Measurable Functions and Simple Approximation

If $E\in \mathcal{A}$, then the preimage of any Borel set $B\subseteq \mathbb{R}$ under ${\mathbf{1}}_E$ is one of $\mathrm{\varnothing }$, $X$, $E$, or $X\setminus E$, because ${\mathbf{1}}_E$ takes only the values $0$ and $1$. Each of those sets lies in $\mathcal{A}$, so ${\mathbf{1}}_E$ is measurable by [L1].

If ${\mathbf{1}}_E$ is measurable, then $E={\mathbf{1}}_E^{-1}((1\mathrm{/}2,\mathrm{\infty }))$. The interval $(1\mathrm{/}2,\mathrm{\infty })$ is Borel, so [L1] gives $E\in \mathcal{A}$.

Steps 1.1 and 1.2 prove the equivalence.

Let

By hypothesis, every member of $\mathcal{G}$ lies in $\mathcal{C}$. Since [given, step 1.1] $\mathcal{B}$ is the sigma-algebra generated by $\mathcal{G}$, one has $\mathcal{B}\subseteq \mathcal{C}$.

Therefore $f^{-1}(B)\in \mathcal{A}$ for every $B\in \mathcal{B}$, and [step 2.1, L1] [L1] says exactly that $f$ is measurable.

By [L1], the threshold set $\{x:f(x)>a\}$ is the preimage [L1, L2] $f^{-1}((a,+\mathrm{\infty }])$. Therefore [L2] gives the equivalence of condition 1 and condition 2.

Suppose the sets $\{f>q\}$ are measurable for every rational $q$. For a [L3, algebra] real $a$,

For every real $a$,

For every real $a$,

The equivalences from steps 2.1 and 2.2 transfer the rational-threshold [step 2.1, step 2.2, step 1.2] reduction of step 1.2 to conditions 3 through 5. Therefore in any one of conditions 2 through 5 it is enough to test only rational thresholds.

Let $B\subseteq {\mathbb{R}}^m$ be Borel. By [L1], the preimage [L1] $f^{-1}(B)$ is a Borel subset of ${\mathbb{R}}^n$.

By the definition of Borel measurability on Euclidean spaces, step 1.1 says [step 1.1] exactly that $f$ is Borel measurable.

Suppose first that $f$ is increasing. For each real $a$, the threshold set [given] $E_a:=\{x:f(x)>a\}$ is upward closed: if $x\in E_a$ and $y>x$, then $f(y)\ge f(x)>a$, so $y\in E_a$. Therefore $E_a$ is one of the four Borel sets $\mathrm{\varnothing }$, $\mathbb{R}$, $(c,\mathrm{\infty })$, or $[c,\mathrm{\infty })$ for some real $c$.

Every set named in step 1.1 is Borel, so [L1] gives that every increasing [step 1.1, L1, algebra] real function is measurable. If $f$ is decreasing, then $-f$ is increasing and hence measurable by the first half, and therefore $f$ is measurable as well.

Suppose $f$ is measurable. Fix $j$ and a real $a$. Then

Conversely, suppose every coordinate $f_j$ is measurable. Let [L1, L3, algebra] $B={\prod }_{j=1}^n(a_j,b_j)$ be a rational open box. Then

By [L1] and [L2], step 1.2 implies that $f$ is measurable. Together with [step 1.1, step 1.2, L1, L2] step 1.1, this proves the equivalence.

Let $C\in \mathcal{C}$. Since $g$ is measurable, [L1] gives [given, L1] $g^{-1}(C)\in \mathcal{B}$.

Since $f$ is measurable, [L1] applied again gives

Scalar multiples are measurable. If $c>0$, then $\{cf>a\}=\{f>a\mathrm{/}c\}$; if $c<0$, then $\{cf>a\}=\{f<a\mathrm{/}c\}$; and if $c=0$, the function is constant. So [L1] gives measurability of $cf$, and in particular of $-f$.

The threshold identities [step 1.1, L1, L2] $\{\max (f,g)>a\}=\{f>a\}\cup \{g>a\},\{\min (f,g)>a\}=\{f>a\}\cap \{g>a\}$ show via [L1] that $\max (f,g)$ and $\min (f,g)$ are measurable. By [L2], this proves measurability of $f^{+}$ and $f^{-}$; replacing $g$ by $-f$ also gives $\mathrm{\mid }f\mathrm{\mid }=\max (f,-f)$. [step 1.1, L1, L2].

Assume $f+g$ is pointwise defined. For every real $a$, [step 2.1, L1] $\{f+g>a\}={\bigcup }_{q\in \mathbb{Q}}(\{f>q\}\cap \{g>a-q\})\mathrm{.}$ The inclusion from right to left is immediate. For the converse, if $f(x)+g(x)>a$ then either $f(x)=+\mathrm{\infty }$, in which case any rational $q>a-g(x)$ works, or $f(x)$ is finite and one may choose a rational $q$ with $a-g(x)<q<f(x)$. Thus [L1] gives measurability of $f+g$. [step 2.1, L1].

Suppose first that $u,v:X\to [0,+\mathrm{\infty }]$ are nonnegative and measurable. [step 3.1, L1, A1] If $a<0$, then $\{uv>a\}=X$. If $a\ge 0$, then $\{uv>a\}={\bigcup }_{q\in \mathbb{Q},q>0}(\{u>q\}\cap \{v>a\mathrm{/}q\})\mathrm{.}$ Again the inclusion from right to left is immediate. For the converse, if $u(x)v(x)>a$, choose a rational $q$ with $0<q<u(x)$ and $a\mathrm{/}q<v(x)$; this is possible because either $u(x)$ is finite positive and the rationals are dense, or $u(x)=+\mathrm{\infty }$, in which case any sufficiently large positive rational works. Hence nonnegative products are measurable by [L1]. [step 3.1, L1, A1].

For general measurable $f$ and $g$, step 2.1 gives measurable nonnegative [step 2.1, step 3.1, step 4.1, L2, A1] functions $f^{+},f^{-},g^{+},g^{-}$. By step 4.1 the four products $f^{+}{g}^{+},f^{-}{g}^{-},f^{+}{g}^{-},f^{-}{g}^{+}$ are measurable. Put $h_{+}:=f^{+}{g}^{+}+f^{-}{g}^{-},h_{-}:=f^{+}{g}^{-}+f^{-}{g}^{+}\mathrm{.}$ At each point, at least one of $h_{+}$ and $h_{-}$ is zero, because at least one of $f^{+},f^{-}$ and at least one of $g^{+},g^{-}$ is zero. So the difference $h_{+}-h_{-}$ is pointwise defined without the forbidden $\mathrm{\infty }-\mathrm{\infty }$ form, and step 3.1 makes it measurable. By the usual sign decomposition, $h_{+}-h_{-}=fg$, with the convention [A1] at the $0\cdot \mathrm{\infty }$ points. [step 2.1, step 3.1, step 4.1, L2, A1].

Steps 1.1 through 5.1 prove all four claims. [step 1.1, step 2.1, step 3.1, step 4.1, step 5.1].

Let $s(x):={\sup }_n{f}_n(x)$ and $i(x):={\inf }_n{f}_n(x)$. Then for every real [L1, given] $a$,

For each $n$, the tail functions [step 1.1, L2] $s_n(x):={\sup }_{k\ge n}{f}_k(x)$ and $i_n(x):={\inf }_{k\ge n}{f}_k(x)$ are measurable by step 1.1. Applying step 1.1 again to the sequences $(s_n)$ and $(i_n)$ and then using [L2] yields measurability of ${\mathrm{lim\hspace{0.17em}sup}}_n{f}_n$ and ${\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$.

Let $u:={\mathrm{lim\hspace{0.17em}sup}}_n{f}_n$ and $v:={\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$. The equality set [step 2.1, L1, L2] $\{u=v\}$ is measurable because

If $f_n\to f$ pointwise, then [L2] gives [step 2.1, step 3.1, L2] $f={\mathrm{lim\hspace{0.17em}sup}}_n{f}_n={\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$. Since step 2.1 has already proved that both limiting functions are measurable, $f$ is measurable.

For fixed $k$, each set [L1, L2] $\{j{2}^{-k}\le f<(j+1)2^{-k}\}$ and $\{f\ge k\}$ is measurable by [L1]. Hence $s_k$ is a measurable real-valued function. Its values belong to the finite set $\{0,2^{-k},2\cdot 2^{-k},\dots ,(k{2}^k-1)2^{-k},k\}$, so [L2] makes $s_k$ a simple function.

Fix $x\in X$. If $f(x)\ge k$, then $s_k(x)=k\le f(x)$ and also [given, algebra] $s_{k+1}(x)\ge k=s_k(x)$. If $f(x)<k$, choose $j$ with $j{2}^{-k}\le f(x)<(j+1)2^{-k}$. Then $s_k(x)=j{2}^{-k}\le f(x)$ and $f(x)-s_k(x)<2^{-k}$. At the finer scale $2^{-k-1}$, the same point lies in one of the two adjacent dyadic cells over that coarse cell, so $s_{k+1}(x)$ is either $j{2}^{-k}$ or $j{2}^{-k}+2^{-k-1}$. Thus $s_k(x)\le s_{k+1}(x)\le f(x)$.

The inequalities of step 1.2 hold for every $x$, so [step 1.2] $0\le s_k\le s_{k+1}\le f$. If $f(x)<+\mathrm{\infty }$, then for all $k>f(x)$ the second case of step 1.2 applies and gives $0\le f(x)-s_k(x)<2^{-k}$, hence $s_k(x)\to f(x)$. If $f(x)=+\mathrm{\infty }$, then $s_k(x)=k$ for every $k$, so $s_k(x)\to +\mathrm{\infty }=f(x)$.

If $f\le M<+\mathrm{\infty }$ on a set $E$, then for every $k>M$ the second case of [step 1.2, step 2.1] step 1.2 applies to every $x\in E$ and gives $0\le f(x)-s_k(x)<2^{-k}$. Therefore

By [L2], the functions $f^{+}$ and $f^{-}$ are measurable and nonnegative. Applying [L1, L2, L3] [L3] to them gives simple functions $u_k\uparrow f^{+}$ and $v_k\uparrow f^{-}$. [L1, L2, L3].

Put $s_k:=u_k-v_k$. Because $u_k$ and $v_k$ are simple, $s_k$ is a [step 1.1, L1] simple real-valued function. At each point, [L1] makes at least one of $u_k$ and $v_k$ equal to $0$, so $\mathrm{\mid }{s}_k\mathrm{\mid }=u_k+v_k\le f^{+}+f^{-}=\mathrm{\mid }f\mathrm{\mid }\mathrm{.}$ Also $s_k\to f^{+}-f^{-}=f$ pointwise because $u_k\to f^{+}$ and $v_k\to f^{-}$. [step 1.1, L1].

Fix a real $a$. On $X\setminus N$, the equality $g=f$ gives

The set $\{f>a\}$ is measurable by [L2]. The set $\{g>a\}\cap N$ is a [step 1.1, L1, L2] subset of the measurable null set $N$, so [L1] makes it measurable. Hence $\{g>a\}$ is measurable for every real $a$.

By [L2], step 2.1 proves that $g$ is measurable.

By [L1], choose simple $\overline{{\mathcal{A}}_0}$-measurable functions [L1, choose] $s_k:X\to \mathbb{R}$ with $\mathrm{\mid }{s}_k\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$ and $s_k(x)\to f(x)$ for every $x\in X$. Write the canonical representation of $s_k$ as $s_k={\sum }_{j=1}^{m_k}{c}_{k,j}{\mathbf{1}}_{E_{k,j}},$ where the $E_{k,j}$ are pairwise disjoint completed measurable level sets.

For each pair $(k,j)$, apply [L2] to $E_{k,j}$ and choose [L2, L3, choose] $A_{k,j}\in {\mathcal{A}}_0$ together with a completed null set $N_{k,j}$ such that $E_{k,j}=A_{k,j}\cup M_{k,j}$ with $M_{k,j}\subseteq N_{k,j}$. Because $A_{k,j}\subseteq E_{k,j}$, the sets $A_{k,j}$ remain pairwise disjoint. Define $t_k:={\sum }_{j=1}^{m_k}{c}_{k,j}{\mathbf{1}}_{A_{k,j}}\mathrm{.}$ Then each $t_k$ is ${\mathcal{A}}_0$-measurable and simple. Let $N:={\bigcup }_{k,j}{N}_{k,j}$. By [L3], $N$ is a completed measurable null set, and for every $x\mathrm{\notin }N$ one has $t_k(x)=s_k(x)$ for all $k$.

Define [step 1.1, step 1.2, L4] $g:={\mathrm{lim\hspace{0.17em}sup}}_{k\to \mathrm{\infty }}{t}_k\mathrm{.}$ By [L4], the function $g$ is ${\mathcal{A}}_0$-measurable. If $x\mathrm{\notin }N$, then step 1.2 gives $t_k(x)=s_k(x)$ for every $k$, and step 1.1 gives $s_k(x)\to f(x)$, so $g(x)={\mathrm{lim\hspace{0.17em}sup}}_k{t}_k(x)={\lim }_k{s}_k(x)=f(x)$. Hence $g=f$ on $X\setminus N$.

The null set $N$ is measurable in the completion by [L3], so step 2.1 says [step 2.1, L3] exactly that $f=g$ almost everywhere. Since $g$ is ${\mathcal{A}}_0$-measurable, it is the required base-measurable representative.

If $g=h\circ f$ for Borel $h$, then for each Borel $C\subseteq \bar{\mathbb{R}}$, $g^{-1}(C)=f^{-1}(h^{-1}(C))\in \sigma (f)$ by [L1].

Conversely suppose $g$ is $\sigma (f)$-measurable. For every rational $q$, [L1] and [L2] say that the family of Borel $A\subseteq \mathbb{R}$ with $f^{-1}(A)=\{g\le q\}$ is nonempty. By [L3], $\mathbb{Q}$ is countable; apply the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ to choose one $A_q$ for every $q$. Define $B_q={\bigcap }_{r\in \mathbb{Q},r>q}{A}_r$. These sets are Borel and increasing in $q$, and $f^{-1}(B_q)={\bigcap }_{r>q}\{g\le r\}=\{g\le q\}$, including when $g$ takes an infinite value.

The family also satisfies $B_q={\bigcap }_{r\in \mathbb{Q},r>q}{B}_r$. Indeed the right side expands to the intersection of all $A_s$ with rational $s>r>q$; every rational $s>q$ has a rational $r$ strictly between $q$ and $s$, and every $s>r>q$ also has $s>q$. Thus the two collections of $A_s$ have the same intersection.