Measurable Level and Equality Sets

From Measurability to Measurable Sets

Measurability of a real-valued function is defined by the measurability of its preimages. Many sets that arise in analysis are precisely such preimages: a level set records where a function lies above or below a fixed number, while an equality set records where two functions take the same value. Comparisons between two functions require a little more care, because the functions need not be combined into a new function to describe the set.

Throughout, let $(X,\mathcal{M},\mu)$ be an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions in this tutorial are finite, real-valued functions. A function $f:X\to\mathbb{R}$ is measurable if $f^{-1}(B)\in\mathcal{M}$ for every Borel set $B\subseteq\mathbb{R}$. We use the half-line tests for real-valued measurability: in particular, if $f$ is measurable, then $\{x:f(x)<a\}$ and $\{x:f(x)\le a\}$ belong to $\mathcal{M}$ for every $a\in\mathbb{R}$. A sigma-algebra is closed under complements and countable unions, and therefore also under countable intersections.

Definition: For a measurable $f:X\to\mathbb{R}$ and $a\in\mathbb{R}$, the strict and non-strict level sets are $\{f<a\}$, $\{f\le a\}$, $\{f>a\}$, and $\{f\ge a\}$. For measurable $f,g:X\to\mathbb{R}$, the comparison sets are $\{f<g\}$, $\{f\le g\}$, $\{f>g\}$, and $\{f\ge g\}$; the equality set is $\{f=g\}$. In each case, for example, $\{f<a\}$ means $\{x\in X:f(x)<a\}$.

Level Sets and Their Endpoints

Taking complements changes a strict inequality into the opposite non-strict inequality. For example, $\{f>a\}=\{f\le a\}^{c}$, not $\{f<a\}^{c}$. The latter complement is $\{f\ge a\}$. This distinction matters at points where $f(x)=a$.

Proposition: If $f:X\to\mathbb{R}$ is measurable on $(X,\mathcal{M},\mu)$, then each of its four level sets at every real number $a$ belongs to $\mathcal{M}$. More explicitly, $$ \{f<a\}=\bigcup_{\substack{q\in\mathbb{Q}\\q<a}}\{f<q\},\qquad \{f\le a\}=\bigcap_{n=1}^{\infty}\{f<a+1/n\},\qquad \{f>a\}=\{f\le a\}^{c},\qquad \{f\ge a\}=\{f<a\}^{c}. $$

Proof. The half-line test gives $\{f<q\}\in\mathcal{M}$ for each rational $q$, so the displayed countable union is measurable. To verify its identity, if $f(x)<a$, density of $\mathbb{Q}$ in $\mathbb{R}$ supplies a rational $q$ with $f(x)<q<a$, and hence $x$ belongs to that union. Conversely, membership in one of its sets gives $f(x)<q<a$, so $f(x)<a$. For the intersection identity, $f(x)\le a$ implies $f(x)<a+1/n$ for every positive integer $n$. If $f(x)>a$, choose $n$ large enough that $1/n<f(x)-a$. Then $f(x)>a+1/n$, so $x\notin\{f<a+1/n\}$. This proves the identity. Finally, the last two identities follow by taking complements, and all four sets are measurable by closure of $\mathcal{M}$ under countable unions, intersections, and complements. $\square$

The rational thresholds in the first identity are countable, which is essential: a sigma-algebra is guaranteed to be closed under countable unions, not arbitrary unions. The intersection identity also displays why the non-strict endpoint is retained: when $f(x)=a$, the inequalities $f(x)<a+1/n$ hold for every $n$.

Comparisons and Equality of Two Functions

For comparisons between measurable functions, rational numbers again provide the needed countable description. A rational number strictly between two distinct real numbers can witness which one is smaller. This avoids assuming at this stage that a difference or any other algebraic combination of measurable functions is measurable.

Theorem: Let $f,g:X\to\mathbb{R}$ be measurable on $(X,\mathcal{M},\mu)$. Then all four comparison sets and the equality set belong to $\mathcal{M}$. In particular, $$ \{f<g\}=\bigcup_{q\in\mathbb{Q}}\bigl(\{f<q\}\cap\{g>q\}\bigr), \qquad \{f=g\}=X\setminus\bigl(\{f<g\}\cup\{g<f\}\bigr). $$ The remaining comparison sets are $\{f\le g\}=\{g<f\}^{c}$, $\{f>g\}=\{g<f\}$, and $\{f\ge g\}=\{f<g\}^{c}$.

Proof. First, for each rational $q$, $\{f<q\}$ and $\{g>q\}$ are measurable by the level-set result, so their intersection is measurable. There are countably many rational numbers, and therefore the union in the formula is measurable. If $f(x)<g(x)$, density of the rationals gives $q\in\mathbb{Q}$ with $f(x)<q<g(x)$, placing $x$ in the union. Conversely, membership in the union means there is a rational $q$ such that $f(x)<q$ and $g(x)>q$; these inequalities imply $f(x)<g(x)$. Thus the first identity holds. Interchanging $f$ and $g$ proves measurability of $\{g<f\}$. At each $x$, exactly one of $f(x)<g(x)$, $f(x)=g(x)$, or $g(x)<f(x)$ holds, because $f(x)$ and $g(x)$ are real numbers. Hence the equality-set identity follows. The other identities follow directly from the definitions and complements; for example, $f(x)\le g(x)$ holds exactly when $g(x)<f(x)$ does not. All these sets are in $\mathcal{M}$. $\square$

These are pointwise set identities. They do not say merely that the relations hold almost everywhere, and they do not require $\mu(X)$ to be finite. If two functions agree outside a null set, their equality set contains the complement of that null set; that is a different statement from equality at every point.

Worked Example: Strict and Non-Strict Levels of a Square

On $\mathbb{R}$ with its Borel sigma-algebra and Lebesgue measure $m$, let $f(x)=x^2$. This is a continuous, hence Borel-measurable, real-valued function. Find the level sets at $a=4$, and check their measures.

The inequality $x^2<4$ is equivalent to $-2<x<2$, so $\{f<4\}=(-2,2)$. The inequality $x^2\le4$ is equivalent to $-2\le x\le2$, so $\{f\le4\}=[-2,2]$. Taking complements gives $\{f>4\}=(-\infty,-2)\cup(2,\infty)$ and $\{f\ge4\}=(-\infty,-2]\cup[2,\infty)$. Equality requires $x^2=4$, which factors as $(x-2)(x+2)=0$; thus $\{f=4\}=\{-2,2\}$.

Check: the interval $(-2,2)$ has length $2-(-2)=4$, so $m(\{f<4\})=4$. The closed interval has the same measure, $m(\{f\le4\})=4$, since adding its two endpoints does not change Lebesgue measure. Each singleton has Lebesgue measure zero, so $m(\{f=4\})=0$. The complements are consistent: the two tails in $\{f>4\}$ exclude the endpoints, whereas $\{f\ge4\}$ includes them.

Worked Example: Comparing Two Affine Functions

On $\mathbb{R}$ with Lebesgue measure, let $f(x)=x$ and $g(x)=2-x$. Find the strict and non-strict comparison sets and the equality set. Both functions are continuous and therefore measurable.

The comparison $f(x)<g(x)$ is $x<2-x$. Adding $x$ to both sides gives $2x<2$, and dividing by the positive number $2$ gives $x<1$. Thus $\{f<g\}=(-\infty,1)$. Replacing the strict inequality by a non-strict one gives $x\le2-x$, hence $2x\le2$ and $x\le1$. Therefore $\{f\le g\}=(-\infty,1]$. Reversing the comparison gives $\{f>g\}=(1,\infty)$ and $\{f\ge g\}=[1,\infty)$.

For equality, solve $x=2-x$: adding $x$ yields $2x=2$, so $x=1$ and $\{f=g\}=\{1\}$. Check: at $x=0$, $f(0)=0<2=g(0)$; at $x=1$, $f(1)=1=g(1)$; and at $x=2$, $f(2)=2>0=g(2)$. These test points confirm the direction of each interval and that the boundary point belongs to both non-strict comparison sets, but neither strict comparison set.

Worked Example: An Equality Set Equal to the Rationals

On $\mathbb{R}$ with its Borel sigma-algebra and Lebesgue measure, define $f(x)=0$ for rational $x$ and $f(x)=1$ for irrational $x$, and let $g(x)=0$ for every $x$. Verify measurability and find the comparison and equality sets.

For any open set $U\subseteq\mathbb{R}$, the preimage $f^{-1}(U)$ is one of $\varnothing$, $\mathbb{Q}$, $\mathbb{R}\setminus\mathbb{Q}$, or $\mathbb{R}$: the possibilities depend on whether $U$ contains $0$, $1$, both, or neither. The rationals are Borel because $\mathbb{Q}$ is a countable union of closed singleton sets. Its complement is Borel as well. Thus these possible preimages are Borel, so $f$ is measurable; the constant function $g$ is measurable too.

On rational inputs, $f(x)=g(x)=0$; on irrational inputs, $f(x)=1>0=g(x)$. Consequently $\{f=g\}=\mathbb{Q}$, $\{f<g\}=\varnothing$, and $\{f>g\}=\mathbb{R}\setminus\mathbb{Q}$. Check: $\mathbb{Q}$ is countable and every singleton has Lebesgue measure zero, so countable subadditivity gives $m(\mathbb{Q})=0$. Since $\mathbb{R}=\mathbb{Q}\,\dot\cup\,(\mathbb{R}\setminus\mathbb{Q})$, with both sets measurable, and $m(\mathbb{R})=\infty$, additivity gives $m(\mathbb{R}\setminus\mathbb{Q})=\infty$. Their Lebesgue measures are not both infinite: $m(\mathbb{Q})=0$, while $m(\mathbb{R}\setminus\mathbb{Q})=\infty$.

Common Endpoint and Measure Pitfalls

A common endpoint error is to solve a non-strict inequality as though it were strict. For example, the incorrect calculation “$x^2\le4$ gives $-2<x<2$” omits the solutions $x=-2$ and $x=2$. Substitution checks the error: $(-2)^2=4$ and $2^2=4$, so both points satisfy $x^2\le4$. The correction is $\{x:x^2\le4\}=[-2,2]$, while $\{x:x^2<4\}=(-2,2)$. Equalities at endpoints can leave the Lebesgue measure unchanged, as in this example, but they still change the set.

A separate mistake is to infer that a measurable set and its complement have the same measure, or that both must have infinite measure. The rational example gives a direct correction: the set and its complement have measures $0$ and $\infty$, respectively. Measurability is a statement about belonging to the sigma-algebra; it does not imply finite measure, positive measure, or any symmetry between a set and its complement.

The countable formulas above are useful whenever a proof needs to restrict attention to where a function satisfies a numerical condition. For example, a claim about $f(x)<g(x)$ can be handled as a measurable-set question without first constructing or proving measurability of a new function. The rational separator is the key: each possible threshold gives a measurable intersection, and the union over rational thresholds remains countable.

Check Your Understanding

Use the set identities and check strictness at any boundary points.

For measurable real-valued $f$, write $\{f\ge a\}$ as a complement of a strict level set. Explain which points with $f(x)=a$ it contains.
Prove directly from density of $\mathbb{Q}$ that $\{f<g\}=\bigcup_{q\in\mathbb{Q}}(\{f<q\}\cap\{g>q\})$.
For $f(x)=x^2$ and $g(x)=1$ on $\mathbb{R}$, find $\{f=g\}$, $\{f<g\}$, and $\{f\le g\}$.
Explain why the equality set of two measurable real-valued functions is measurable, using only the two strict comparison sets and sigma-algebra operations.
For the rational-versus-irrational function in the worked example, state the measures of its equality set with the zero function and of the strict greater-than set.

Measurable Level and Equality Sets

What You'll Learn