Measurable Functions Under Absolute Value, Maximum, and Minimum
Measurability is preserved by many familiar operations on real numbers. In this tutorial we establish this for absolute values and for the pointwise maximum and minimum of two functions, then use these operations to express and check concrete piecewise formulas. Throughout the general results, \((X,\mathcal{M},\mu)\) is an arbitrary measure space: it is not assumed finite, sigma-finite, or complete. All functions in those results are real-valued and all function identities hold everywhere, not merely almost everywhere.
A real-valued function \(f:X\to\mathbb{R}\) is measurable if \(f^{-1}(U)=\{x\in X:f(x)\in U\}\) belongs to \(\mathcal{M}\) for every open set \(U\subseteq\mathbb{R}\). We use the earlier results that sums and differences of real-valued measurable functions are measurable, and that composing a measurable map with a continuous real-valued map preserves measurability. Neither result requires the measure space to have finite measure.
Why These Operations Preserve Measurability
The absolute-value map \(s\mapsto |s|\) is continuous: the triangle inequality gives \(||s|-|t||\le |s-t|\) for all real \(s,t\). Thus, if \(f:X\to\mathbb{R}\) is measurable, the continuous-composition result implies that \(x\mapsto |f(x)|\) is measurable.
For maxima and minima, the key is to write each operation using sums, differences, and absolute value. For all \(s,t\in\mathbb{R}\),
To check the identities, first suppose \(s\ge t\). Then \(|s-t|=s-t\), so the first expression is \((s+t+s-t)/2=s\), and the second is \((s+t-s+t)/2=t\). These are the maximum and minimum, respectively. If \(s<t\), then \(|s-t|=t-s\); the first expression becomes \((s+t+t-s)/2=t\), and the second becomes \((s+t-t+s)/2=s\), again giving the maximum and minimum. The equality case \(s=t\) is included in the first case and gives \(s=t\) for both expressions.
Proof. The function \(|f|\) is measurable by continuity of absolute value and the continuous-composition result. Since \(f\) and \(g\) are measurable, their difference \(f-g\) is measurable. Therefore \(|f-g|\) is measurable by the same absolute-value argument. The sums \(f+g+|f-g|\) and \(f+g-|f-g|\) are measurable by closure under sums and differences. Multiplication by the constant \(1/2\) preserves measurability, again by the arithmetic rules. Applying the two pointwise identities proved above gives \(\max(f,g)\) and \(\min(f,g)\), so both are measurable. Every operation here is applied to finite real values, hence the resulting functions are real-valued everywhere. \(\square\)
There is also a direct way to read measurability from strict level sets. For every \(a\in\mathbb{R}\),
These are exact set identities: a maximum is strictly above \(a\) precisely when at least one entry is, while a minimum is strictly above \(a\) precisely when both entries are. The half-line test for real-valued measurability says that a function is measurable if its sets \(\{h>a\}\) are measurable for every real \(a\). Thus the identities provide a second verification, using only unions and intersections of measurable sets. The strict inequality matters: replacing \(>\) with \(\ge\) gives different level sets, although those too can be handled by the appropriate level-set test.
Worked Examples
Worked Example: Computing All Three Operations on a Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the power set of \(X\), and let \(\mu\) be counting measure. This is a finite, complete measure space. Every real-valued function on it is measurable because every subset of \(X\) belongs to \(\mathcal{M}\). Define \(f\) and \(g\) by the table and compute \(|f|\), \(\max(f,g)\), and \(\min(f,g)\).
| Point | \(f\) | \(g\) | \(|f|\) | \(\max(f,g)\) | \(\min(f,g)\) |
|---|---|---|---|---|---|
| \(a\) | \(-2\) | \(3\) | \(2\) | \(3\) | \(-2\) |
| \(b\) | \(1\) | \(1\) | \(1\) | \(1\) | \(1\) |
| \(c\) | \(4\) | \(-2\) | \(4\) | \(4\) | \(-2\) |
Check the formulas at each point. At \(a\), \(f(a)-g(a)=-2-3=-5\), so \(|f(a)-g(a)|=5\); the formula gives \(\max=(-2+3+5)/2=3\) and \(\min=(-2+3-5)/2=-2\). At \(b\), the difference is \(1-1=0\), giving \(\max=(1+1+0)/2=1\) and \(\min=(1+1-0)/2=1\). At \(c\), the difference is \(4-(-2)=6\), giving \(\max=(4-2+6)/2=4\) and \(\min=(4-2-6)/2=-2\). The absolute values are respectively \(|-2|=2\), \(|1|=1\), and \(|4|=4\). These calculations agree with the direct comparisons in the table, including the tie at \(b\).
Check: At every point, the computed maximum is at least each input and the computed minimum is at most each input. At \(a\), for example, \(-2\le3\) and the outputs are \(3\) and \(-2\); at \(c\), \(-2\le4\) and the outputs are \(4\) and \(-2\).
Worked Example: Two Crossing Affine Functions
On \((\mathbb{R},\mathcal{L},m)\), Lebesgue measure on the Lebesgue sigma-algebra, define \(f(x)=x-1\) and \(g(x)=2-x\). This measure space is complete and sigma-finite, but has infinite total measure. The affine functions are continuous and hence measurable. Find their maximum and minimum, first using the formulas and then by locating where their order changes.
The difference is \(f(x)-g(x)=(x-1)-(2-x)=2x-3\). Therefore the formulas give
The functions are equal when \(2x-3=0\), that is, at \(x=3/2\). If \(x\le3/2\), then \(2x-3\le0\), so \(f(x)\le g(x)\); if \(x\ge3/2\), then \(f(x)\ge g(x)\). Thus the same functions can be written pointwise as
At the shared endpoint \(x=3/2\), both branches give \(1/2\), since \(2-3/2=1/2\) and \(3/2-1=1/2\). At \(x=0\), the inputs are \(f(0)=-1\) and \(g(0)=2\); the formula gives \(\max=(1+|-3|)/2=(1+3)/2=2\) and \(\min=(1-3)/2=-1\). At \(x=2\), the inputs are \(f(2)=1\) and \(g(2)=0\); the formula gives \(\max=(1+|1|)/2=1\) and \(\min=(1-|1|)/2=0\).
Check: At \(x=0\), the piecewise expressions give maximum \(2-0=2\) and minimum \(0-1=-1\), matching the formula. At \(x=2\), they give maximum \(2-1=1\) and minimum \(2-2=0\), again matching. The formulas established above guarantee measurability everywhere, including at the crossing point.
Worked Example: Clipping a Real Number to an Interval
On \((\mathbb{R},\mathcal{L},m)\), consider \(q(x)=\min(2,\max(0,x))\). This combines a maximum and a minimum, so it is measurable by the theorem. Compute its piecewise form and check the boundary values.
If \(x\le0\), then \(\max(0,x)=0\), and consequently \(q(x)=\min(2,0)=0\). If \(0\le x\le2\), then \(\max(0,x)=x\), and \(\min(2,x)=x\). If \(x\ge2\), then \(\max(0,x)=x\), and \(\min(2,x)=2\). Thus
The formulas agree where the intervals overlap: at \(x=0\), the first two branches both give \(0\); at \(x=2\), the last two both give \(2\). Direct evaluation checks the construction: \(q(-1)=\min(2,\max(0,-1))=\min(2,0)=0\); \(q(1)=\min(2,\max(0,1))=\min(2,1)=1\); and \(q(3)=\min(2,\max(0,3))=\min(2,3)=2\). In each case the result lies in \([0,2]\), as clipping requires.
Check: Substituting \(-1,1,3\) into the piecewise formula gives \(0,1,2\), respectively, the same values as direct evaluation. The endpoint checks also show why overlapping branch definitions cause no conflict in this example.
Worked Example: A Tent Function from Absolute Value and Maximum
On \((\mathbb{R},\mathcal{L},m)\), define \(T(x)=\max(0,1-|x|)\). Since \(x\mapsto |x|\) is measurable and maximum preserves measurability, \(T\) is measurable. To find its explicit form, note that \(1-|x|\ge0\) exactly when \(|x|\le1\), or equivalently \(-1\le x\le1\). Outside this interval the maximum selects zero. Inside it the maximum selects \(1-|x|\), which is \(1+x\) for \(-1\le x\le0\) and \(1-x\) for \(0\le x\le1\). Hence
At the junctions the values agree: \(T(-1)=\max(0,1-1)=0\), \(T(0)=\max(0,1-0)=1\), and \(T(1)=\max(0,1-1)=0\). As additional checks, \(T(-1/2)=\max(0,1-1/2)=1/2\) and \(T(2)=\max(0,1-2)=0\); the piecewise formula gives \(1+(-1/2)=1/2\) and \(0\), respectively.
A Common Sign Error and the Scope of the Result
A frequent error is to use the minus sign in the maximum formula. For the concrete inputs \(s=2\) and \(t=5\), the incorrect calculation \((s+t-|s-t|)/2=(2+5-|2-5|)/2=(7-3)/2=2\) gives the minimum, not the maximum. The correction is to use the plus sign for the maximum: \((2+5+|2-5|)/2=(7+3)/2=5\). The identity can be checked by first deciding whether \(s\ge t\) or \(s<t\); the absolute value changes sign exactly at that comparison.
These results concern real-valued functions and the maximum or minimum of two values at each point. They do not assume that the functions are integrable, bounded, or continuous. They also do not, by themselves, assert a result for a supremum or infimum over an arbitrary family of functions. For the operations treated here, the practical procedure is precise: verify that the inputs are measurable, express the desired output using absolute value, sums, differences, maxima, or minima, and check any displayed piecewise form at its boundary points.
Check Your Understanding
For each question, distinguish the pointwise algebra from the measurability conclusion.
- For real-valued measurable \(f,g\), derive the formula for \(\max(f,g)\) by considering separately the cases \(f(x)\ge g(x)\) and \(f(x)<g(x)\).
- On \((\mathbb{R},\mathcal{L},m)\), write \(\min(x,4-x)\) in piecewise form and check that both branches agree where \(x=4-x\).
- Use strict level sets to express \(\{\min(f,g)>a\}\) in terms of \(f\) and \(g\). Why must both functions exceed \(a\)?
- Show that \(r(x)=\max(-1,\min(1,x))\) is measurable on \((\mathbb{R},\mathcal{L},m)\), and compute its values at \(-3\), \(0\), and \(2\).
- For \(s=-4\) and \(t=1\), compute both the maximum and minimum from the absolute-value formulas, showing which sign gives each answer.