Tail Integrals and the Layer-Cake Formula

From Function Values to Threshold Sets

A nonnegative number can be recovered by measuring how many thresholds it exceeds: the length of the interval of thresholds below it is exactly that number. Applying this pointwise observation to a measurable function turns its integral into an integral of measures of level sets. The resulting identity is useful even when the integral is infinite; no integrability assumption is needed.

Throughout, let $(X,\mathcal{M},\mu)$ be a sigma-finite measure space, and let $f:X\to[0,\infty]$ be measurable. Write $m$ for Lebesgue measure on $[0,\infty)$. Integrals and equalities below are understood in the extended nonnegative sense, so $+\infty$ is allowed. The formula concerns the given pointwise representative of $f$; changing it on a $\mu$-null set does not change either side.

Pointwise identity: For every $a\in[0,\infty]$, $$ a=\int_0^\infty \chi_{\{t<a\}}\,dt. $$ For finite $a$, the indicator is one on an interval of length $a$ and zero outside it. For $a=+\infty$, it is one for every finite $t$, and the integral is $+\infty$.

In particular, applying this identity with $a=f(x)$ gives $f(x)=\int_0^\infty\chi_{\{t<f(x)\}}\,dt$ for each $x\in X$. To integrate this equality over $X$, we must ensure the indicator is measurable on the product space and justify changing the order of integration. Those are precisely the roles of product measurability and Tonelli’s theorem.

The Layer-Cake Formula

Theorem (Layer-Cake Formula): Let $(X,\mathcal{M},\mu)$ be sigma-finite and let $f:X\to[0,\infty]$ be measurable. Then $t\mapsto\mu(\{x\in X:f(x)>t\})$ is a measurable extended-valued function on $[0,\infty)$, and $$ \int_X f\,d\mu=\int_0^\infty \mu(\{x\in X:f(x)>t\})\,dt. $$ Both sides may equal $+\infty$.

Proof. First define $D=\{(x,t)\in X\times[0,\infty):f(x)>t\}$. This set belongs to $\mathcal{M}\otimes\mathcal{B}_{[0,\infty)}$. Indeed, using the positive rationals $\mathbb{Q}_+$, it can be written as $$ D=\bigcup_{q\in\mathbb{Q}_+}\bigl(\{x:f(x)>q\}\times[0,q)\bigr). $$ If $f(x)>t$, there is a rational $q$ strictly between $t$ and $f(x)$; this also holds when $f(x)=+\infty$. Conversely, if $t<q<f(x)$, then $f(x)>t$. Each set in the countable union is product-measurable, so $D$ is product-measurable, and therefore $\chi_D$ is a nonnegative measurable function on the product space.

Both factors are sigma-finite: $\mu$ is sigma-finite by assumption, and Lebesgue measure on $[0,\infty)$ is sigma-finite because this space is the union of intervals of finite length. Tonelli’s theorem therefore applies to $\chi_D$ on $(X\times[0,\infty),\mathcal{M}\otimes\mathcal{B}_{[0,\infty)},\mu\times m)$. Integrating first in $t$, the pointwise identity gives $$ \int_X\left(\int_0^\infty \chi_D(x,t)\,dt\right)d\mu(x) =\int_X f(x)\,d\mu(x). $$ Integrating first in $x$, the section at a fixed $t$ is $\{x:f(x)>t\}$, and hence $$ \int_0^\infty\left(\int_X\chi_D(x,t)\,d\mu(x)\right)dt =\int_0^\infty\mu(\{x:f(x)>t\})\,dt. $$ Tonelli equates these iterated integrals, including when their common value is infinite. It also gives measurability of the inner integral as a function of $t$. Thus the asserted identity and measurability follow. $\square$

The use of the strict inequality $f(x)>t$ matches the pointwise interval $[0,f(x))$. In particular, if $f(x)$ is finite, the threshold $t=f(x)$ is not counted. Choosing a non-strict threshold instead would not change the integral over $t$, since for each fixed $x$ the two indicator functions differ at only one value of $t$; the formula above uses the strict version throughout.

Worked Examples

Worked Example: A Two-Level Function on a Finite Measure Space

Let $X=[0,1]$, let $\mathcal{M}$ be the Lebesgue sigma-algebra, let $\mu=m$, and set $f=2$ on $[0,1/3]$ and $f=1$ on $(1/3,1]$. Compute its integral from its tails and check it directly.

The function is nonnegative and measurable, and $\mu(X)=1$, so the theorem applies. For $0\le t<1$, both parts of $X$ have $f(x)>t$, so $\mu(\{f>t\})=1$. For $1\le t<2$, only $[0,1/3]$ has $f(x)>t$, so the tail measure is $1/3$. For $t\ge2$, neither value exceeds $t$, so the tail measure is zero. The endpoint $t=1$ belongs to the second case because the value $1$ is not strictly greater than $1$; $t=2$ belongs to the last case.

\int_0^\infty \mu(\{f>t\})\,dt =\int_0^1 1\,dt+\int_1^2 \frac13\,dt =1+\frac13 =\frac43.

Directly, the integral of this simple function is $$ \int_X f\,dm=2\,m([0,1/3])+1\,m((1/3,1]) =2\cdot\frac13+1\cdot\frac23 =\frac43. $$ The direct computation agrees with the tail calculation.

Worked Example: Tails Under Counting Measure

Take $X=\{1,2,3,4\}$ with its full sigma-algebra and counting measure $\mu$. Define $f(1)=0$, $f(2)=1/2$, $f(3)=3/2$, and $f(4)=3$. Compute both sides of the formula.

For $0\le t<1/2$, exactly the three points $2,3,4$ satisfy $f(n)>t$, so the tail measure is $3$. For $1/2\le t<3/2$, exactly $3,4$ satisfy the strict inequality, so it is $2$. For $3/2\le t<3$, only $4$ satisfies it, so it is $1$. For $t\ge3$, no point satisfies it, so it is $0$. These intervals account for the boundary cases: at $t=1/2$, the value $1/2$ is not counted, and at $t=3/2$, the value $3/2$ is not counted.

\int_0^\infty\mu(\{f>t\})\,dt =3\cdot\frac12+2\cdot\left(\frac32-\frac12\right) +1\cdot\left(3-\frac32\right) =\frac32+2+\frac32 =5.

For counting measure, the integral is the sum of the values: $$ \int_X f\,d\mu=0+\frac12+\frac32+3=5. $$ This independently checks the tail calculation.

Worked Example: A Function with Infinite Integral

Let $X=\mathbb{R}$ with Lebesgue measure $m$, and define $f(x)=1/x$ for $0<x\le1$, with $f(x)=0$ otherwise. This is a nonnegative measurable function. Find its tail measures and use them to determine its integral.

If $0\le t<1$, every $x\in(0,1]$ satisfies $1/x>t$, so $\mu(\{f>t\})=1$. If $t\ge1$, the inequality $1/x>t$ on $(0,1]$ is equivalent to $0<x<1/t$, whose Lebesgue measure is $1/t$. At $t=1$, the set is $(0,1)$, also of measure $1$; at every $t\ge1$, the strict endpoint $x=1/t$ is excluded but does not affect the measure.

\int_0^\infty\mu(\{f>t\})\,dt =\int_0^1 1\,dt+\int_1^\infty\frac1t\,dt =+\infty.

For a check directly on $X$, on each interval $(2^{-(k+1)},2^{-k}]$, where $k$ is a nonnegative integer, we have $1/x\ge2^k$. The interval has length $2^{-k}-2^{-(k+1)}=2^{-(k+1)}$, so its contribution to the integral is at least $2^k\cdot2^{-(k+1)}=1/2$. The first $N$ such disjoint intervals therefore contribute at least $N/2$. Since this holds for every positive integer $N$, $\int_{\mathbb{R}}f\,dm=+\infty$, in agreement with the tail calculation.

Truncation and a Common Pitfall

Corollary (Truncated Layer-Cake Formula): Under the hypotheses of the theorem, for every finite $a\ge0$, $$ \int_X \min(f,a)\,d\mu=\int_0^a\mu(\{f>t\})\,dt. $$

Proof. For each $x\in X$, the length of the set of thresholds $t\in[0,a]$ for which $t<f(x)$ is $\min(f(x),a)$. Equivalently, $$ \min(f(x),a)=\int_0^a\chi_{\{t<f(x)\}}\,dt. $$ The same product-measurable indicator $ \chi_D $ used in the theorem is nonnegative. Tonelli’s theorem applies on $X\times[0,a]$, since both factors are sigma-finite. Integrating first in $t$ gives the left side. Integrating first in $x$ gives $\int_0^a\mu(\{f>t\})\,dt$. This proves the formula, including $a=0$, when both sides are zero. $\square$

Truncation restricts the threshold range to a finite interval and caps the function at $a$; it does not assert that the original integral is finite. In particular, the layer-cake identity is not a statement that every tail measure is finite. For example, on a space with infinite measure, a set $\{f>t\}$ may have infinite measure for some or many thresholds, and the tail integral can then be infinite.

Common mistake: extending a nonzero tail past the largest function value. On $X=[0,1]$ with Lebesgue measure, let $f=2\chi_{[0,1]}$. A mistaken calculation might say that $\mu(\{f>t\})=1$ for every $t>0$, giving an infinite tail integral. But the inequality $2>t$ holds only for $t<2$. Thus the tail measure is $1$ for $0\le t<2$ and $0$ for $t\ge2$, so $$ \int_0^\infty\mu(\{f>t\})\,dt=\int_0^2 1\,dt=2. $$ The direct integral is also $\int_0^1 2\,dm=2$. The correction is to determine the threshold set using the actual strict inequality $f(x)>t$, including where the tail becomes zero.

The layer-cake formula converts information about the sizes of level sets into information about the integral of a function. It applies to every nonnegative measurable function on a sigma-finite measure space, without requiring boundedness or finite integral. The examples show why the extended-value formulation matters: both the function and the integral of its tails may legitimately take the value $+\infty$.

Check Your Understanding

Use the strict threshold convention in each question, and state whenever an integral may be infinite.

For $f=3\chi_E$ on a measure space with $\mu(E)=2$, determine $\mu(\{f>t\})$ for every $t\ge0$, then compute its tail integral.
On a finite set with counting measure and function values $0,1,2$, determine the tail measure on each interval of thresholds and verify the layer-cake formula.
Explain why the set $D=\{(x,t):f(x)>t\}$ is product-measurable using positive rational numbers.
For $f(x)=1/x$ on $(0,1]$, compute the tail measure at $t=1$ and at $t=4$, taking care with the strict inequality.
Use the truncated formula to express $\int_X\min(f,2)\,d\mu$ as an integral over a bounded threshold interval.

Tail Integrals and the Layer-Cake Formula

What You'll Learn