Tonelli’s Theorem and Distribution Tails
The previous tutorial established the Nonnegative Iterated-Integration Theorem: for a nonnegative product-measurable function on a product of sigma-finite measure spaces, the product integral equals either iterated integral, even if the common value is infinite. This result is commonly called Tonelli’s Theorem. Its nonnegativity hypothesis matters: no absolute-integrability assumption is needed, but the integral may be infinite.
Tonelli’s Theorem is often used to turn an integral into a question about the sizes of level sets. A nonnegative function can be viewed as the accumulation of its superlevel sets, the sets where it exceeds a given threshold. This leads to the layer-cake formula. The formula is useful in its own right, and its proof reveals an important distinction: unlike the product-space version of Tonelli’s Theorem, the layer-cake formula holds on an arbitrary measure space, with no sigma-finiteness assumption.
The distribution tail is nonincreasing: if \(t_1\leq t_2\), then \(\{f>t_2\}\subseteq\{f>t_1\}\), so \(\mu(\{f>t_2\})\leq\mu(\{f>t_1\})\). A monotone extended-real-valued function is measurable, so this tail can be integrated with respect to Lebesgue measure on \([0,\infty)\).
The Layer-Cake Formula
Proof. First let \(s\) be a nonnegative measurable simple function. If \(s\) has no positive values, both sides are zero. Otherwise, write its distinct positive values in increasing order as \(0<a_1<\cdots<a_m\), and let \(A_j=\{x:s(x)=a_j\}\). Put \(a_0=0\). For \(a_{k-1}\leq t<a_k\), the superlevel set is $$ \{s>t\}=\bigcup_{j=k}^m A_j. $$ For \(t\geq a_m\), it is empty. Since the \(A_j\) are disjoint, integrating this finite step function of \(t\) gives $$ \int_0^\infty\mu(\{s>t\})\,dt =\sum_{k=1}^m(a_k-a_{k-1})\sum_{j=k}^m\mu(A_j). $$ All terms are nonnegative, so the finite sums can be regrouped even when some measures are infinite. The coefficient of \(\mu(A_j)\) after regrouping is $$ \sum_{k=1}^j(a_k-a_{k-1})=a_j. $$ Thus the right side is \(\sum_{j=1}^m a_j\mu(A_j)=\int_Xs\,d\mu\), proving the formula for simple functions.
Now let \(f\) be any nonnegative measurable function. By the Increasing Simple Approximation theorem, there are nonnegative measurable simple functions \(s_n\) with \(s_n\uparrow f\) pointwise. The Monotone Convergence Theorem gives $$ \int_Xs_n\,d\mu\longrightarrow\int_Xf\,d\mu. $$ For each \(t\geq0\), the sets \(\{s_n>t\}\) increase to \(\{f>t\}\): if \(f(x)>t\), convergence from below ensures that \(s_n(x)>t\) for some \(n\). Continuity from below of the measure therefore gives \(\mu(\{s_n>t\})\uparrow\mu(\{f>t\})\). Applying the Monotone Convergence Theorem to these nonnegative measurable functions of \(t\), with Lebesgue measure on \([0,\infty)\), yields $$ \int_0^\infty\mu(\{s_n>t\})\,dt \longrightarrow \int_0^\infty\mu(\{f>t\})\,dt. $$ The simple-function case equates the integrals on the left with \(\int_Xs_n\,d\mu\) for every \(n\). Taking limits proves the stated identity. This argument used monotone convergence, not a product-space application of Tonelli, so it applies whether or not \(\mu\) is sigma-finite.
Worked Example: Recovering the Integral of a Linear Function
Take \(X=[0,1]\) with Lebesgue measure and \(f(x)=x\). For \(0\leq t<1\), the set \(\{x\in[0,1]:x>t\}\) is \((t,1]\), of measure \(1-t\). For \(t\geq1\), it is empty. Therefore the layer-cake formula gives $$ \int_0^1 x\,dx =\int_0^1(1-t)\,dt =\left[t-\frac{t^2}{2}\right]_0^1 =\frac12. $$ The endpoints do not affect the integral over \(t\), and the result agrees with direct integration.
Worked Example: A Layer-Cake Identity Without Sigma-Finiteness
Let \(X\) be an uncountable set, equip it with the counting measure, and choose a point \(x_0\in X\). Counting measure assigns to a set its number of elements, with value \(+\infty\) for an infinite set. This measure is not sigma-finite: a countable union of finite sets is countable, so it cannot cover \(X\). Nevertheless, for \(f=\mathbf{1}_{\{x_0\}}\), the integral is $$ \int_X f\,d\mu=\mu(\{x_0\})=1. $$ For \(0\leq t<1\), the superlevel set \(\{f>t\}\) is \(\{x_0\}\), of measure \(1\); for \(t\geq1\), it is empty. Hence $$ \int_0^\infty\mu(\{f>t\})\,dt =\int_0^1 1\,dt=1. $$ This example illustrates why the layer-cake formula’s proof must not rely on the sigma-finiteness hypothesis in the product-space version of Tonelli’s Theorem.
Positive Moments from Tail Probabilities
The layer-cake formula also gives a useful expression for positive powers of a function. The weight \(p\,t^{p-1}\) accounts for the change in height when the function is raised to the power \(p\). The result applies to every positive real exponent, and, like the layer-cake formula, does not require sigma-finiteness.
Proof. Apply the Layer-Cake Formula to the nonnegative measurable function \(f^p\). This gives $$ \int_X f^p\,d\mu =\int_0^\infty\mu(\{f^p>s\})\,ds. $$ For \(s\geq0\), since \(p>0\) and the power function is increasing on \([0,\infty]\), \(\{f^p>s\}=\{f>s^{1/p}\}\). In the Lebesgue integral on the right, make the substitution \(s=t^p\), so \(ds=p\,t^{p-1}\,dt\). The change of variables for nonnegative integrals remains valid if the integrals are infinite; it can be applied first on bounded intervals away from zero and then extended by monotone convergence. The result is $$ \int_X f^p\,d\mu =p\int_0^\infty t^{p-1}\mu(\{f>t\})\,dt, $$ as required. The point \(t=0\) does not affect the Lebesgue integral, including when \(0<p<1\) and the weight is unbounded near zero. The proof used the layer-cake formula on \(X\), not Tonelli’s Theorem on a product space, so it requires no sigma-finiteness of \(\mu\).
Worked Example: Computing a Positive Moment
On \(X=[0,1]\) with Lebesgue measure, take \(f(x)=x\) and any \(p>0\). As above, \(\mu(\{f>t\})=1-t\) for \(0\leq t<1\), and the tail is zero for \(t\geq1\). The positive-moment formula gives $$ \int_0^1 x^p\,dx =p\int_0^1t^{p-1}(1-t)\,dt =p\left(\frac1p-\frac1{p+1}\right) =\frac1{p+1}. $$ The integral is finite for every \(p>0\), as the expression confirms.
Worked Example: Testing Integrability of a Singular Function
On \((0,1)\) with Lebesgue measure, let \(f(x)=x^{-\alpha}\), where \(\alpha>0\). For \(0\leq t<1\), every \(x\in(0,1)\) satisfies \(f(x)>t\), so the superlevel set has measure \(1\). For \(t\geq1\), $$ x^{-\alpha}>t \quad\Longleftrightarrow\quad 0<x<t^{-1/\alpha}. $$ Since \(t^{-1/\alpha}\leq1\) for \(t\geq1\), this superlevel set has measure \(t^{-1/\alpha}\). The layer-cake formula therefore gives $$ \int_0^1 x^{-\alpha}\,dx =1+\int_1^\infty t^{-1/\alpha}\,dt. $$ The tail integral is finite exactly when \(1/\alpha>1\), or \(0<\alpha<1\). In that case it equals \(\alpha/(1-\alpha)\), so the full integral is \(1/(1-\alpha)\). When \(\alpha\geq1\), the tail integral diverges, and so does the original integral. Thus the behavior near \(x=0\) is detected by the measure of the high-level sets.
What Tonelli Does—and a Common Pitfall
The product-space Nonnegative Iterated-Integration Theorem from the previous tutorial remains the central Tonelli result for exchanging integration order. If the two factor measure spaces are sigma-finite and the function on their product is nonnegative and measurable, either iterated integral equals the product integral; no finiteness assumption is needed. This theorem is especially useful when the sections have a simpler description than the original function.
The layer-cake and positive-moment formulas are related in spirit, but their proofs above do not require forming a product measure or interchanging two integrals. That distinction prevents a common mistake: invoking product-space Tonelli when a factor space has not been shown to be sigma-finite. In such a setting, the product-space theorem may not apply as stated. The layer-cake identity still applies on an arbitrary measure space because it was proved directly from simple approximation and monotone convergence.
Another useful point is that these formulas compare extended nonnegative quantities. If the tail integral diverges, the corresponding function integral is infinite; this is a valid conclusion, not a failure of the theorem. Conversely, finiteness of the tail integral proves integrability. For powers, the weighted tail tells us exactly which decay in \(\mu(\{f>t\})\) is sufficient for a finite moment.
Check Your Understanding
Use the statements and proofs in this tutorial to answer the following questions.
- What is the superlevel set of a function \(f\) at height \(t\), and why is its measure nonincreasing as a function of \(t\)?
- Which approximation and convergence theorem allow the layer-cake formula to extend from simple functions to arbitrary nonnegative measurable functions?
- Why does the layer-cake formula apply to counting measure on an uncountable set, even though that measure is not sigma-finite?
- For \(p>0\), what weight appears in the tail formula for \(\int_X f^p\,d\mu\), and which substitution produces it?
- For \(f(x)=x^{-\alpha}\) on \((0,1)\), for which positive values of \(\alpha\) is the integral finite?