Truncation and Tail Control in L1

Clipping an Integrable Function

An integrable function need not be bounded: it may take very large values on sets of very small measure. A useful way to approximate it is to clip its values at a fixed height. The resulting functions are bounded and remain integrable, and their errors become small in \(L^1\). The key is not merely that the sets where the function is large become small; it is that the integral of the function’s magnitude over those sets becomes small.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite or sigma-finite. Functions in this tutorial are real-valued and measurable, and integrable means \(\int_X |f|\,d\mu<\infty\). For a measurable set \(E\), \(\chi_E\) denotes its indicator. Equalities defining truncations and errors below hold everywhere. Integral conclusions concern the chosen functions, and are unchanged if functions are modified on a \(\mu\)-null set.

Thus \(T_K(f)\) is measurable, since it is formed from measurable functions using minimum and maximum, and \(|T_K(f)|\le K\) everywhere. Also \(|T_K(f)|\le |f|\) everywhere: on \(\{|f|\le K\}\) there is equality, and on \(\{|f|>K\}\) the truncated magnitude is \(K<|f|\). In particular, \(\int |T_K(f)|\,d\mu\le\int |f|\,d\mu<\infty\), even when \(\mu(X)=\infty\).

Tail Integrals Vanish

The tail above level \(K\) is the set \(\{|f|>K\}\); the inequality is strict, so points with \(|f|=K\) are not in the tail. Its weighted tail integral is \(\int_{\{|f|>K\}}|f|\,d\mu\), which measures both the size of the function there and the size of the set.

We use the Monotone Convergence Theorem in its standard form: if nonnegative measurable functions increase pointwise to a limit, their integrals increase to the integral of that limit, with \(+\infty\) allowed. We also use linearity for integrable functions and monotonicity of the nonnegative integral, established earlier in the course.

Proof. For each positive integer \(n\), set \(u_n=|f|\chi_{\{|f|\le n\}}\). The sets \(\{|f|\le n\}\) increase with \(n\), so \(u_n\le u_{n+1}\) everywhere. Since \(f\) is real-valued, every \(x\in X\) has \(|f(x)|\le n\) for all sufficiently large \(n\); hence \(u_n(x)\to |f(x)|\) everywhere. The Monotone Convergence Theorem gives \(\int u_n\,d\mu\to\int |f|\,d\mu\), a finite limit by integrability. Pointwise, \(|f|=u_n+|f|\chi_{\{|f|>n\}}\). Both terms are integrable, so linearity yields \(\int_{\{|f|>n\}}|f|\,d\mu=\int|f|\,d\mu-\int u_n\,d\mu\to0\). For real \(K\ge n\), \(\{|f|>K\}\subseteq\{|f|>n\}\). Monotonicity of the integral therefore bounds the tail integral at \(K\) by the one at \(n\), proving the asserted limit for all real \(K\to\infty\). Given \(\varepsilon>0\), choose \(n\) so the integer-level tail is less than \(\varepsilon\); the same bound holds for every \(K\ge n\). \(\square\)

The Truncation Error in \(L^1\)

For every \(x\), clipping changes a value only when its magnitude exceeds \(K\). On that tail, the difference is the excess beyond \(K\), not the whole value. Writing \(a^+=\max(a,0)\) for the positive part of a real number, we have the pointwise identity and bound

At \(|f|=K\), both sides of the identity are zero; outside the strict tail, the error is zero. Inside the tail, the error is \(|f|-K\), which is at most \(|f|\).

Proof. Boundedness and integrability of \(T_K(f)\) were checked above. For the error, integrate the pointwise inequality to obtain \(\|f-T_K(f)\|_1\le\int_{\{|f|>K\}}|f|\,d\mu\). The right side tends to zero by the tail-control proposition. Since the norm is nonnegative, this proves convergence to zero. Finally, for each fixed \(x\), \(K\ge |f(x)|\) implies \(T_K(f)(x)=f(x)\), so the truncations converge pointwise everywhere as well. \(\square\)

The pointwise and \(L^1\) conclusions are different statements. The pointwise conclusion follows because each fixed value is eventually unchanged; the \(L^1\) conclusion requires control of the integral of the error over all of \(X\). No uniform convergence is asserted: an unbounded function can have an error with unbounded supremum for every finite \(K\).

Worked Truncation Calculations

Why the Weighted Tail Matters

A common mistake is to infer that the tail integral is small merely because the tail set has small measure. For the singular function \(f(x)=x^{-1/2}\) on \((0,1)\), the set \(\{|f|>K\}=(0,K^{-2})\) has measure \(K^{-2}\), while its integral is \(2K^{-1}\). The measure and the integral are different quantities. For example, at \(K=4\), the set has measure \(1/16\) but its integral is \(1/2\).

Another tempting but incorrect equality is to identify the clipping error with the entire tail integral. At \(K=4\) in the same example, that would give an error of \(1/2\). The correct computation is \(\int_0^{1/16}(x^{-1/2}-4)\,dx=1/2-4(1/16)=1/4\). The error subtracts the clipped value \(K\) on the tail, so it is bounded by, and can be strictly smaller than, the weighted tail.

Clipping provides a reliable \(L^1\) approximation even on an infinite-measure space, but boundedness by itself does not guarantee integrability there: the constant function \(1\) on \(\mathbb{R}\) is bounded and has infinite integral with respect to Lebesgue measure. In the present construction integrability is guaranteed by \(|T_K(f)|\le|f|\), not by boundedness alone. The tail estimate is also a useful preparation for limit arguments: it isolates the part of an integrable function where large values occur and quantifies its contribution without confusing pointwise convergence with convergence in \(L^1\).