Integrals of Maxima and Minima

Pointwise Maxima and Minima

Given two nonnegative functions, their maximum is at least as large as either function, while their minimum is at most as large as either. Integrating these pointwise comparisons gives useful bounds, but the integral of a maximum is generally not the maximum of the two integrals. A second fact is more precise: at every point, maximum plus minimum equals the sum of the original values. Additivity of the nonnegative integral turns this into an exact identity, even when some integrals are infinite.

Throughout, let $(X,\mathcal{M},\mu)$ be an arbitrary measure space; no assumption of finite measure, sigma-finiteness, or completeness is made. The functions $f,g:X\to[0,\infty]$ are measurable, and all pointwise identities below hold everywhere. Their nonnegative integrals are extended values in $[0,\infty]$. We use the established properties of this integral: if $u\le v$, then $\int u\,d\mu\le\int v\,d\mu$; if $c\ge0$, then $\int cu\,d\mu=c\int u\,d\mu$; and for nonnegative measurable $u,v$, $\int(u+v)\,d\mu=\int u\,d\mu+\int v\,d\mu$. These statements allow infinite values.

Definition: The pointwise maximum and minimum of $f$ and $g$ are the functions $f\vee g$ and $f\wedge g$ defined by $(f\vee g)(x)=\max\{f(x),g(x)\}$ and $(f\wedge g)(x)=\min\{f(x),g(x)\}$ for every $x\in X$. The definitions include the value $+\infty$.

These functions are measurable. Indeed, for every real $t$, $\{f\vee g>t\}=\{f>t\}\cup\{g>t\}$, and $\{f\wedge g>t\}=\{f>t\}\cap\{g>t\}$. The sets on the right are measurable because $f$ and $g$ are measurable, and a sigma-algebra is closed under finite unions and intersections. The superlevel-set characterization of measurability for extended nonnegative functions then gives the claim. Thus their nonnegative integrals are defined.

Bounds for the Two Integrals

Proposition: On any measure space $(X,\mathcal{M},\mu)$, for nonnegative measurable $f,g:X\to[0,\infty]$, the following inequalities hold in the extended nonnegative reals: $$ \max\left\{\int f\,d\mu,\int g\,d\mu\right\} \le \int(f\vee g)\,d\mu \le \int f\,d\mu+\int g\,d\mu, $$ $$ 0\le\int(f\wedge g)\,d\mu \le \min\left\{\int f\,d\mu,\int g\,d\mu\right\}. $$ In addition, $\int(f\vee g)\,d\mu\ge \tfrac12\bigl(\int f\,d\mu+\int g\,d\mu\bigr)$, with the right side interpreted in $[0,\infty]$.

Proof. At every $x$, $f(x)\le(f\vee g)(x)$ and $g(x)\le(f\vee g)(x)$. Monotonicity of the integral gives both lower bounds for the maximum integral. Also, $(f\vee g)(x)\le f(x)+g(x)$, since both values are nonnegative. Integrating and using additivity gives the upper bound. The minimum is nonnegative and satisfies $(f\wedge g)(x)\le f(x)$ and $(f\wedge g)(x)\le g(x)$, so monotonicity proves its stated bounds.

For the additional lower bound, first suppose both $\int f\,d\mu$ and $\int g\,d\mu$ are finite. Pointwise, $f+g\le2(f\vee g)$. Monotonicity, additivity and positive homogeneity therefore give $\int f\,d\mu+\int g\,d\mu\le2\int(f\vee g)\,d\mu$. If either original integral is infinite, monotonicity and $f\le f\vee g$, $g\le f\vee g$ imply that $\int(f\vee g)\,d\mu=\infty$; the desired inequality then holds with both sides interpreted as extended values. $\square$

The upper bound for the maximum and the upper bound for the minimum answer different questions: the maximum cannot have integral larger than the sum of the two integrals, while the minimum cannot have integral larger than either one. The additional lower bound says that the maximum must capture at least half the combined integral. None of these claims requires the measure space to have finite measure.

Worked Example: A Finite Counting Measure

Let $X=\{1,2,3\}$, let $\mathcal{M}$ be all subsets of $X$, and let $\mu$ be counting measure. Set $f(1)=1,\ f(2)=4,\ f(3)=2$ and $g(1)=3,\ g(2)=2,\ g(3)=5$. Every function on this finite measurable space is measurable. Counting measure gives the integral as the sum of the three values.

At the three points, the maxima are $3,4,5$, and the minima are $1,2,2$. Thus $$ \int f\,d\mu=1+4+2=7,\qquad \int g\,d\mu=3+2+5=10, $$ $$ \int(f\vee g)\,d\mu=3+4+5=12,\qquad \int(f\wedge g)\,d\mu=1+2+2=5. $$ The bounds check directly: $10\le12\le7+10=17$, and $0\le5\le\min\{7,10\}=7$. Check: the maximum and minimum integrals sum to $12+5=17$, the same as $7+10=17$.

The Exact Sum Identity

Theorem: Let $f,g:X\to[0,\infty]$ be measurable on an arbitrary measure space $(X,\mathcal{M},\mu)$. Then $$ \int(f\vee g)\,d\mu+\int(f\wedge g)\,d\mu = \int f\,d\mu+\int g\,d\mu. $$ The equality is an equality in $[0,\infty]$, so it remains valid if either side is infinite.

Proof. Fix $x\in X$. If both $f(x)$ and $g(x)$ are finite, the larger and smaller of the two numbers are the same two numbers in a possibly different order; hence $(f\vee g)(x)+(f\wedge g)(x)=f(x)+g(x)$. If at least one of $f(x),g(x)$ is $+\infty$, then the maximum is $+\infty$, and both sides of this pointwise identity are $+\infty$. Therefore the identity holds everywhere. The maximum and minimum are nonnegative measurable functions, as shown above. Integrating the pointwise identity and applying additivity of the nonnegative integral to both sums proves the result, including when the values are infinite. $\square$

When both original integrals are finite, the theorem also permits subtraction of finite real numbers: $$ \int(f\vee g)\,d\mu = \int f\,d\mu+\int g\,d\mu-\int(f\wedge g)\,d\mu. $$ This rearrangement is not justified if the total is infinite: an expression such as $\infty-\infty$ is undefined. The sum identity itself has no such problem, because it adds nonnegative extended values.

Worked Example: Overlapping Indicator Functions

Take $X=[0,1]$ with its Lebesgue sigma-algebra and Lebesgue measure $m$. Let $A=[0,3/4]$, $B=[1/4,1]$, $f=2\chi_A$, and $g=\chi_B$. These functions are measurable and nonnegative. The three regions $[0,1/4)$, $[1/4,3/4]$, and $(3/4,1]$ have measures $1/4,\ 1/2,\ 1/4$, respectively. Endpoint choices do not change these measures, since a finite set has Lebesgue measure zero.

On the first region, $f=2$ and $g=0$; on the overlap, $f=2$ and $g=1$; on the last region, $f=0$ and $g=1$. The maximum has values $2,2,1$ on these regions, so $$ \int(f\vee g)\,dm=2\cdot\frac14+2\cdot\frac12+1\cdot\frac14 =\frac12+1+\frac14=\frac74. $$ The minimum has values $0,1,0$, giving $$ \int(f\wedge g)\,dm=0\cdot\frac14+1\cdot\frac12+0\cdot\frac14=\frac12. $$ Separately, $\int f\,dm=2\cdot m(A)=2\cdot\frac34=\frac32$, and $\int g\,dm=m(B)=\frac34$. Check: $\frac74+\frac12=\frac94$, while $\frac32+\frac34=\frac94$. The two sides of the exact identity agree.

Worked Example: An Infinite Value Under a Dirac Measure

Let $X=\mathbb{R}$, $\mathcal{M}=\mathcal{B}_{\mathbb{R}}$, and $\mu=\delta_0$, the Dirac measure at $0$. Define $f(0)=+\infty$, $f(x)=0$ for $x\ne0$, and let $g(x)=3$ for every $x$. Both functions are Borel measurable: $f$ is supported on the Borel set $\{0\}$, and $g$ is constant. The Dirac integral of a nonnegative measurable function is its value at $0$, with the value $+\infty$ allowed.

Consequently, $\int f\,d\delta_0=+\infty$ and $\int g\,d\delta_0=3$. At $0$, the maximum is $+\infty$ and the minimum is $3$, so $\int(f\vee g)\,d\delta_0=+\infty$ and $\int(f\wedge g)\,d\delta_0=3$. Check: the left side of the sum identity is $+\infty+3=+\infty$, and the right side is $+\infty+3=+\infty$. No subtraction of infinite quantities is needed.

A Common Mistake and Its Correction

A tempting but false rule is $\int(f\vee g)\,d\mu=\max\{\int f\,d\mu,\int g\,d\mu\}$. For a concrete counterexample, use counting measure on $X=\{1,2\}$, and set $f(1)=4,\ f(2)=0,\ g(1)=0,\ g(2)=3$. Then $\int f\,d\mu=4$ and $\int g\,d\mu=3$, so the maximum of the two integrals is $4$. But the pointwise maximum has values $4,3$, and therefore $\int(f\vee g)\,d\mu=4+3=7$, not $4$. The error is treating the integral as if it preserved pointwise maximum. The valid bounds give $4\le7\le4+3=7$, and the exact identity is verified by $\int(f\wedge g)\,d\mu=0$: $7+0=4+3$.

These results are useful whenever two nonnegative contributions are combined. The bounds control their maximum without requiring the functions to have disjoint supports or any special relationship. The exact identity also records the overlap: a larger minimum contributes more to the sum of the two integrals, leaving correspondingly less for the maximum when the integrals are finite. For infinite integrals, retain the identity in its additive form and do not rearrange it by subtracting an infinite quantity.

Check Your Understanding

In each question, assume the functions are nonnegative and measurable on the stated measure space. Keep track of whether an integral could be infinite.

On an arbitrary measure space, prove that $\int(f\wedge g)\,d\mu\le\int f\,d\mu$ using a pointwise comparison.
If $\int f\,d\mu=5$ and $\int g\,d\mu=8$, what lower and upper bounds follow for $\int(f\vee g)\,d\mu$?
For counting measure on $\{1,2\}$, take $f=(2,6)$ and $g=(5,1)$. Compute both integrals, the maximum integral and the minimum integral, then check the sum identity.
Why does the exact sum identity remain valid when $\int f\,d\mu=+\infty$, while the rearrangement involving $\int f\,d\mu+\int g\,d\mu-\int(f\wedge g)\,d\mu$ may not be meaningful?
Give an example showing that the integral of the maximum need not equal the maximum of the two integrals.

What You'll Learn