Nonnegative Measurable Functions

Nonnegative Functions as the Starting Point for Integration

The integral of a nonnegative function can be defined without first subtracting positive and negative contributions. This is why nonnegative functions are the natural starting point for the next stage of Lebesgue integration. Earlier in the course, measurable functions were treated as real-valued; for integration, it is also useful to allow a function to take the value $+\infty$. We will distinguish those two settings carefully.

Throughout, $(X,\mathcal{F})$ is a measurable space. A set such as $\{x:f(x)>a\}$ is called a strict superlevel set of $f$. The Threshold Characterization of Measurability from earlier in the course gives a convenient test for real-valued functions: their strict superlevel sets are measurable. We will use the same threshold language when the function is allowed to take the value $+\infty$.

Definition: A function $f:X\to[0,\infty)$ is a finite-valued nonnegative measurable function if it is measurable as a real-valued function. An extended nonnegative measurable function is a function $f:X\to[0,\infty]$ such that $$ \{x\in X:f(x)>a\}\in\mathcal{F} $$ for every real number $a$. The value $+\infty$ is allowed, but $-\infty$ is not.

For $a<0$, every point of $X$ satisfies $f(x)>a$, so the corresponding strict superlevel set is $X$. For $a\geq 0$, these sets describe where the function is larger than a nonnegative threshold. In particular, the definition does not require $f$ to be strictly positive: it may vanish on some or all of $X$.

Level Sets and the Infinite-Value Set

The threshold definition immediately provides measurable sets that will be used repeatedly. For example, the set where $f$ is at least a fixed nonnegative number can be written using strict superlevel sets: $$ \{x:f(x)\geq a\}=\bigcap_{n=1}^{\infty}\{x:f(x)>a-1/n\}. $$ Indeed, if $f(x)\geq a$, then $f(x)>a-1/n$ for every $n$. If $f(x)<a$, choose $n$ so large that $1/n<a-f(x)$; then $f(x)\leq a-1/n$, so $x$ is excluded from the intersection. The right-hand side is measurable by closure of a sigma-algebra under countable intersections.

Theorem (Measurability of the Infinite-Value Set): If $f:X\to[0,\infty]$ is extended nonnegative measurable, then $$ \{x\in X:f(x)=+\infty\}=\bigcap_{n=1}^{\infty}\{x\in X:f(x)>n\} $$ is measurable.

Proof. Each set $\{x:f(x)>n\}$ is measurable by the definition of extended nonnegative measurability. Their countable intersection is therefore measurable. If $f(x)=+\infty$, then $f(x)>n$ for every positive integer $n$, so $x$ belongs to the intersection. Conversely, suppose $f(x)$ is finite. There is a positive integer $n$ with $n\geq f(x)$, and then $f(x)>n$ is false. Thus $x$ does not belong to the intersection. These two implications prove the set identity and the result. $\square$

Worked Example: A Function That Is Infinite at One Point

Let $X=[0,2]$ with its Borel sigma-algebra, and define $$ f(x)= \begin{cases} +\infty,&x=1,\\ 0,&x\neq 1. \end{cases} $$ For $a<0$, the strict superlevel set $\{f>a\}$ is all of $X$. For $a\geq 0$, it is $\{1\}$, since $+\infty>a$ while $0>a$ is false. Both $X$ and $\{1\}$ are Borel sets, so $f$ is extended nonnegative measurable. Its infinite-value set is exactly $\{1\}$, in agreement with the theorem.

This example also separates measurability from finiteness: the function is measurable even though it is not finite-valued at every point.

Truncation Produces Finite-Valued Measurable Functions

An extended-valued function can be studied through bounded functions obtained by cutting off its values above a chosen level. For $M>0$, define the truncation $f\wedge M$ by $$ (f\wedge M)(x)=\min\{f(x),M\}. $$ This function never exceeds $M$, and it is finite-valued even when $f$ takes the value $+\infty$. The following theorem explains why truncation is a safe operation for measurability.

Theorem (Measurability of Truncations): If $f:X\to[0,\infty]$ is extended nonnegative measurable and $M>0$, then $f\wedge M:X\to[0,M]$ is a finite-valued measurable function.

Proof. For any real $a$, consider the strict superlevel set of $f\wedge M$. If $a<0$, this set is $X$, because $f\wedge M\geq 0$. If $0\leq a<M$, then $$ (f\wedge M)(x)>a \quad\Longleftrightarrow\quad f(x)>a. $$ To verify the equivalence, if $f(x)>a$ and $M>a$, then both entries in $\min\{f(x),M\}$ exceed $a$, so their minimum exceeds $a$. Conversely, if their minimum exceeds $a$, then $f(x)>a$. If $a\geq M$, the strict superlevel set is empty because $f\wedge M\leq M$. Thus every strict superlevel set is either $X$, empty, or a measurable strict superlevel set of $f$. The Threshold Characterization of Measurability now gives measurability of $f\wedge M$. Its values lie in $[0,M]$, so it is finite-valued. $\square$

Worked Example: Truncating an Unbounded Function

On $X=[0,2]$ with Lebesgue measurability, define $f(x)=1/x$ for $x>0$, and set $f(0)=+\infty$. For $a<0$, $\{f>a\}=X$. For $a\geq0$, if $a=0$ the set is $(0,2]\cup\{0\}=X$; if $a>0$, then $$ \{x:f(x)>a\}=\{0\}\cup\bigl((0,2]\cap(0,1/a)\bigr). $$ This is a Lebesgue measurable set, so $f$ is extended nonnegative measurable.

For $M=2$, the truncation equals $2$ at $x=0$ and whenever $0<x<1/2$; it equals $1/x$ when $1/2\leq x\leq2$. In particular, $f\wedge2$ is finite-valued and bounded by $2$, despite the infinite value of $f$ at the origin.

Increasing Truncations Recover the Function

Truncations do more than produce measurable bounded functions: as the cutoff grows, they approximate the original function from below. This gives a pointwise description that applies whether $f$ is finite everywhere or takes the value $+\infty$ somewhere.

Theorem (Increasing Truncation Approximation): Let $f:X\to[0,\infty]$ be extended nonnegative measurable, and for each positive integer $n$ define $f_n=f\wedge n$. Then each $f_n$ is finite-valued and measurable, $$ f_n(x)\leq f_{n+1}(x)\quad\text{for every }x\in X, $$ and $f_n(x)\to f(x)$ pointwise in $[0,\infty]$.

Proof. Each $f_n$ is measurable and finite-valued by the Measurability of Truncations Theorem. Since $n\leq n+1$, taking the minimum with $f(x)$ gives $$ \min\{f(x),n\}\leq\min\{f(x),n+1\}, $$ so $f_n(x)\leq f_{n+1}(x)$.

Fix $x\in X$. If $f(x)=L<\infty$, then for every integer $n\geq L$, $f_n(x)=\min\{L,n\}=L$. Thus the sequence is eventually equal to $f(x)$, and so converges to it. If $f(x)=+\infty$, then $f_n(x)=\min\{+\infty,n\}=n$, which tends to $+\infty$. Hence $f_n(x)\to f(x)$ in either case. $\square$

Worked Example: Increasing Truncations of a Function with an Infinite Value

Return to $X=[0,2]$ and define $f(0)=+\infty$, while $f(x)=3$ for $x>0$. This function is measurable: for $a<3$, its strict superlevel set is $X$; for $a\geq3$, it is $\{0\}$ if $a<+\infty$, with the inequality $3>a$ false. More explicitly, at $a=3$ the set is $\{0\}$, and for $a>3$ it is also $\{0\}$. These are measurable.

For $n=1$, $f_1$ equals $1$ everywhere. For $n=2$, $f_2$ equals $2$ everywhere. For every $n\geq3$, $f_n(0)=n$, while $f_n(x)=3$ for $x>0$. Thus at every $x>0$, the truncations eventually equal $3=f(x)$, while at $0$ they increase without bound to $+\infty=f(0)$.

Pointwise Maxima and Minima

Two other basic constructions preserve nonnegative measurability. For extended nonnegative measurable functions $f$ and $g$, define $\max(f,g)$ and $\min(f,g)$ pointwise. Their superlevel sets have simple descriptions: $$ \{\max(f,g)>a\}=\{f>a\}\cup\{g>a\}, \qquad \{\min(f,g)>a\}=\{f>a\}\cap\{g>a\}. $$ The first identity holds because a maximum exceeds $a$ exactly when at least one entry exceeds $a$; the second holds because a minimum exceeds $a$ exactly when both entries do.

Theorem (Measurability of Pointwise Maxima and Minima): If $f,g:X\to[0,\infty]$ are extended nonnegative measurable, then $\max(f,g)$ and $\min(f,g)$ are extended nonnegative measurable.

Proof. Fix a real number $a$. By the displayed identities, the strict superlevel set of the maximum is a union of two measurable sets, and the strict superlevel set of the minimum is an intersection of two measurable sets. Both are in $\mathcal{F}$. This is true for every real $a$, so the definition of extended nonnegative measurability applies to both functions. $\square$

Worked Example: Level Sets of a Maximum and a Minimum

On $X=[0,3]$ with Lebesgue measure, let $f(x)=x$ and $g(x)=3-x$. Both are nonnegative measurable functions. At each $x$, the larger value is $\max\{x,3-x\}$, and the smaller is $\min\{x,3-x\}$. For the threshold $a=2$, $$ \{\max(f,g)>2\}=[0,1)\cup(2,3], \qquad \{\min(f,g)>2\}=\varnothing. $$ Indeed, $x>2$ holds on $(2,3]$, and $3-x>2$ holds on $[0,1)$, giving the maximum's superlevel set by union. Both inequalities would have to hold for the minimum to exceed $2$, which cannot happen, giving the empty set by intersection. These sets are measurable, as the theorem predicts.

Why Truncation Matters

A nonnegative measurable function may be unbounded or may take the value $+\infty$, but its truncations are ordinary finite-valued measurable functions. The Increasing Truncation Approximation therefore lets us work with finite functions while retaining pointwise information about the original one. This will be useful when the nonnegative integral is defined through simple functions lying below the function.

One should not confuse pointwise truncation with changing the function on a negligible set. If $f=+\infty$ on a measurable set, then each truncation is finite there, but the truncations increase to $+\infty$ at every point of that set. No assumption that the infinite-value set has measure zero has been made. Nor does measurability by itself imply that an integral is finite; that question requires additional information about both the function and the measure.

Key takeaway: Nonnegative measurable functions may be finite-valued or extended-valued. Their level sets are measurable, their finite truncations remain measurable, and the increasing sequence $f\wedge n$ converges pointwise to $f$.

Check Your Understanding

Use the level-set definition and truncation arguments to check your understanding.

Why does the definition of extended nonnegative measurability require the sets $\{f>a\}$ to be measurable for every real $a$?
Show that $\{f\geq a\}=\bigcap_{n=1}^{\infty}\{f>a-1/n\}$ for a nonnegative measurable $f$ and $a\geq0$.
What is the infinite-value set of an extended nonnegative measurable function in terms of its strict superlevel sets?
For $0\leq a<M$, why does $\{f\wedge M>a\}=\{f>a\}$?
Explain why $f\wedge n$ converges to $+\infty$ at a point where $f$ equals $+\infty$.
Write the strict superlevel set of $\min(f,g)$ as an intersection of superlevel sets.