Positive Homogeneity of the Integral

Scaling an Integral

Multiplying a nonnegative function by a constant changes the height of every value by the same factor. The corresponding integral identity is fundamental, but its proof must respect the definition of the nonnegative integral as a supremum and the possibility that the integral is infinite. In particular, a positive finite scalar and the zero scalar require different arguments: multiplying $+\infty$ by a positive finite number gives $+\infty$, while the expression $0\cdot(+\infty)$ is not defined by ordinary extended-real arithmetic.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is made. Functions $f:X\to[0,\infty]$ are measurable, and the scalar $c$ is finite and nonnegative. Equalities between functions in the theorem are pointwise. Integrals are nonnegative extended real numbers and may equal $+\infty$.

Definition: A nonnegative simple function is a measurable function with finite range in $[0,\infty)$. If $\varphi=\sum_{j=1}^k a_j\chi_{E_j}$, where $E_j\in\mathcal{M}$ are disjoint and the coefficients $a_j$ are finite and nonnegative, its integral is $$ \int\varphi\,d\mu=\sum_{\{j:\,a_j>0\}}a_j\mu(E_j). $$ Thus a zero coefficient contributes zero and is omitted, even if its set has infinite measure. The nonnegative integral of a measurable $f:X\to[0,\infty]$ is $$ \int f\,d\mu=\sup\left\{\int\varphi\,d\mu:\ \varphi\text{ is nonnegative simple and }0\le\varphi\le f\right\}. $$

The omission of zero-coefficient terms is important: the formula for a simple function must not require the undefined product $0\cdot\infty$. The zero function has integral zero, since its only simple minorant is the zero function. We will use the simple-function integral formula and the supremum definition directly, together with monotonicity of the nonnegative integral established earlier.

A Scaling Fact for Supremums

The proof for a positive scalar will compare two collections of simple-function integrals. The elementary fact below explains why multiplying every member of such a collection by the same positive finite number multiplies its supremum by that number, even when the supremum is infinite.

Lemma: Let $S$ be a nonempty subset of $[0,\infty]$, and let $0<c<\infty$. Then $\sup\{cs:s\in S\}=c\sup S$, where $c(+\infty)=+\infty$.

Proof. Put $A=\sup S$. If $A=+\infty$, then $S$ is unbounded: for every finite $R\ge0$, some $s\in S$ satisfies $s>R/c$, and therefore $cs>R$. Thus $\sup\{cs:s\in S\}=+\infty=cA$. If $A<\infty$, then $cs\le cA$ for every $s\in S$, so $\sup\{cs:s\in S\}\le cA$. For the reverse inequality, fix $\varepsilon>0$. By the definition of supremum there is an $s\in S$ with $s>A-\varepsilon/c$; otherwise $A-\varepsilon/c$ would be an upper bound smaller than $A$. Hence $cs>cA-\varepsilon$. This holds for every $\varepsilon>0$, so $\sup\{cs:s\in S\}\ge cA$. The two inequalities prove the identity. $\square$

Positive Homogeneity

Theorem: Let $(X,\mathcal{M},\mu)$ be any measure space, let $f:X\to[0,\infty]$ be measurable, and let $c$ be a finite nonnegative scalar. For $c>0$, define $(cf)(x)=c f(x)$, with $c(+\infty)=+\infty$. For $c=0$, define $0f$ to be the function identically zero, including at points where $f(x)=+\infty$. Then $$ \int cf\,d\mu=c\int f\,d\mu\quad\text{if }c>0, \qquad \int 0f\,d\mu=0\quad\text{if }c=0. $$ In the second identity the right side is defined to be $0$, not as an extended-real product $0\cdot\int f\,d\mu$.

Proof. First suppose $0<c<\infty$. Multiplication by $c$ preserves measurability, and $(cf)(x)=c f(x)$ pointwise, with the stated convention at $+\infty$. Let $S_f$ be the set of integrals of all nonnegative simple functions $\varphi$ satisfying $\varphi\le f$. If $\varphi$ is one such minorant, then $c\varphi$ is a nonnegative simple function and $c\varphi\le cf$. From the simple-function integral definition, scaling each positive coefficient by $c$ gives $\int c\varphi\,d\mu=c\int\varphi\,d\mu$. In particular, every value $c\int\varphi\,d\mu$, for $\int\varphi\,d\mu\in S_f$, is the integral of a simple minorant of $cf$.

Conversely, let $\psi$ be any nonnegative simple function with $\psi\le cf$. The function $\psi/c$ is nonnegative simple. At every point, $\psi(x)\le c f(x)$ implies $\psi(x)/c\le f(x)$; this remains true when $f(x)=+\infty$. The simple-function formula gives $\int\psi\,d\mu=c\int(\psi/c)\,d\mu$. Consequently the integrals of the simple minorants of $cf$ are exactly the values $c s$ for $s\in S_f$. Taking their supremum and applying the lemma yields $$ \int cf\,d\mu=\sup\{cs:s\in S_f\} =c\sup S_f =c\int f\,d\mu. $$ If $c=0$, then $0f$ is identically zero by definition. Its only simple minorant is zero, whose integral is zero, so $\int 0f\,d\mu=0$. This proves both cases, including when $\int f\,d\mu=+\infty$. $\square$

Worked Applications

Worked Example: Rescaling an Indicator on the Real Line

Take $(\mathbb{R},\mathcal{L},m)$, with Lebesgue measure $m$, and let $f=\chi_{[0,2]}$. Compute the integral of $f$ and of $\tfrac32 f$.

The function $f$ is simple, with coefficient $1$ on $[0,2]$, so its integral is $1\cdot m([0,2])=1\cdot2=2$. The rescaled function has value $\tfrac32$ on $[0,2]$ and zero outside it. Its simple-function integral is therefore $$ \int_{\mathbb{R}}\tfrac32 f\,dm=\tfrac32\,m([0,2])=\tfrac32\cdot2=3. $$ The theorem gives the same result: $\tfrac32\int f\,dm=\tfrac32\cdot2=3$. Check: the scaled function has height $\tfrac32$ across a set of length $2$, so height times length is $\tfrac32\cdot2=3$. Its zero value off the interval contributes nothing; no term of the form $0\cdot m(\mathbb{R}\setminus[0,2])$ is needed.

Worked Example: Rescaling on a Finite Atomic Space

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be its power set, and define $\mu(\{a\})=2$, $\mu(\{b\})=\tfrac12$, and $\mu(\{c\})=0$. For $f(a)=1$, $f(b)=4$, $f(c)=7$, compute $\int f\,d\mu$ and $\int \tfrac52 f\,d\mu$.

Using the three singleton sets as the disjoint level sets, the simple-function formula gives $$ \int f\,d\mu =1\cdot2+4\cdot\tfrac12+7\cdot0 =2+2+0=4. $$ The coefficient $7$ is positive, but its set has measure zero, so that term is the well-defined product $7\cdot0=0$. After scaling, the values at $a,b,c$ are $\tfrac52,10,\tfrac{35}{2}$, respectively. Hence $$ \int \tfrac52 f\,d\mu =\tfrac52\cdot2+10\cdot\tfrac12+\tfrac{35}{2}\cdot0 =5+5+0=10. $$ The homogeneity formula independently gives $\tfrac52\cdot4=10$. Check: the three pointwise values scale as $1\mapsto\tfrac52$, $4\mapsto10$, and $7\mapsto\tfrac{35}{2}$; substituting them into the weighted sum gives $5+5+0=10$, as above.

Worked Example: Infinite Integral and the Zero Scalar

Let $\mathbb{N}=\{1,2,\ldots\}$ carry counting measure $\mu$, so $\mu(E)$ is the number of elements of $E$, with value $+\infty$ for infinite $E$. Let $f(n)=1$ for every $n$. Find the integrals of $2f$ and $0f$.

The function $f$ itself is simple, and its positive coefficient is on all of $\mathbb{N}$. Thus $\int f\,d\mu=1\cdot\mu(\mathbb{N})=+\infty$. One can also verify this from below: for each positive integer $N$, the simple function $\chi_{\{1,\ldots,N\}}\le f$ has integral $1\cdot N=N$, and these lower bounds are unbounded as $N$ increases. For $2f$, the positive-scalar theorem gives $2\int f\,d\mu=2(+\infty)=+\infty$. Directly, $2f$ is simple with coefficient $2$ on $\mathbb{N}$, so its integral is $2\cdot\mu(\mathbb{N})=+\infty$; its finite truncations have integrals $2N$, also unbounded.

For the zero scalar, $0f$ is identically zero, and its integral is zero. In particular, the statement is not obtained by evaluating $0\cdot(+\infty)$: that product is undefined in extended-real arithmetic. Check: the zero function has only the zero simple minorant, while every positive integer $N$ is a lower bound for neither its integral nor any of its minorants; its integral is exactly $0$.

Why the Zero Case Must Be Separate

For $c>0$, scaling the simple minorants of $f$ gives a one-to-one correspondence with the simple minorants of $cf$: divide by $c$ to go back. The correspondence is unavailable at $c=0$, because division by zero is impossible, and a zero multiple erases all information about $f$. The proof therefore uses the definition directly: $0f$ is the zero function, whose integral is zero.

A common mistake is to write, for a function with infinite integral, $$ \int 0f\,d\mu=0\int f\,d\mu=0\cdot(+\infty)=0. $$ The last equality is not a valid extended-real calculation: $0\cdot(+\infty)$ is undefined. The correct argument is that $0f$ is identically zero, so the supremum of the integrals of its simple minorants is zero. Alternatively, one may state the theorem in two cases, as above, without using a product on the right when $c=0$.

Positive homogeneity applies on arbitrary measure spaces and does not require $\int f\,d\mu$ to be finite. When that integral is finite, the identity is an ordinary finite scaling rule. When it is infinite, every positive finite rescaling still has infinite integral. This property is specific to scalar multiplication; it does not by itself establish a rule for integrating sums of functions.

Check Your Understanding

In each question, state whether the scalar is positive or zero and keep the possibility of an infinite integral in view.

For a nonnegative measurable $f$ on an arbitrary measure space with $\int f\,d\mu=7$, what is $\int 3f\,d\mu$? Justify the value using the theorem.
If $\int f\,d\mu=+\infty$, what is $\int \tfrac14 f\,d\mu$, and why is the scalar-scaling lemma valid in this case?
Explain why $0f$ has integral zero even if $f(x)=+\infty$ at some points.
Let $E\in\mathcal{M}$ have infinite measure and consider the simple function $0\chi_E$. Explain how the simple-function integral is defined without using $0\cdot\mu(E)$.
For counting measure on $\mathbb{N}$, compute $\int 5\chi_{\{1,2,3\}}\,d\mu$, and verify the result by positive homogeneity.