Truncating Unbounded Functions Before Estimation

Why Truncate Before Estimating?

An estimate proved for bounded functions may be difficult to apply directly to an unbounded measurable function. Truncation provides a controlled approximation: clip the function at a finite height, prove the estimate for the clipped functions, and then identify a limit that recovers the quantity of interest. The last step matters. For a nonnegative function, increasing upper truncations recover its integral by the Monotone Convergence Theorem. For a signed function, the positive and negative parts must be tracked separately; cancellation between them cannot be used to define an integral when both parts have infinite integral.

Throughout the general statements, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions are measurable and may take extended real values unless stated otherwise. For an extended real-valued function $f$, define its positive and negative parts by $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Thus $f=f^+-f^-$ wherever this difference is defined, and $|f|=f^++f^-$.

Definition: For an integer $N\ge1$, the upper truncation of a nonnegative measurable function $g:X\to[0,\infty]$ is $g\wedge N=\min(g,N)$. The two-sided truncation of a measurable $f:X\to[-\infty,\infty]$ is $f_N=\max(-N,\min(f,N))=\min(f^+,N)-\min(f^-,N)$. Each truncation is measurable and bounded. For every $x$, $(f\wedge N)(x)\uparrow f(x)$ when $g\ge0$, and $f_N(x)\to f(x)$ in the extended-real sense.

Measurability follows because these truncations are compositions of measurable functions with continuous clipping maps on the real line, with the endpoint values at $+\infty$ and $-\infty$ assigned their clipped limits. The limit statements follow directly: if $f(x)$ is finite, then $f_N(x)=f(x)$ once $N\ge |f(x)|$; if $f(x)=+\infty$, then $f_N(x)=N$; and if $f(x)=-\infty$, then $f_N(x)=-N$.

The Limit That Recovers the Integral

We use the Monotone Convergence Theorem, a standard result established earlier in the course: on any measure space, if nonnegative measurable functions $g_N$ increase pointwise to $g$, then $\int g_N\,d\mu\to\int g\,d\mu$, with $+\infty$ allowed. Apply it separately to the two nonnegative sequences $\min(f^+,N)$ and $\min(f^-,N)$. This separate application is the basic recovery principle for signed functions.

Proposition: For every extended real-valued measurable $f$ on $(X,\mathcal{M},\mu)$, $$ \int f^+\,d\mu=\lim_{N\to\infty}\int\min(f^+,N)\,d\mu,\qquad \int f^-\,d\mu=\lim_{N\to\infty}\int\min(f^-,N)\,d\mu. $$ Both equalities are in $[0,\infty]$. Consequently, $\int |f|\,d\mu$ is recovered by adding these two limits. If $f$ is integrable, meaning $\int |f|\,d\mu<\infty$, then $\int f\,d\mu=\lim_{N\to\infty}\int f_N\,d\mu$.

Proof. For each $N$, $\min(f^+,N)$ is measurable, nonnegative, and at most $\min(f^+,N+1)$. At each $x$, these values increase to $f^+(x)$, including when $f^+(x)=+\infty$. The Monotone Convergence Theorem gives the first equality. The same reasoning applied to $f^-$ gives the second. Since $|f|=f^++f^-$, additivity of the nonnegative integral gives the asserted recovery of $\int|f|\,d\mu$. If $f$ is integrable, both limits are finite. For each $N$, both truncated parts have finite integrals, and additivity for integrable functions gives $\int f_N\,d\mu=\int\min(f^+,N)\,d\mu-\int\min(f^-,N)\,d\mu$. Taking limits of these two finite convergent sequences yields $\int f^+\,d\mu-\int f^-\,d\mu=\int f\,d\mu$. $\square$

The proposition does not say that every $f_N$ is integrable: boundedness alone does not imply integrability on a space of infinite measure. Its final conclusion requires $f$ to be integrable, which ensures that the two truncated parts have finite integrals. In general, the safe limit procedure is to recover the positive and negative integrals separately.

Worked Example: Truncating an Integrable Singularity

Let $X=[0,1]$ with the Lebesgue sigma-algebra and Lebesgue measure $m$. Define $f(0)=+\infty$ and $f(x)=x^{-1/2}$ for $0<x\le1$. For integer $N\ge1$, set $f_N=\min(f,N)$. Compute $\int f_N\,dm$ and use truncation to recover $\int f\,dm$.

The functions are measurable, nonnegative, bounded by $N$, and increase pointwise to $f$. For $0\le x\le N^{-2}$, $f_N(x)=N$; for $N^{-2}<x\le1$, $f_N(x)=x^{-1/2}$. The single value at $x=0$ does not change the Lebesgue integral. Therefore,

\int_0^1 f_N\,dm =N\cdot N^{-2}+\int_{N^{-2}}^1x^{-1/2}\,dx =\frac1N+\left[2\sqrt{x}\right]_{N^{-2}}^1 =\frac1N+2\left(1-\frac1N\right) =2-\frac1N.

By the Monotone Convergence Theorem, $\int_0^1 f\,dm=\lim_{N\to\infty}(2-1/N)=2$. Check: for $N=2$, the truncated integral is $2-\tfrac12=\tfrac32$; direct calculation gives $2\cdot\tfrac14+\int_{1/4}^1x^{-1/2}\,dx=\tfrac12+1=\tfrac32$. The limit also agrees with the improper integral $\int_0^1x^{-1/2}\,dx=2$.

Passing Bounds from Truncations to the Original Function

A common proof strategy is to establish a bound at every truncation level using a result that applies to bounded functions, then pass to the limit. For nonnegative functions, the direction of the limit is especially useful: if the integrals of the increasing truncations are bounded by the same constant, their limit gives the corresponding bound for the original function.

Proposition: Let $g:X\to[0,\infty]$ be measurable on $(X,\mathcal{M},\mu)$. If there is a finite constant $C$ such that $\int(g\wedge N)\,d\mu\le C$ for every integer $N\ge1$, then $\int g\,d\mu\le C$.

Proof. The functions $g\wedge N$ are nonnegative, measurable, and increase pointwise to $g$. The Monotone Convergence Theorem therefore gives $\int g\,d\mu=\lim_{N\to\infty}\int(g\wedge N)\,d\mu$. Every term of this convergent, nondecreasing sequence is at most $C$, so its limit is at most $C$. This proves the claim. $\square$

A useful signed version begins with an integrable majorant. Suppose $f$ and $g$ are measurable on $(X,\mathcal{M},\mu)$, $g\ge0$, $\int g\,d\mu<\infty$, and $|f|\le g$ $\mu$-a.e. Then $f^+\le g$ and $f^-\le g$ a.e., so monotonicity of the nonnegative integral gives $\int f^+\,d\mu\le\int g\,d\mu$ and $\int f^-\,d\mu\le\int g\,d\mu$. Thus $f$ is integrable. Applying the inequalities $-|f|\le f\le |f|$ and monotonicity for integrable functions yields $|\int f\,d\mu|\le\int|f|\,d\mu\le\int g\,d\mu$. Truncation is one way to establish such estimates, but it does not remove the need to verify the hypotheses of the estimate being applied.

Worked Example: Recovering Both Sides of a Signed Integral

On $X=(0,2)$ with Lebesgue measure, define $f(x)=x^{-1/2}$ for $0<x<1$, and $f(x)=-(x-1)^{-1/2}$ for $1<x<2$; assign any finite value at $x=1$. Let $f_N=\max(-N,\min(f,N))$. Compute the positive-part, negative-part, and signed truncated integrals.

On each of the intervals $(0,1)$ and $(1,2)$, the magnitude has the same form after shifting the variable. For either part, the region within distance $N^{-2}$ of the singular endpoint contributes $N\cdot N^{-2}=1/N$, and the rest contributes $\int_{N^{-2}}^1u^{-1/2}\,du=2(1-1/N)$. Thus each truncated part has integral $2-1/N$. In particular,

\int f_N^+\,dm=2-\frac1N,\qquad \int f_N^-\,dm=2-\frac1N,\qquad \int f_N\,dm=\left(2-\frac1N\right)-\left(2-\frac1N\right)=0.

The positive and negative truncations increase separately, so their integrals converge to $2$ and $2$. Hence $f$ is integrable, $\int |f|\,dm=2+2=4$, and $\int f\,dm=2-2=0$. Check: for $N=2$, each part has integral $2-\tfrac12=\tfrac32$, so the signed truncated integral is $\tfrac32-\tfrac32=0$. Directly integrating the positive and negative halves gives $2$ and $-2$, also yielding signed integral zero.

Worked Example: An Unbounded Function on a Finite Atomic Space

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}$ be all subsets of $X$, and give each point $k$ mass $2^{-k}$. This defines a finite measure, since $\sum_{k\ge1}2^{-k}=1$. For $f(k)=k$, compute the integral of $f_N(k)=\min(k,N)$ and recover $\int f\,d\mu$.

Each $f_N$ is measurable and bounded by $N$, and $f_N(k)\uparrow k$ for every $k$. By the definition of integration with respect to this atomic measure,

\int f_N\,d\mu =\sum_{k=1}^{N}\frac{k}{2^k}+N\sum_{k=N+1}^{\infty}\frac1{2^k} =\left(2-\frac{N+2}{2^N}\right)+\frac{N}{2^N} =2-\frac{2}{2^N}.

Here the finite-sum identity follows by induction on $N$, and the tail is a geometric series with sum $2^{-N}$. The Monotone Convergence Theorem gives $\int f\,d\mu=2$. Check: for $N=2$, the truncated values contribute $\tfrac12+2\cdot\tfrac14+2\sum_{k=3}^{\infty}2^{-k}=\tfrac12+\tfrac12+\tfrac12=\tfrac32$; the formula gives $2-\tfrac{2}{4}=\tfrac32$. Independently, the nonnegative series $\sum_{k\ge1}k2^{-k}$ sums to $2$, consistent with the truncation limit.

A Pitfall: Symmetric Truncation Can Hide Infinite Parts

Pointwise convergence $f_N\to f$ does not by itself justify passing a limit through an integral. In particular, signed truncations may cancel positive and negative contributions at every $N$, even when the original function has neither a defined signed integral nor integrable absolute value.

Worked Example: The False Comfort of Cancellation

On $(-1,1)$ with Lebesgue measure, define $f(x)=1/x$ for $x\ne0$, and assign $f(0)=0$. For integer $N\ge1$, let $f_N=\max(-N,\min(f,N))$. Each $f_N$ is bounded and integrable on this finite-measure space. For $x>0$, its integral is

\int_0^1 f_N\,dm =N\cdot\frac1N+\int_{1/N}^1\frac{dx}{x} =1+\ln N.

On $(-1,0)$, the integral is $-(1+\ln N)$, by the substitution $u=-x$. Therefore $\int_{-1}^1 f_N\,dm=0$ for every $N$, and $f_N\to f$ pointwise. But $\int f^+\,dm=\int_0^1 dx/x=+\infty$ and $\int f^-\,dm=\int_{-1}^0(-1/x)\,dx=+\infty$. The signed Lebesgue integral is undefined because it would require subtracting two infinite values; moreover $\int|f|\,dm=+\infty$. Correction: recover the positive and negative parts separately. Their truncated integrals both tend to $+\infty$, so the zero values of the signed truncated integrals do not represent an integral of $f$.

Another frequent error is to assume a bounded truncation is automatically integrable. For example, on $\mathbb{R}$ with Lebesgue measure, the constant function $g(x)=1$ is already bounded but has $\int g\,dm=+\infty$. A truncation estimate must therefore include whatever makes its integrals meaningful, such as finite measure of the domain, finite support, or an integrable majorant. The truncation controls the values of the function; it does not control the measure of the set on which those values occur.

Check Your Understanding

In each question, specify whether the functions are nonnegative or signed and identify the limit that would justify the integral statement.

For a measurable $g:X\to[0,\infty]$, why does $g\wedge N\uparrow g$ hold even where $g=+\infty$?
If $\int(g\wedge N)\,d\mu\le C$ for every $N$, which theorem gives $\int g\,d\mu\le C$, and what hypotheses must be checked?
For integrable signed $f$, express $\int f\,d\mu$ as a difference of limits involving $f^+$ and $f^-$.
Why does boundedness of $f_N$ fail to guarantee that $\int|f_N|\,d\mu$ is finite on an arbitrary measure space?
For the truncations of $1/x$ on $(-1,1)$, what are the limits of the positive-part and negative-part integrals, and why does their equality not define $\int f\,dm$?