Why a Pointwise Approximation Needs a Finite Range
Simple functions are the basic finite-valued objects used to define the Lebesgue integral. A useful approximation theorem must therefore do more than round a measurable function to a dyadic grid: at each stage, the approximation must take only finitely many values. This requirement matters for unbounded functions, whose direct dyadic roundings can have infinitely many values.
Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption on \(\mu\) is needed. A function \(f:X\to[0,\infty]\) is understood to be measurable, and may take the value \(+\infty\). A simple function here means a measurable function with finite range contained in \([0,\infty)\). The approximations below converge at every point, not merely \(\mu\)-almost everywhere.
The cap at \(n\) is part of the construction. It ensures that the values of \(s_n\) belong to the finite set \(\{j2^{-n}:j=0,1,\ldots,n2^n\}\). The dyadic grid becomes finer as \(n\) increases, while the cap also increases.
The Level-Set Formula
Proof. Fix \(x\in X\) and write \(t=f(x)\), allowing \(t=+\infty\). The integer \(\lfloor 2^n\min(t,n)\rfloor\) is the number of integers \(j\in\{1,\ldots,n2^n\}\) satisfying \(j\le 2^n\min(t,n)\). Since each threshold \(j2^{-n}\) in this range is at most \(n\), the condition \(j\le 2^n\min(t,n)\) is equivalent to \(f(x)\ge j2^{-n}\). Thus the right-hand side at \(x\) is \(2^{-n}\lfloor 2^n\min(t,n)\rfloor=Q_n(t)=s_n(x)\). This holds for every \(x\), proving the formula. For each \(j\), the level set \(\{f\ge j2^{-n}\}\) is measurable: it equals \(\bigcap_{r=1}^{\infty}\{f>j2^{-n}-1/r\}\). The formula is a finite sum of measurable indicators, so \(s_n\) is measurable. Its values belong to \(\{j2^{-n}:j=0,\ldots,n2^n\}\), a finite set. Hence it is simple. \(\square\)
The level-set formula is also a direct way to see what the quantizer does. At a point where \(f(x)=t\), it adds one dyadic increment for each threshold \(j2^{-n}\) that does not exceed \(t\), up to the cap \(n\). Equality at a threshold counts: the sets use \(\ge\), not \(>\).
Proof of Monotone Simple Approximation
Proof. Fix \(x\in X\), and put \(t=f(x)\). We first prove \(Q_n(t)\le Q_{n+1}(t)\), considering all possible values of \(t\).
If \(t\le n\), let \(r=\lfloor 2^nt\rfloor\). Then \(2^nt\ge r\), so \(2^{n+1}t\ge 2r\). Since \(2r\) is an integer, \(\lfloor 2^{n+1}t\rfloor\ge2r\). The cap at \(n+1\) does not alter \(t\), and therefore $$ Q_{n+1}(t)\ge 2^{-(n+1)}2r=2^{-n}r=Q_n(t). $$ If \(n<t\le n+1\), then \(Q_n(t)=n\). Also \(t>n\) implies \(2^{n+1}t>2^{n+1}n\), so \(Q_{n+1}(t)\ge n\). Finally, if \(t>n+1\), including \(t=+\infty\), then \(Q_n(t)=n\) and \(Q_{n+1}(t)=n+1\). These cases prove \(s_n(x)\le s_{n+1}(x)\) for every \(x\).
If \(t<+\infty\), then for every integer \(n\ge t\) the cap does not change \(t\). The defining property of the floor gives $$ 0\le t-Q_n(t)<2^{-n}. $$ As \(n\to\infty\), the right-hand side tends to zero, so \(Q_n(t)\to t\). If \(t=+\infty\), then \(Q_n(t)=n\) for every \(n\), and this sequence tends to \(+\infty\). Thus \(s_n(x)\to f(x)\) at every point, as required. \(\square\)
The proof uses only measurability of the level sets and elementary properties of the dyadic grid. In particular, it does not depend on the measure of \(X\), or on integration or a convergence theorem for integrals.
Worked Example: Three Values, Including Infinity
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and let \(\mu\) be counting measure. This is a finite measure space with \(\mu(X)=3\). Define the measurable extended-real-valued function by \(f(a)=3/4\), \(f(b)=5/2\), and \(f(c)=+\infty\). Compute \(s_n\) for \(n=1,2,3\).
At \(n=1\), the cap is \(1\) and the grid spacing is \(1/2\). Thus $$ s_1(a)=\tfrac12\lfloor 2\cdot\tfrac34\rfloor=\tfrac12,\qquad s_1(b)=\tfrac12\lfloor 2\cdot1\rfloor=1,\qquad s_1(c)=\tfrac12\lfloor 2\cdot1\rfloor=1. $$ At \(n=2\), the cap is \(2\) and the spacing is \(1/4\): $$ s_2(a)=\tfrac14\lfloor 4\cdot\tfrac34\rfloor=\tfrac34,\qquad s_2(b)=\tfrac14\lfloor 4\cdot2\rfloor=2,\qquad s_2(c)=\tfrac14\lfloor 4\cdot2\rfloor=2. $$ At \(n=3\), the cap is \(3\) and the spacing is \(1/8\): $$ s_3(a)=\tfrac18\lfloor 8\cdot\tfrac34\rfloor=\tfrac34,\qquad s_3(b)=\tfrac18\lfloor 8\cdot\tfrac52\rfloor=\tfrac52,\qquad s_3(c)=\tfrac18\lfloor 8\cdot3\rfloor=3. $$ So the values at \(a\) are \(1/2,3/4,3/4\); at \(b\) they are \(1,2,5/2\); and at \(c\) they are \(1,2,3\). Each list is nondecreasing. For \(a\) and \(b\), the values converge to \(3/4\) and \(5/2\); for \(c\), they increase without bound to \(+\infty=f(c)\). This checks both monotonicity and the asserted pointwise limits in this example.
Worked Example: The Identity Function on the Unit Interval
Take \(X=[0,1]\), its Lebesgue sigma-algebra \(\mathcal{L}\), Lebesgue measure \(m\), and \(f(x)=x\). The function is measurable and finite-valued, and the cap does not affect its values. For every \(n\ge1\), the level-set formula reduces to $$ s_n(x)=\sum_{j=1}^{2^n}2^{-n}\chi_{[j2^{-n},\,1]}(x). $$ Threshold equality is included: for example, \(x=1/2\) belongs to the set at threshold \(1/2\).
At \(n=1\), \(s_1\) is \(0\) on \([0,1/2)\), \(1/2\) on \([1/2,1)\), and \(1\) at \(x=1\). At \(n=2\), the grid spacing is \(1/4\), so its values on \([0,1)\) are \(0,1/4,1/2,3/4\) on the successive intervals \([0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)\), with value \(1\) at \(1\).
For a pointwise check, take \(x=3/8\). At \(n=1\), only the threshold \(1/2\) could contribute, and \(3/8<1/2\), so \(s_1(3/8)=0\). At \(n=2\), the threshold \(1/4\) contributes and \(1/2\) does not, so \(s_2(3/8)=1/4\). At \(n=3\), the thresholds \(1/8,1/4,3/8\) contribute and \(1/2\) does not, so \(s_3(3/8)=3/8\). This agrees with the formula \(s_n(3/8)=2^{-n}\lfloor 2^n(3/8)\rfloor\). The theorem guarantees that these values increase to \(3/8\); more generally, \(0\le x-s_n(x)<2^{-n}\) whenever \(x\) is not already represented exactly on the grid, and the difference is zero when it is.
Worked Example: An Unbounded Function on a Countable Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) contain every subset, and let \(\mu\) be counting measure. This measure space is sigma-finite but not finite. Define \(f(k)=k\). Every function on this measurable space is measurable. Since \(k\) is an integer, the quantizer gives $$ s_n(k)=2^{-n}\lfloor 2^n\min(k,n)\rfloor=\min(k,n). $$ For instance, at \(k=5\), the values \(s_1(5),s_2(5),s_3(5),s_4(5),s_5(5),s_6(5)\) are \(1,2,3,4,5,5\). At \(k=2\), they are \(1,2,2,2,2,2\). For each fixed \(k\), \(s_n(k)=k\) whenever \(n\ge k\), so \(s_n(k)\to k=f(k)\). The measure space need not be finite, and the function need not be bounded, for the pointwise approximation theorem to apply.
A Common Mistake and Its Correction
A tempting formula is \(2^{-n}\lfloor 2^nf(x)\rfloor\), without a cap. For \(f(x)=x\) on \([0,\infty)\), this formula at \(n=1\) takes the distinct values \(0,1/2,1,3/2,\ldots\), so it has infinite range and is not a simple function. If \(f\) can equal \(+\infty\), the expression also attempts to take the floor of \(+\infty\).
The correction is to use \(s_n(x)=2^{-n}\lfloor 2^n\min(f(x),n)\rfloor\). For the same function \(f(x)=x\), the first approximant is \(s_1(x)=0\) for \(0\le x<1/2\), \(s_1(x)=1/2\) for \(1/2\le x<1\), and \(s_1(x)=1\) for \(x\ge1\). It has only the three values \(0,1/2,1\). For \(x=5\), the corrected approximants are \(s_1(5)=1\), \(s_2(5)=2\), \(s_3(5)=3\), \(s_4(5)=4\), and \(s_5(5)=5\). The cap makes every stage simple, while the increasing caps allow the sequence to recover arbitrarily large finite values and the value \(+\infty\).
The result matters because it reduces questions about nonnegative measurable functions to questions about finite-valued measurable functions, with the approximation increasing at every point. That is the form needed when simple functions serve as the starting point for defining the Lebesgue integral.
Check Your Understanding
Use the capped quantizer \(Q_n(t)=2^{-n}\lfloor2^n\min(t,n)\rfloor\) in your reasoning.
- For \(t=7/8\), compute \(Q_1(t)\), \(Q_2(t)\), and \(Q_3(t)\). Which of these equals \(t\)?
- Explain why each set \(\{f\ge j2^{-n}\}\) in the level-set formula is measurable when \(f:X\to[0,\infty]\) is measurable.
- For \(f(k)=k\) on \(\mathbb{N}\), give \(s_4(6)\) and \(s_6(6)\), and state the pointwise limit at \(k=6\).
- Why does the uncapped dyadic rounding of \(f(x)=x\) on \([0,\infty)\) fail to be a simple function at \(n=1\)?
- What does the capped construction give at a point \(x\) where \(f(x)=+\infty\), and why does that sequence converge to \(f(x)\)?