Worked Approximations for Indicators and Step Profiles

From General Approximation to Explicit Formulas

A simple function has finite range, so approximating a profile that takes many values requires grouping points into measurable sets and assigning one value to each group. For a bounded profile, dyadic range quantization gives such groups directly. For an unbounded profile, quantization alone can produce infinitely many values; truncating first keeps each approximant simple. The examples below make these choices explicit and check the behavior at endpoints and threshold values.

Unless an example specifies otherwise, a simple function means a measurable function with finite range. Increasing means pointwise nondecreasing, and all pointwise limits below hold everywhere on the stated domain, not merely almost everywhere. For the general construction, let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed. Let $f:X\to[0,\infty)$ be a finite-valued measurable function.

Proposition: For $n\ge1$, define $$ \varphi_n(x)=2^{-n}\left\lfloor 2^n\min\{f(x),2^n\}\right\rfloor. $$ Then each $\varphi_n$ is a nonnegative measurable simple function, $\varphi_n\le\varphi_{n+1}$ everywhere, and $\varphi_n(x)\to f(x)$ for every $x\in X$.

Proof. Fix $n$. The function $\varphi_n$ takes values in the finite set $\{j2^{-n}:j=0,1,\ldots,2^{2n}\}$. For $0\le j<2^{2n}$, its level set at $j2^{-n}$ is $\{j2^{-n}\le f<(j+1)2^{-n}\}$; its level set at $2^n$ is $\{f\ge2^n\}$. These sets are measurable because $f$ is measurable, so $\varphi_n$ is measurable and simple.

Fix $x$ and write $t=f(x)$. If $t\le2^n$, then $\varphi_n(x)=2^{-n}\lfloor2^nt\rfloor$. The dyadic grid at level $n+1$ refines the grid at level $n$: if $k=\lfloor2^nt\rfloor$, then $\lfloor2^{n+1}t\rfloor\ge2k$. Therefore $\varphi_{n+1}(x)\ge\varphi_n(x)$. If $t>2^n$, then $\varphi_n(x)=2^n$. Since $\min\{t,2^{n+1}\}\ge2^n$, quantizing this value in multiples of $2^{-(n+1)}$ gives $\varphi_{n+1}(x)\ge2^n=\varphi_n(x)$. Finally, for this fixed finite $t$, choose $n$ large enough that $2^n>t$. Then $$ 0\le t-\varphi_n(x)=t-2^{-n}\lfloor2^nt\rfloor<2^{-n}. $$ As $n\to\infty$, the right-hand side tends to zero, proving pointwise convergence. $\square$

Indicators: Exact Simple Functions and Nested Sets

If $E\in\mathcal{M}$, then $\chi_E$, equal to $1$ on $E$ and $0$ on $E^c$, is already a simple function. Thus the constant sequence $\chi_E,\chi_E,\ldots$ is an exact approximation on any measure space $(X,\mathcal{M},\mu)$. A different construction is useful when one wants to see the indicator emerge from an increasing sequence of smaller sets.

Worked Example: Approximating the Indicator of an Open Interval

On $([0,1],\mathcal{L}|_{[0,1]},m)$, let $E=(0,1)$. For $n\ge2$, set $E_n=[1/n,1-1/n]$ and $\psi_n=\chi_{E_n}$. Each $\psi_n$ is simple and measurable. For $n=2$, $E_2=\{1/2\}$; for every $n\ge2$, both endpoints move outward as $n$ increases, so $E_n\subseteq E_{n+1}$.

If $x\in(0,1)$, then both $1/n\le x$ and $x\le1-1/n$ hold for all sufficiently large $n$. Hence $x\in E_n$ eventually and $\psi_n(x)\to1$. If $x=0$ or $x=1$, then $x\notin E_n$ for every $n$, so $\psi_n(x)\to0$. Points outside $[0,1]$ are not in the domain. Thus $\psi_n\uparrow\chi_{(0,1)}$ everywhere on $[0,1]$.

For $n\ge2$, the interval length is $1-2/n$, including the degenerate case $n=2$. Therefore $\int_{[0,1]}\psi_n\,dm=m(E_n)=1-2/n$, which tends to $1=m((0,1))=\int_{[0,1]}\chi_{(0,1)}\,dm$. Check: at $n=2,3,4$, the integrals are respectively $0$, $1/3$, and $1/2$; the formula gives each value directly. The omitted parts of $(0,1)$ have total length $2/n$, agreeing with the difference between $1$ and $1-2/n$.

The general lesson is that an indicator does not need value quantization: it already has only two values. The nested-set construction above is specific to this interval. For an arbitrary measurable $E$, the exact simple function $\chi_E$ is available even when no particular increasing family of convenient subsets has been chosen.

A Bounded Step Profile: The Identity Function

Worked Example: Dyadic Approximants to x on [0,1]

On $([0,1],\mathcal{L}|_{[0,1]},m)$, define $\varphi_n(x)=2^{-n}\lfloor2^nx\rfloor$. For $0\le k<2^n$, this function equals $k/2^n$ on $[k/2^n,(k+1)/2^n)$, and it equals $1$ at $x=1$. These are finitely many measurable pieces, so $\varphi_n$ is simple. Refinement of the dyadic grid gives $\varphi_n\le\varphi_{n+1}$, and $0\le x-\varphi_n(x)<2^{-n}$ for $x<1$, while the error at $x=1$ is zero. Consequently $\varphi_n(x)\to x$ everywhere.

At $n=2$, the values on the four half-open intervals are $0,1/4,1/2,3/4$, and the value at $1$ is $1$. The endpoint is a singleton of Lebesgue measure zero, so $$ \int_{[0,1]}\varphi_2\,dm =0\cdot\frac14+\frac14\cdot\frac14+\frac12\cdot\frac14+\frac34\cdot\frac14 =\frac{6}{16}=\frac38. $$ In general, $$ \int_{[0,1]}\varphi_n\,dm =\sum_{k=0}^{2^n-1}\frac{k}{2^n}\frac1{2^n} =\frac{(2^n-1)2^n}{2\cdot4^n} =\frac{2^n-1}{2^{n+1}}. $$ For $n=2$, the general formula again gives $3/8$. The integrals tend to $1/2$, the area under $x$ on $[0,1]$. As a second check, on each dyadic interval the difference $x-\varphi_n(x)$ rises linearly from zero toward $2^{-n}$; summing the triangular areas gives $2^n(2^{-2n}/2)=2^{-(n+1)}$. This is exactly $1/2-(2^n-1)/2^{n+1}$.

An Unbounded Profile: Truncate, Then Quantize

Worked Example: Simple Approximants to the Reciprocal Function

Take $((0,1],\mathcal{L}|_{(0,1]},m)$ and $f(x)=1/x$. This function is finite and measurable at every point of its domain, but it is unbounded. Apply the proposition: $$ \varphi_n(x)=2^{-n}\left\lfloor2^n\min\{1/x,2^n\}\right\rfloor. $$ The proposition gives finite range, measurability, monotonicity, and $\varphi_n(x)\to1/x$ everywhere on $(0,1]$.

For a concrete calculation, take $n=1$. If $0<x\le1/2$, then $1/x\ge2$, so $\varphi_1(x)=2$. If $1/2<x\le2/3$, then $3/2\le1/x<2$, so $\varphi_1(x)=3/2$. If $2/3<x\le1$, then $1\le1/x<3/2$, so $\varphi_1(x)=1$. The strict inequalities at $1/2$ and $2/3$ matter: $x=1/2$ belongs to the first set, while $x=2/3$ belongs to the second. The interval lengths are $1/2$, $1/6$, and $1/3$, respectively, and therefore $$ \int_{(0,1]}\varphi_1\,dm =2\cdot\frac12+\frac32\cdot\frac16+1\cdot\frac13 =1+\frac14+\frac13=\frac{19}{12}. $$ As a check, the three disjoint pieces have total measure $1/2+1/6+1/3=1$, and their contributions sum to $19/12$.

The target integral is infinite. For $k\ge0$, on $I_k=(2^{-(k+1)},2^{-k}]$ we have $1/x\ge2^k$, and $m(I_k)=2^{-(k+1)}$. Thus $\int_{I_k}(1/x)\,dm\ge1/2$. These intervals are disjoint and, for any positive integer $N$, the integral over their first $N$ members is at least $N/2$. Hence $\int_{(0,1]}(1/x)\,dm=\infty$. The Monotone Convergence Theorem applies to $\varphi_n$: the functions are nonnegative and measurable, increase pointwise to $f$, and so $\int\varphi_n\,dm\to\int f\,dm=\infty$. No integrability bound on $f$ is required for this theorem.

Why the Truncation Matters

A common mistake is to quantize $1/x$ without truncating it, using $2^{-n}\lfloor2^n/x\rfloor$ on $(0,1]$. For any fixed $n$, this function has unbounded range: as $x$ approaches zero, $2^n/x$ becomes arbitrarily large. It is measurable, but it is not a simple function because its range is not finite. The correction is to cap the input at $2^n$ before quantizing. The resulting range has at most $2^{2n}+1$ values, and the cap rises with $n$, allowing the pointwise limit to recover every finite value $1/x$.

Monotonicity is also a substantive condition when using Monotone Convergence. For example, on $((0,1],\mathcal{L}|_{(0,1]},m)$, the nonnegative measurable functions $g_n=n\chi_{(0,1/n)}$ tend pointwise to zero, but $\int g_n\,dm=n(1/n)=1$ for every $n$. Thus their integrals do not tend to the integral of the limit. This sequence is not increasing: at $x=3/4$, $g_1(x)=1$ and $g_2(x)=0$. The explicit approximants above avoid that failure by ensuring the required pointwise ordering.

Check Your Understanding

For each question, specify the domain and check both the endpoint behavior and the finite-range requirement.

On $[0,1]$, define $E_n=[1/n,1-1/n]$ for $n\ge2$. What is $m(E_n)$, and at which points does $\chi_{E_n}$ converge to $1$?
Compute $\int_{[0,1]}\varphi_3\,dm$ for $\varphi_n(x)=2^{-n}\lfloor2^nx\rfloor$.
For $f(x)=1/x$ on $(0,1]$, list the values of $\varphi_1$ on its three level sets and verify which set contains $x=2/3$.
Explain why quantizing $1/x$ without a truncation does not produce a simple function on $(0,1]$.
State the pointwise error bound for the general truncated dyadic construction once $2^n>f(x)$.