infty }\}$. Then $F\subseteq \{f\ge n\}$ for every $n\ge 1$, so[L1, given] $n{\chi }_F\le f\mathrm{.}$ By [L1], $n\mu (F)=\int n{\chi }_{F}d\mu \le \int fd\mu <+\mathrm{\infty }\mathrm{.}$

If $\mu (F)>0$, the inequality in step 1.1 would fail for large $n$. [step 1.1, L2] ∎ Therefore $\mu (F)=0$, so the indicator ${\chi }_F$ has integral $0$ and hence vanishes almost everywhere by [L2]. Equivalently, $f<+\mathrm{\infty }$ almost everywhere.

Let $s={\sum }_j{c}_j{\chi }_{A_j}$ be a simple minorant of $f{\chi }_E$. If $c_j>0$,[L1, L2, given] then $A_j\subseteq E$, so $\mu (A_j)=0$. Since $A\mapsto {\int }_{A}sd\mu$ is a measure by [L1], every positive-coefficient term contributes $0$, and the zero-coefficient terms contribute $0$ as well. Hence $\int sd\mu =0$.

Taking the supremum over all such simple minorants in [L3] gives[step 1.1, L2, L3] ∎ ${\int }_{E}fd\mu =\int f{\chi }_{E}d\mu =0.$

Let $N$ be a measurable null set outside which $f_n\uparrow f$, and define $g_n:=f_n{\chi }_{X\setminus N}$ and $g:=f{\chi }_{X\setminus N}$.

Each difference $f_n-g_n=f_n{\chi }_N$ and $f-g=f{\chi }_N$ is supported on the null set $N$.

If $E:=\{f\ge t\}$, then $t{\chi }_E\le f$. Applying [L1] gives[L1, given] $t\mu (E)=\int t{\chi }_{E}d\mu \le \int fd\mu \mathrm{.}$

Dividing by the positive number $t$ yields[step 1.1, algebra] ∎ $\mu (\{f\ge t\})\le t^{-1}\int fd\mu \mathrm{.}$