Two Sources of Approximation Error
Approximating a measurable function by a finite-valued function involves two distinct operations. Quantization replaces values in a bounded interval by nearby grid points; its error is controlled by the grid spacing. Truncation first clips values outside that interval; its error depends on how far the original function lies beyond the cutoff. For a bounded function, the grid spacing alone can give a uniform error bound. For an unbounded function, clipping generally leaves a tail error that no fixed cutoff can control uniformly.
Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed. Let \(f:X\to\mathbb{R}\) be real-valued and measurable. Fix a grid spacing \(\delta>0\) and a cutoff \(M=N\delta\), where \(N\) is a positive integer. The alignment of the cutoff with the grid ensures that both endpoints of the retained interval are grid points.
The clipping map sends every real number into \([-M,M]\). On that interval, rounding down to the grid gives a value at most \(\delta\) below the input. Since clipping and the grid map have Borel level sets, their composition with measurable \(f\) is measurable. More explicitly, each level set of \(s_{M,\delta}\) is an inverse image under \(f\) of an interval, possibly with an endpoint included or excluded; such sets belong to \(\mathcal{M}\). Therefore \(s_{M,\delta}\) is a finite-valued measurable function, that is, a simple function.
The Quantization Error on the Retained Range
Proof. Put \(k=\lfloor t/\delta\rfloor\). The definition of the floor gives \(k\le t/\delta<k+1\). Multiplying by the positive number \(\delta\) gives \(k\delta\le t<(k+1)\delta\). Subtracting \(k\delta\) proves \(0\le t-k\delta<\delta\). If \(t=M=N\delta\), the floor is \(N\) and the error is zero; the same inequality still holds. For \(|f(x)|\le M\), truncation leaves \(f(x)\) unchanged, so the result applies with \(t=f(x)\). \(\square\)
The error is strictly less than \(\delta\), not necessarily bounded above by a smaller constant. Values just below a grid point are rounded down to the preceding grid point, and their errors approach \(\delta\). At a grid point, the error is zero. The strict inequality is useful: it is valid at every retained value, including both endpoints.
Worked Example: Quantizing a Bounded Function
On \(X=[0,1]\) with Lebesgue measure \(m\), let \(f(x)=2x-1\), choose \(M=1\), and choose \(\delta=\tfrac12\). Here \(M=2\delta\), and \(f\) already takes values in \([-1,1]\), so truncation does not change it.
The grid values are \(-1,-\tfrac12,0,\tfrac12,1\). The quantized function is \(s_{1,1/2}=-1\) on \([0,\tfrac14)\), \(-\tfrac12\) on \([\tfrac14,\tfrac12)\), \(0\) on \([\tfrac12,\tfrac34)\), \(\tfrac12\) on \([\tfrac34,1)\), and \(1\) at \(x=1\). For example, \(f(\tfrac14)=-\tfrac12\), so the value at that grid boundary is \(-\tfrac12\), not \(-1\).
On the first interval the error is \(2x\), which is nonnegative and less than \(\tfrac12\). On the second it is \(2x-\tfrac12\), again in \([0,\tfrac12)\). On the third it is \(2x-1\), and on the fourth it is \(2x-\tfrac32\); each ranges from zero up to, but not including, \(\tfrac12\). At \(x=1\), both \(f\) and \(s_{1,1/2}\) equal \(1\), so the error is zero. Thus \(|f-s_{1,1/2}|<\tfrac12\) everywhere.
Check: At \(x=\tfrac38\), \(f(x)=2(\tfrac38)-1=-\tfrac14\), and rounding down gives \(s(x)=-\tfrac12\). The error is \(|-\tfrac14-(-\tfrac12)|=\tfrac14<\tfrac12\). As \(x\) approaches \(\tfrac14\) from below, \(f(x)\) approaches \(-\tfrac12\) from below while \(s(x)=-1\); the error approaches \(\tfrac12\). Hence the supremum error is \(\tfrac12\), although it is not attained.
The Combined Bound and the Truncation Tail
Proof. Measurability and finite-valuedness were established above. If \(x\in A\), then \(T_M(f)(x)=f(x)\), so the quantization proposition gives \(|f(x)-s_{M,\delta}(x)|<\delta\). Also \((|f(x)|-M)^+=0\), proving the stated bound there.
If \(x\in A^c\), then \(f(x)>M\) or \(f(x)<-M\). In the first case \(T_M(f)(x)=M=N\delta\), so \(s_{M,\delta}(x)=M\) and the error is \(f(x)-M=|f(x)|-M\). In the second case \(T_M(f)(x)=-M=-N\delta\), so \(s_{M,\delta}(x)=-M\) and the error is \(|f(x)+M|=|f(x)|-M\). On \(A^c\), \(\chi_A(x)=0\) and \((|f(x)|-M)^+=|f(x)|-M\), which proves the bound and the exact tail formula. These cases include every \(x\), including the boundary points \(|f(x)|=M\), which belong to \(A\). \(\square\)
The formula isolates the two errors instead of hiding them in one estimate: the retained range contributes less than one grid spacing, while the discarded tail contributes exactly the amount by which \(|f|\) exceeds the cutoff. In particular, a coarser but sometimes convenient bound is \(|f-s_{M,\delta}|\le\delta+(|f|-M)^+\) everywhere. It is less sharp on the tail, where the grid error is actually zero because the cutoff itself is a grid point.
Worked Example: An Unbounded Function on a Half-Line
On \([0,\infty)\) with Lebesgue measure, let \(f(x)=x\), \(M=2\), and \(\delta=\tfrac12\). The cutoff is aligned with the grid because \(2=4(\tfrac12)\). For \(k=0,1,2,3\), the approximation equals \(k/2\) when \(x\in[k/2,(k+1)/2)\); it equals \(2\) for \(x\in[2,\infty)\). The endpoint \(x=2\) is included in the last interval, where the approximation is exactly \(2\).
For \(0\le x<2\), the error is \(x-k/2\) on the interval indexed by \(k\), so it lies in \([0,\tfrac12)\). At \(x=2\), the error is zero. For \(x>2\), the error is \(x-2\), exactly as the tail formula gives. For instance, at \(x=5\), the truncated and quantized value is \(2\), and the error is \(5-2=3\).
Check: Given any proposed uniform error bound \(K\ge0\), choose \(x=K+3\). Then \(x>2\), and the error is \(x-2=K+1>K\). Thus this fixed-cutoff approximation has no finite uniform error bound. On the retained interval its error is still less than \(\tfrac12\); it is the unbounded tail, not the quantizer, that causes the failure.
Exact Tail Estimates for a Singular Function
Worked Example: Locating a Strict Error Tail
On \(X=(0,1]\) with Lebesgue measure, let \(f(x)=1/x\), and take \(M=4\), \(\delta=1\). The truncated quantizer equals \(4\) on \((0,\tfrac14]\), since \(1/x\ge4\) there. On \((\tfrac12,1]\) it equals \(1\); on \((\tfrac13,\tfrac12]\) it equals \(2\); and on \((\tfrac14,\tfrac13]\) it equals \(3\). The interval endpoints follow from rounding down: for example, \(1/(\tfrac13)=3\), so \(x=\tfrac13\) belongs to the level with value \(3\).
On the retained range \(x\in[\tfrac14,1]\), the error is less than \(1\). On the truncated tail \(0<x<\tfrac14\), it is exactly \(1/x-4\). For the strict tail event where the error exceeds \(2\), the retained range contributes no points, and on the truncated tail we solve \(1/x-4>2\). This is equivalent to \(1/x>6\), hence to \(0<x<\tfrac16\). The boundary \(x=\tfrac16\) is excluded because its error is exactly \(2\). Therefore $$ m\bigl(\{x\in(0,1]:|f(x)-s_{4,1}(x)|>2\}\bigr) =m((0,\tfrac16))=\tfrac16. $$ Check: At \(x=\tfrac18\), \(f(x)=8\), the approximation is \(4\), and the error is \(4>2\). At \(x=\tfrac16\), \(f(x)=6\), the approximation is \(4\), and the error is \(2\), confirming the strict endpoint convention.
When Uniform Error Control Is Possible
If \(|f(x)|\le M\) for every \(x\), the theorem gives a uniform error strictly below \(\delta\). More precisely, \(\sup_{x\in X}|f(x)-s_{M,\delta}(x)|\le\delta\); the supremum need not itself be strictly less than \(\delta\), as the first worked example illustrates. If \(f\) is only essentially bounded, meaning \(|f|\le M\) outside a \(\mu\)-null set, the same error bound holds \(\mu\)-almost everywhere, not necessarily everywhere.
Conversely, if \(f\) is unbounded in absolute value at points of \(X\), then for every finite aligned cutoff \(M\), the error of this truncated approximation is unbounded: on \(\{|f|>M\}\) it equals \(|f|-M\). If \(f\) is essentially unbounded, the same argument shows that the error is essentially unbounded. This conclusion concerns this fixed-cutoff construction; increasing the cutoff can reduce the tail error at individual points, but no finite cutoff controls an unbounded tail uniformly.
Common mistake. It is incorrect to claim that clipping first guarantees \(|f-s_{M,\delta}|<\delta\) on all of \(X\). For \(f(x)=x\), \(M=2\), and \(\delta=\tfrac12\), at \(x=5\) the approximation is \(2\), so the error is \(3\), not less than \(\tfrac12\). The quantization estimate applies to the clipped value \(T_M(f)\), not to the difference between the original \(f\) and its clipped approximation. The correct estimate separates the retained range from the tail, and on the tail the error is exactly \((|f|-M)^+\).
These bounds provide a practical way to choose approximation parameters. First decide how small the grid error should be on the range that is retained; then choose a cutoff appropriate to the values of the function outside that range. A small grid spacing alone cannot compensate for a large truncation tail.
Check Your Understanding
In each question, distinguish the quantization error on the retained range from the truncation error outside it.
- For an aligned cutoff \(M=N\delta\), prove directly from the floor definition that a value \(t\in[-M,M]\) is rounded down by an amount in \([0,\delta)\).
- For \(f(x)=x\) on \([0,\infty)\), \(M=3\), and \(\delta=1\), give the exact error at \(x=2.4\) and at \(x=7\).
- For \(f(x)=1/x\) on \((0,1]\), with \(M=2\) and \(\delta=1\), solve exactly for the set where the approximation error is strictly greater than \(1\).
- Explain why an essentially bounded function can have an almost-everywhere uniform error bound even if its values on a null set exceed the chosen cutoff.
- If \(f\) is unbounded in absolute value, show that the fixed-cutoff approximation \(s_{M,\delta}\) has unbounded pointwise error.