Proof of Approximation of Measurable Functions by Simple Functions

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above.

If $X$ is empty, then the unique map $X\to\mathbb{R}$ is measurable and its set of values is empty, hence finite, so it is a simple function and the constant sequence with that term satisfies every condition in the three claims vacuously. We therefore assume throughout that $X$ is nonempty.

For a real number $y$ we write $\lfloor y\rfloor$ for the integer part of $y$, the unique integer $n$ with $n\le y<n+1$. We shall use twice the inequality $m\le 2^{m}$, valid for every $m\in\mathbb{N}$ by induction: $1\le 2$, and if $m\le 2^{m}$ then $m+1\le 2^{m}+1\le 2^{m}+2^{m}=2^{m+1}$, since $1\le 2^{m}$.

We also record once that for a measurable $f:X\to[0,\infty]$ and a real $c$ the set $A_{c}=\{x\in X:c\le f(x)\}$ belongs to $\mathcal{F}$: for $0<c$ this is part of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §markov, and for $c\le 0$ we have $A_{c}=X$ because $f$ is nonnegative. A real-valued measurable map with nonnegative values is measurable as a map into $[0,\infty]$, by the agreement of the two readings recorded in Measure Spaces and the Lebesgue Integral: Standing Notation §measurable, so the same applies to it.

Claim 1. Fix $m\in\mathbb{N}$, put $M=m2^{m}\in\mathbb{N}$ and $t_{i}=i/2^{m}$ for integers $i$ with $0\le i\le M$, and define $s_{m}:X\to\mathbb{R}$ by

(a) The values of $s_{m}$. Let $x$ satisfy $f(x)<\infty$ and put $k=\lfloor 2^{m}f(x)\rfloor$, so that $k\le 2^{m}f(x)<k+1$. Since $0\le f(x)$ we have $0\le 2^{m}f(x)$, and $k<0$ would give $k+1\le 0\le 2^{m}f(x)$, contradicting $2^{m}f(x)<k+1$; hence $0\le k$. If $k\le M$ then $s_{m}(x)=\min\{m,t_{k}\}$, which is $t_{k}$ or $t_{M}=m$; if $M<k$ then $m=t_{M}<t_{k}$ and $s_{m}(x)=m=t_{M}$. Together with the value $m=t_{M}$ taken where $f(x)=\infty$, this shows that $s_{m}$ takes its values in $V_{m}=\{t_{0}\}\cup\{t_{i}:i\in[M]\}$. The set $[M]$ has $M$ elements by claim 1 of Basic Properties of Finite Sets, so its image under $i\mapsto t_{i}$ is finite by claim 4 of that lemma, and adjoining the single element $t_{0}$ leaves a finite set by claim 2 of that lemma (or the same set, if $t_{0}$ already belongs to the image). The set of values of $s_{m}$ is a subset of the finite set $V_{m}$, hence finite by claim 3 of that lemma.

(b) Measurability. Let $0\le i<M$. We show $s_{m}^{-1}(\{t_{i}\})=A_{t_{i}}\setminus A_{t_{i+1}}$. If $f(x)<\infty$ and $k=\lfloor 2^{m}f(x)\rfloor$, then $k<M$ implies $t_{k}<m$ and hence $s_{m}(x)=t_{k}$, while $M\le k$ implies $m\le t_{k}$ and hence $s_{m}(x)=m=t_{M}$; as the numbers $t_{0},\dots,t_{M}$ are distinct, $s_{m}(x)=t_{i}$ holds exactly when $k=i$, that is exactly when $i\le 2^{m}f(x)<i+1$, that is exactly when $t_{i}\le f(x)<t_{i+1}$. If $f(x)=\infty$ then $s_{m}(x)=t_{M}\ne t_{i}$, and also $x\in A_{t_{i+1}}$. This proves the asserted identity, and the right-hand side belongs to $\mathcal{F}$.

Next, $s_{m}^{-1}(\{t_{M}\})=A_{m}$. Indeed if $f(x)=\infty$ both sides contain $x$; and if $f(x)<\infty$ then, by the description just given, $s_{m}(x)=t_{M}$ holds exactly when $M\le k$, which holds exactly when $M\le 2^{m}f(x)$ (if $M\le k$ then $M\le k\le 2^{m}f(x)$; conversely $M$ is an integer with $M\le 2^{m}f(x)$, so $M\le\lfloor 2^{m}f(x)\rfloor=k$), that is exactly when $m\le f(x)$. So $s_{m}^{-1}(\{t_{M}\})\in\mathcal{F}$.

Since $s_{m}$ takes its values among $t_{0},\dots,t_{M}$, for every Borel subset $B$ of the real line the set $s_{m}^{-1}(B)$ is the union of those finitely many sets $s_{m}^{-1}(\{t_{i}\})$ for which $t_{i}\in B$, hence belongs to $\mathcal{F}$. Thus $s_{m}$ is measurable and, by (a), a simple function; it is nonnegative because each $t_{i}$ is.

(c) $s_{m}\le f$. If $f(x)=\infty$ this is clear. Otherwise $s_{m}(x)\le t_{k}=k/2^{m}\le f(x)$, since $k\le 2^{m}f(x)$.

(d) Monotonicity in $m$. If $f(x)=\infty$ then $s_{m}(x)=m\le m+1=s_{m+1}(x)$. Otherwise write $u=f(x)$, $k=\lfloor 2^{m}u\rfloor$ and $k'=\lfloor 2^{m+1}u\rfloor$. The integer $2k$ satisfies $2k\le 2\cdot 2^{m}u=2^{m+1}u<k'+1$, so $2k<k'+1$ and therefore $2k\le k'$, both being integers. Dividing by $2^{m+1}$ gives $k/2^{m}\le k'/2^{m+1}$. Since also $m\le m+1$, taking the smaller of the two entries on each side gives $s_{m}(x)\le s_{m+1}(x)$.

(e) The pointwise least upper bound. Fix $x$. If $f(x)=\infty$, then $s_{m}(x)=m$ for every $m$, and by The Archimedean Property of the Real Numbers the set $\{m:m\in\mathbb{N}\}$ has no real upper bound, so its least upper bound in $[0,\infty]$ is $\infty=f(x)$. Suppose $f(x)=u<\infty$. By (c) the number $u$ is an upper bound of $\{s_{m}(x):m\in\mathbb{N}\}$. Let $c$ be a real number with $c<u$. By The Archimedean Property of the Real Numbers there is $m\in\mathbb{N}$ with $u<m$ and $1/(u-c)<m$, hence $1/m<u-c$. For this $m$ we have $k/2^{m}\le u<m$, so $s_{m}(x)=k/2^{m}$; and $2^{m}u<k+1$ gives $u-1/2^{m}<k/2^{m}$, while $1/2^{m}\le 1/m<u-c$ because $m\le 2^{m}$. Therefore $s_{m}(x)=k/2^{m}>u-(u-c)=c$, so $c$ is not an upper bound. Hence $u$ is the least upper bound, as asserted.

Claim 2. Fix $m\in\mathbb{N}$ and define $s_{m}(x)=\frac{1}{2^{m}}\lfloor 2^{m}f(x)\rfloor$, which is meaningful because $f$ is real-valued. From $k\le 2^{m}f(x)<k+1$ with $k=\lfloor 2^{m}f(x)\rfloor$ we get at once

which is the assertion. As in (a) above, $0\le k$; and $k\le 2^{m}f(x)\le 2^{m}K$, so $k\le M'=\lfloor 2^{m}K\rfloor$ because $M'$ is the greatest integer not exceeding $2^{m}K$. Hence $s_{m}$ takes its values in $\{0\}\cup\{i/2^{m}:i\in[M']\}$ when $1\le M'$, and in $\{0\}$ otherwise; in either case this set is finite, exactly as in (a). The preimages are $s_{m}^{-1}(\{i/2^{m}\})=A_{i/2^{m}}\setminus A_{(i+1)/2^{m}}$ for $0\le i\le M'$, by the computation of (b) with the truncation absent, and these belong to $\mathcal{F}$; so $s_{m}$ is a nonnegative simple function, as in (b).

Claim 3. Let $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$ be the positive and negative parts of Integrable Function and the Lebesgue Integral, so that $f=f^{+}-f^{-}$ and $|f|=f^{+}+f^{-}$ pointwise; they are measurable by claims 1, 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and take nonnegative real values. Apply claim 1 to $f^{+}$ and to $f^{-}$, obtaining sequences $(p_{m})$ and $(q_{m})$ of nonnegative simple functions, nondecreasing in $m$, with $p_{m}\le f^{+}$ and $q_{m}\le f^{-}$ pointwise and with pointwise least upper bounds $f^{+}$ and $f^{-}$. Put $s_{m}=p_{m}-q_{m}$, measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

Fix $x$. If $0\le f(x)$ then $f^{-}(x)=0$, so $0\le q_{m}(x)\le 0$ and $s_{m}(x)=p_{m}(x)$; if $f(x)<0$ then $f^{+}(x)=0$, so $p_{m}(x)=0$ and $s_{m}(x)=-q_{m}(x)$. Hence, writing $V$ and $V'$ for the sets of values of $p_{m}$ and of $q_{m}$, which are finite and nonempty, the set of values of $s_{m}$ is contained in the union of $V$ and the image of $V'$ under $v\mapsto -v$. That image is finite by claim 4 of Basic Properties of Finite Sets, the union of the two finite sets is finite by Peeling an Element off a Finite Set, and Unions of Finite Sets, and a subset of it is finite by claim 3 of Basic Properties of Finite Sets. So each $s_{m}$ is a simple function.

In the first case $|s_{m}(x)|=p_{m}(x)\le f^{+}(x)=|f(x)|$, and in the second $|s_{m}(x)|=q_{m}(x)\le f^{-}(x)=|f(x)|$; so $|s_{m}|\le|f|$ pointwise.

Finally, in the first case $(s_{m}(x))_{m}=(p_{m}(x))_{m}$ is nondecreasing with least upper bound $f^{+}(x)=f(x)$, which is a real number, so it converges to $f(x)$ by A Bounded Monotone Sequence of Real Numbers Converges §nondecreasing; in the second case $(q_{m}(x))_{m}$ is nondecreasing with least upper bound $f^{-}(x)$, so it converges to $f^{-}(x)$ by the same result, and hence $(s_{m}(x))_{m}=(-q_{m}(x))_{m}$ converges to $-f^{-}(x)=f(x)$ by claim 3 of Arithmetic of Limits of Real Sequences. In both cases $(s_{m}(x))_{m}$ converges to $f(x)$.