Arithmetic with Extended-Real-Measurable Functions

Arithmetic Requires More Care at Infinity

For real-valued measurable functions, sums and products are measurable. With extended-real-valued functions, the same conclusion holds only where the arithmetic operation is defined. In particular, $+\infty+(-\infty)$ and $0\cdot(+\infty)$ are not extended-real numbers. The point is not merely to check measurability: first identify the points where the formula has a value.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions in this tutorial take values in $[-\infty,\infty]$ and are $\mathcal{M}$-measurable, meaning measurable as maps into the extended real line with its Borel sigma-algebra. Statements about pointwise arithmetic and measurability are everywhere statements, not merely almost-everywhere statements. The endpoint level-set criterion established earlier implies that $\{f=+\infty\}$, $\{f=-\infty\}$, and $\{f\in\mathbb{R}\}$ are measurable for every such $f$.

Definition: The domain of a pointwise arithmetic expression is the set of $x\in X$ at which its scalar arithmetic is defined in $[-\infty,\infty]$. A function defined on a measurable subset $D\subseteq X$ is measurable on $D$ with respect to the trace sigma-algebra $\{E\cap D:E\in\mathcal{M}\}$. An expression undefined on $X\setminus D$ is not, by itself, a function on all of $X$.

For addition, the scalar sum is defined unless one term is $+\infty$ and the other is $-\infty$. For multiplication, the product is defined unless one term is zero and the other is an infinity. Where defined, the usual sign rules apply: two infinities with the same sign have product $+\infty$, and two with opposite signs have product $-\infty$. In particular, $+\infty+2=+\infty$, $(-\infty)+(-\infty)=-\infty$, and $(-3)(+\infty)=-\infty$.

Defined Sums and Products

Theorem: Let $f,g:X\to[-\infty,\infty]$ be measurable on $(X,\mathcal{M},\mu)$. Define $$ D_+=X\setminus\bigl(\{f=+\infty,\ g=-\infty\}\cup\{f=-\infty,\ g=+\infty\}\bigr). $$ Then $f+g$ is defined and measurable on $D_+$. It is an extended-real-measurable function on all of $X$ if $D_+=X$. Define $$ D_\times=X\setminus\bigl(\{f=0,\ |g|=\infty\}\cup\{|f|=\infty,\ g=0\}\bigr), $$ where $\{|g|=\infty\}=\{g=+\infty\}\cup\{g=-\infty\}$. Then $fg$ is defined and measurable on $D_\times$, and is measurable on all of $X$ if $D_\times=X$.

Proof. All the sets appearing in $D_+$ and $D_\times$ are measurable: singleton level sets of measurable extended-real functions are measurable, and finite intersections, unions, and complements of measurable sets are measurable. Thus both domains are measurable.

For the sum, partition $D_+$ into the measurable sets determined by whether $f$ and $g$ are finite or infinite. On $\{f,g\in\mathbb{R}\}$, the ordinary sum of the restricted real-valued measurable functions is measurable by the real-valued sum theorem. On $\{f=+\infty,\ g\in\mathbb{R}\}$ and $\{f\in\mathbb{R},\ g=+\infty\}$, the sum is constantly $+\infty$; the corresponding sets with $-\infty$ give the constant value $-\infty$. On $\{f=g=+\infty\}$ it is $+\infty$, and on $\{f=g=-\infty\}$ it is $-\infty$. These cases exhaust $D_+$, because the two opposite-infinity cases have been removed. A function defined measurably on each of finitely many measurable pieces is measurable on their union: the inverse image of any Borel set is the union of its inverse images on those pieces. Hence $f+g$ is measurable on $D_+$.

For the product, partition $D_\times$ according to whether each factor is finite or infinite, and, for finite factors, according to whether they are zero, positive, or negative. On the piece where both factors are finite, the real-valued product theorem gives measurability. If exactly one factor is infinite, the other is finite and nonzero on $D_\times$. Split that piece into where the finite factor is positive and where it is negative; on each resulting measurable piece the product is respectively the infinity with the same sign as the infinite factor, or the infinity with the opposite sign. If both factors are infinite, split into the four measurable sign combinations; on each piece the product is a constant $+\infty$ or $-\infty$ according to the sign rule. The excluded cases are exactly zero times infinity. The finite-piece argument therefore proves measurability on $D_\times$, as claimed. $\square$

The proof uses the earlier real-valued sum and product results only on the region where both functions are finite. The remaining regions require no real arithmetic: the output is a fixed infinity on each of finitely many measurable pieces.

Worked Example: A Sum with Both Signs of Infinity

Let $X=[0,1]$ with the Lebesgue sigma-algebra and Lebesgue measure $m$. Partition $X$ into $A=[0,\tfrac13)$, $B=[\tfrac13,\tfrac23)$, and $C=[\tfrac23,1]$. Define $f=+\infty$ on $A$, $f=-\infty$ on $B$, and $f=2$ on $C$; define $g=5$ on $A$, $g=-\infty$ on $B$, and $g=-3$ on $C$. These are measurable because they are constant on measurable pieces.

On $A$, $f+g=+\infty+5=+\infty$. On $B$, $f+g=-\infty+(-\infty)=-\infty$. On $C$, both values are finite and $f+g=2+(-3)=-1$. There is no point with opposing infinities, so $D_+=X$, and the sum is a measurable function on all of $X$. Check: its level set $\{f+g>0\}$ is exactly $A$, since $+\infty>0$, while $-\infty$ and $-1$ are not greater than zero. Its level set $\{f+g=-1\}$ is exactly $C$. Both are measurable.

Worked Example: A Product with an Infinite Factor

On the same measurable space, let $A=[0,\tfrac12)$. Set $f=+\infty$ on $A$ and $f=-2$ on $A^c$; set $g=3$ on $A$ and $g=-4$ on $A^c$. On $A$, the finite factor is nonzero and positive, so $fg=(+\infty)3=+\infty$. On $A^c$, both factors are finite and $(-2)(-4)=8$. Thus $D_\times=X$ and $$ fg(x)= \begin{cases} +\infty,&x\in A,\\ 8,&x\in A^c. \end{cases} $$ The function is measurable because each of its two values is assigned on a measurable set. Check: $\{fg>10\}=A$, since $+\infty>10$ and $8$ is not greater than $10$; also $\{fg=8\}=A^c$. These level sets agree with the formula and are measurable.

Order Operations Always Have Values

Unlike addition and multiplication, the pointwise maximum and minimum of two extended-real numbers are always defined. Write $f\vee g=\max(f,g)$ and $f\wedge g=\min(f,g)$. Their measurability follows from level-set identities, without excluding any points.

Proposition: If $f,g:X\to[-\infty,\infty]$ are measurable on $(X,\mathcal{M},\mu)$, then $f\vee g$ and $f\wedge g$ are extended-real-measurable on $X$.

Proof. Fix any real number $a$. Pointwise, $\max(f(x),g(x))>a$ holds exactly when at least one of $f(x)>a$ and $g(x)>a$ holds. Therefore $$ \{f\vee g>a\}=\{f>a\}\cup\{g>a\}. $$ Likewise, $\min(f(x),g(x))>a$ holds exactly when both $f(x)>a$ and $g(x)>a$ hold, so $$ \{f\wedge g>a\}=\{f>a\}\cap\{g>a\}. $$ The right-hand sets are measurable because $f$ and $g$ are measurable. The extended-real half-line criterion for measurability now gives measurability of both functions. The identities also hold when either value is an endpoint infinity: they are equivalences of the order relation, not computations involving infinity arithmetic. $\square$

Worked Example: Maximum and Minimum with Endpoint Values

On $X=[0,1]$ with Lebesgue measure, let $A=[0,\tfrac12)$, and define $f=+\infty$ on $A$, $f=-\infty$ on $A^c$, and $g(x)=x$ everywhere. Both functions are measurable. On $A$, $\max(f,g)=+\infty$ and $\min(f,g)=x$. On $A^c$, $\max(f,g)=x$ and $\min(f,g)=-\infty$. Hence both order operations are defined everywhere.

For a direct level-set check at the threshold $a=\tfrac14$, $A\subseteq\{f\vee g>\tfrac14\}$ because the maximum is $+\infty$ there. On $A^c$, the maximum is $x$, so it is greater than $\tfrac14$ exactly for $x\in(\tfrac14,1]$. Thus $\{f\vee g>\tfrac14\}=[0,1]$. For the minimum, on $A$ it equals $x$, and on $A^c$ it equals $-\infty$, so $\{f\wedge g>\tfrac14\}=(\tfrac14,\tfrac12)$. The endpoint $\tfrac12$ is excluded because it belongs to $A^c$, where the minimum is $-\infty$. These sets are measurable, as the proposition predicts.

Differences, Scalar Multiplication, and the Main Pitfall

Subtraction is covered by the sum rule after negating the second function. Negation preserves extended-real measurability: it exchanges the endpoint level sets and reverses the finite level tests. Consequently $f-g$ is defined and measurable precisely on the region where it is not forced to contain an indeterminate difference of equal infinities: $+\infty-(+\infty)$ and $-\infty-(-\infty)$ are undefined. At all other points, subtraction follows the usual extended-real rules. In particular, finite minus an infinity is defined, as is $+\infty-(-\infty)=+\infty$.

For a real constant $c\ne0$, the product $cf$ is defined everywhere and measurable, by the product theorem with the constant function $g(x)=c$. If $c=0$, the scalar expression $0f$ is defined only where $f$ is finite; at points where $f=\pm\infty$, it is the indeterminate product $0\cdot(\pm\infty)$. This distinction matters when writing formulas involving extended-real-valued functions.

Common mistake: treating infinity as an ordinary number. If $f=+\infty$ and $g=-\infty$ on a measurable set $E$, it is incorrect to write $f+g=0$ on $E$ by informal cancellation. The scalar sum has no value there, so $f+g$ is not defined on $E$. For a concrete example, take $X=[0,1]$, $E=[0,\tfrac12)$, $f=+\infty$ and $g=-\infty$ on $E$, with $f=g=0$ on $E^c$. The formula defines a sum equal to $0$ on $E^c$, but defines nothing on $E$. It therefore does not define an extended-real-valued function on $X$.

One may separately define a new function by assigning, for example, the value $0$ on $E$ and the value $0$ on $E^c$. That new function is measurable, but its value on $E$ is an explicit extension, not the sum $f+g$. The same warning applies to setting $0\cdot\infty=0$: an assigned value may produce a measurable extension, but it does not make the original product defined under the standard extended-real arithmetic rules.

The practical test is therefore two-stage: first find the measurable set on which the scalar operation is defined; then use the finite-valued arithmetic theorem and measurable case divisions to establish measurability there. Maxima and minima need no such domain check, because order comparisons remain defined at both infinities.

Check Your Understanding

For each question, distinguish the domain of the arithmetic expression from the measurability of any function it defines.

For measurable extended-real-valued $f,g$, write the set on which $f+g$ is undefined.
Give the domain of $fg$ when $f=0$ on $E$, $f=2$ on $E^c$, $g=+\infty$ on $E$, and $g=3$ on $E^c$, where $E\in\mathcal{M}$.
Explain why $\{f\vee g>a\}=\{f>a\}\cup\{g>a\}$ for real $a$, including when one value is $+\infty$.
For $f=+\infty$ on $E$ and $g=-\infty$ on $E$, can $f+g$ be assigned the value $0$ there by extended-real arithmetic? What would such an assignment represent?
State the domain of $f-g$ in terms of the level sets where $f$ and $g$ equal $+\infty$ or $-\infty$.

Arithmetic with Extended-Real-Measurable Functions

What You'll Learn