Dyadic Simple Approximants for Nonnegative Functions

Quantize the Values and Cut Off the Range

A nonnegative measurable function may take infinitely many values, or even the value \(+\infty\), so it is not itself necessarily a simple function. A dyadic approximant replaces its values by multiples of \(2^{-n}\) and caps those values at \(n\). The grid becomes finer as \(n\) increases, while the cap moves outward. These two operations together give a finite-valued function at each stage, even when the original function is unbounded.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Let \(f:X\to[0,\infty]\) be an extended-real-valued measurable function. All the pointwise equalities and inequalities below hold everywhere on \(X\). We use the floor \(\lfloor r\rfloor\), the greatest integer less than or equal to the finite real number \(r\).

The cap is part of the construction, not an optional convention: it limits the range to finitely many values. The level sets specify exactly which grid value is assigned. For \(k=0,\ldots,n2^n-1\), set $$ E_{n,k}=\left\{x\in X:\frac{k}{2^n}\le f(x)<\frac{k+1}{2^n}\right\}, $$ and set \(E_{n,*}=\{x\in X:f(x)\ge n\}\). The lower endpoint is included and the upper endpoint is excluded; thus a value exactly on a grid boundary belongs to the interval starting at that boundary.

These sets are measurable. Indeed, for \(a\le b\), the set \(\{a\le f<b\}\) is \(\{f\ge a\}\setminus\{f\ge b\}\), and the level sets of an extended-real-valued measurable function are measurable. Also \(\{f\ge n\}\in\mathcal{M}\). The sets \(E_{n,k}\) and \(E_{n,*}\) are pairwise disjoint and cover \(X\): every value in \([0,n)\) lies in exactly one of the half-open grid intervals, and every value in \([n,\infty]\) lies in the cap set. Consequently, $$ s_n=\sum_{k=0}^{n2^n-1}\frac{k}{2^n}\chi_{E_{n,k}}+n\chi_{E_{n,*}}. $$ This finite sum exhibits \(s_n\) as a measurable simple function. Some of its level sets may be empty; that does not affect the finite-valued property.

Basic Properties of the Approximants

Proof. The level sets in the displayed representation are measurable and form a finite partition of \(X\), so that representation makes \(s_n\) a measurable simple function. If \(f(x)<n\), the defining property of the floor gives $$ \lfloor 2^n f(x)\rfloor\le 2^n f(x)<\lfloor 2^n f(x)\rfloor+1. $$ Multiplying by \(2^{-n}>0\) yields \(s_n(x)\le f(x)<s_n(x)+2^{-n}\), which gives both the lower bound and the strict error bound. If \(f(x)\ge n\), including the case \(f(x)=+\infty\), then \(s_n(x)=n\le f(x)\) and \(s_n(x)\ge0\). Thus the asserted bounds hold at every point. \(\square\)

Proof. Fix \(x\in X\) and write \(t=f(x)\). If \(t=+\infty\), then \(s_n(x)=n\) for every \(n\), so the sequence increases to \(+\infty\). Suppose next that \(t<+\infty\). If \(t<n\), then the cap does not apply at either index \(n\) or \(n+1\). Since \(2^{n+1}t=2(2^nt)\), the floor identity $$ \lfloor 2^{n+1}t\rfloor=2\lfloor 2^nt\rfloor+r,\qquad r\in\{0,1\}, $$ shows that \(s_{n+1}(x)\ge s_n(x)\). If \(t\ge n\), then \(s_n(x)=n\). When \(n\le t<n+1\), quantization at level \(n+1\) gives \(s_{n+1}(x)\ge n\); when \(t\ge n+1\), the cap gives \(s_{n+1}(x)=n+1\). Hence the sequence is increasing in this case as well.

For convergence, fix finite \(t\). For every integer \(n>t\), the cap does not apply, and the floor estimate already proved gives \(0\le t-s_n(x)<2^{-n}\). Given any \(\varepsilon>0\), choose \(N\) so large that \(2^{-N}<\varepsilon\) and \(N>t\). For \(n\ge N\), we have \(0\le t-s_n(x)<2^{-n}\le2^{-N}<\varepsilon\), so \(s_n(x)\to t\). The infinite case was established at the start. Thus \(s_n(x)\uparrow f(x)\) for every \(x\in X\). \(\square\)

These are pointwise statements about simple functions; they do not require an integral or any assumption on the size of \(\mu(X)\). The construction supplies a specific sequence, rather than a claim that every nonnegative measurable function has such an approximation. The general approximation result is a separate theorem.

Worked Calculations

Why the Cutoff Matters

A common mistake is to quantize without truncating and then call the result a simple function. For instance, on \([0,\infty)\), consider the incorrect proposed approximant \(r_1(x)=\lfloor2x\rfloor/2\). It takes the values \(0,1/2,1,3/2,\ldots\) on infinitely many nonempty intervals, so its range is infinite. It is not a simple function under the finite-valued definition.

The correction is to use the cap: \(s_1(x)=\lfloor2x\rfloor/2\) for \(x<1\), and \(s_1(x)=1\) for \(x\ge1\). This function has only the possible values \(0,1/2,1\). The cutoff does not interfere with convergence at a fixed finite \(x\): once \(n>x\), the cap no longer applies there, and the quantization error is less than \(2^{-n}\). At a point where \(f(x)=+\infty\), the approximant instead equals \(n\) at stage \(n\) and tends to \(+\infty\).

Another endpoint error is to use overlapping or incomplete bins. The convention \([k/2^n,(k+1)/2^n)\) assigns every finite value below \(n\) to exactly one bin, including values exactly on grid boundaries. The separate set \(\{f\ge n\}\) then handles the cutoff boundary and all larger values. Keeping both conventions explicit makes the level sets a measurable partition and the resulting formula unambiguous.