Range Quantization for Bounded Measurable Functions

Turning a Bounded Range into Finitely Many Values

A bounded measurable function can take infinitely many values, so it need not itself be a simple function. A direct way to produce a finite-valued approximation is to divide an interval containing its range into finitely many smaller intervals, then replace each function value by a selected value from the interval containing it. This is called range quantization. The construction applies to negative as well as nonnegative values, and its error can be bounded at every point of the domain.

We work first on an arbitrary measure space \((X,\mathcal{M},\mu)\), with no assumption that \(\mu\) is finite, sigma-finite, or complete. A real-valued function \(f:X\to\mathbb{R}\) is measurable if \(f^{-1}(B)\in\mathcal{M}\) for every Borel set \(B\subseteq\mathbb{R}\). A finite-valued measurable function is a measurable function whose range is a finite set; it is a simple function. The bounds used below are pointwise bounds on all of \(X\), not merely bounds almost everywhere.

The Quantization Theorem

Proof. For \(k=0,\ldots,N-1\), define $$ E_k=\{x\in X:a+kh\le f(x)<a+(k+1)h\}, $$ and define \(E_N=\{x\in X:f(x)=b\}\). Each interval in the definition of \(E_k\), and the singleton \(\{b\}\), is a Borel subset of \(\mathbb{R}\). Since \(f\) is measurable, all these sets belong to \(\mathcal{M}\). The sets \(E_0,\ldots,E_N\) are pairwise disjoint and their union is \(X\): each value \(f(x)<b\) belongs to exactly one of the consecutive half-open intervals, while \(f(x)=b\) belongs to \(E_N\). Define $$ q=\sum_{k=0}^{N-1}(a+kh)\chi_{E_k}+b\chi_{E_N}. $$ This is a finite linear combination of indicators of measurable sets, so \(q\) is measurable and has only finitely many values. If \(x\in E_k\) with \(k<N\), then \(a+kh\le f(x)<a+(k+1)h\), and \(q(x)=a+kh\); hence \(0\le f(x)-q(x)<h\). If \(x\in E_N\), then \(f(x)=q(x)=b\), so the error is zero. This proves the assertion at every point. \(\square\)

The endpoint convention matters. The half-open intervals assign every value below \(b\) to exactly one interval, and the separate set \(E_N\) assigns the upper endpoint. The approximant always rounds down, so it never exceeds \(f\). Other choices of representative are possible, but their error estimates and endpoint rules must be checked separately.

Negative Values and Arbitrary Measurable Sets

Nothing in the construction requires the range interval to start at zero. In particular, one should not shift a function to be nonnegative unless there is a reason to do so: partitioning an interval that includes negative values works directly. Also, the quantization is performed in the range, not by subdividing the domain. The sets in the domain are just the measurable preimages of the chosen range intervals, and they need not be intervals or have positive measure.

Arbitrarily Accurate Uniform Approximation

Proof. Since \(f\) is bounded, choose real numbers \(a<b\) with \(a\le f(x)\le b\) for every \(x\in X\). If \(f\) is constant, take \(q=f\), which is already finite-valued and measurable. Otherwise, choose an integer \(N\ge1\) large enough that \((b-a)/N<\varepsilon\); such an integer exists by the Archimedean property of the real numbers. Apply the theorem with this \(N\). It gives a finite-valued measurable \(q\) satisfying \(0\le f(x)-q(x)<(b-a)/N<\varepsilon\) for every \(x\), hence \(|f(x)-q(x)|<\varepsilon\) everywhere. Applying this construction for each \(\varepsilon=1/n\), \(n\ge1\), gives a sequence \(q_n\) with \(|f(x)-q_n(x)|<1/n\) for all \(x\); this is uniform convergence by its definition. \(\square\)

The corollary only requires boundedness of the actual values of \(f\). If \(f\) is bounded only almost everywhere, the theorem does not immediately apply as written: the exceptional set may contain points where \(f\) is unbounded. Changing \(f\) on a null set is a separate operation, and changing a measurable function on an arbitrary subset of a null set need not preserve measurability unless the measure space is complete or the change is otherwise known to be measurable.

Endpoint Pitfalls and What the Bound Says

A frequent mistake is to use intervals that exclude both endpoints, then assume they still cover the range. For instance, on \(X=[0,1]\) with Lebesgue measure, consider \(f(x)=x\) and the proposed intervals \([0,\tfrac12)\) and \([\tfrac12,1)\). Their union is \([0,1)\), not \([0,1]\). The value \(f(1)=1\) belongs to neither interval, so a function defined by choosing a representative on those two sets is not defined at \(x=1\). The correction is to include the upper endpoint, for example by using \([\tfrac12,1]\) as the last interval, or by treating \(\{1\}\) separately as in the theorem. A single missing endpoint matters even though \(\{1\}\) has Lebesgue measure zero: the theorem’s bound is pointwise, not merely almost everywhere.

There is a second distinction between a pointwise strict bound and a strict bound on the supremum. The theorem gives \(f(x)-q(x)<h\) at every point, but the supremum of these errors can equal \(h\), because errors may approach \(h\) without attaining it. Thus one should safely conclude \(\sup_{x\in X}|f(x)-q(x)|\le h\), not always a strict inequality. For a requested tolerance \(\varepsilon\), choose \(h<\varepsilon\); then the supremum is at most \(h\), hence strictly below \(\varepsilon\). This is why the corollary chooses the interval width strictly smaller than the desired tolerance.

Finite-valued approximations are useful because they reduce questions about a bounded measurable function to finitely many measurable level sets and assigned values. The range partition provides that reduction while preserving a pointwise error estimate, including on exceptional points and at the endpoints. No integration theorem or assumption about the size of the underlying measure is needed for the construction.