What a Bounded Multiplier Guarantees
An integrable function can be multiplied by many other functions without losing integrability, but measurability alone is not enough: the size of the multiplier matters. The useful condition is a finite bound that holds almost everywhere. Under that condition, the product's absolute value is controlled pointwise, and this control passes directly to its integral.
Unless an example specifies otherwise, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; it need not be finite, sigma-finite or complete. Functions in the statements below are finite-valued real- or complex-valued measurable functions. A function \(f\) is integrable when \(\int_X |f|\,d\mu<\infty\), and we write \(\|f\|_1=\int_X|f|\,d\mu\). Integral and norm statements are unchanged by changes on a \(\mu\)-null set. The product of two measurable real- or complex-valued functions is measurable.
The first inequality controls the full \(L^1\) norm of the product. The second controls the absolute value of its signed or complex integral; it follows from the first and the integral triangle inequality. The bound on \(g\) need only hold almost everywhere, not at every point.
Proof. Since \(|g|\le C\) outside a measurable null set and \(|f|\ge0\), we have \(|g(x)f(x)|\le C|f(x)|\) for almost every \(x\in X\). Both sides are measurable and nonnegative. Monotonicity of the nonnegative Lebesgue integral, together with invariance of integrals under changes on null sets, gives $$ \int_X|gf|\,d\mu\le C\int_X|f|\,d\mu<\infty. $$ Thus \(gf\) is integrable by the definition of integrability. The integral triangle inequality gives \(\left|\int_Xgf\,d\mu\right|\le\int_X|gf|\,d\mu\), and combining these inequalities proves the second bound. This argument also covers \(C=0\): then \(g=0\) almost everywhere, so \(gf=0\) almost everywhere and both integrals are zero. \(\square\)
If \(g\) is essentially bounded, its essential supremum \(\|g\|_\infty\) is the infimum of the constants \(C\) for which \(|g|\le C\) almost everywhere. Applying the theorem with such bounds (or with \(\|g\|_\infty+1/n\) and then letting \(n\to\infty\)) yields the concise norm estimate $$ \|gf\|_1\le\|g\|_\infty\|f\|_1. $$ The value \(\|g\|_\infty\) is an almost-everywhere bound: the countable family of bounds \(\|g\|_\infty+1/n\) implies \(|g|\le\|g\|_\infty\) outside a countable union of null sets.
Worked Example: A Signed Product on the Unit Interval
Take \([0,1]\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\). Let \(f(x)=x\) and \(g(x)=2x-1\). Both are measurable and \(f\) is integrable. For every \(x\in[0,1]\), \(|2x-1|\le1\), so the theorem applies with \(C=1\).
First compute the bound's right-hand side: $$ \int_0^1|f(x)|\,dx=\int_0^1x\,dx=\left[\frac{x^2}{2}\right]_0^1=\frac12, \qquad \int_0^1|g(x)f(x)|\,dx\le\frac12. $$ The signed product integral is $$ \int_0^1(2x-1)x\,dx =\int_0^1(2x^2-x)\,dx =\left[\frac{2x^3}{3}-\frac{x^2}{2}\right]_0^1 =\frac23-\frac12=\frac16. $$ To compute the absolute product integral, split at the zero \(x=\frac12\) of \(g\). On \([0,\frac12]\), \(x|2x-1|=x-2x^2\); on \([\frac12,1]\), it equals \(2x^2-x\). Hence $$ \int_0^1|gf|\,dx =\left[\frac{x^2}{2}-\frac{2x^3}{3}\right]_0^{1/2} +\left[\frac{2x^3}{3}-\frac{x^2}{2}\right]_{1/2}^{1} =\left(\frac18-\frac1{12}\right) +\left(\frac16+\frac1{24}\right) =\frac1{24}+\frac5{24}=\frac14. $$ Check: \(\left|\int_0^1gf\,dx\right|=\frac16\le\frac14=\int_0^1|gf|\,dx\le\frac12=C\int_0^1|f|\,dx\), as required.
The Multiplier as an \(L^1\) Operation
For fixed \(g\), the theorem says that multiplication sends each integrable \(f\) to another integrable function. In the usual \(L^1(\mu)\) space, whose elements identify functions equal almost everywhere, this operation is well-defined: if \(f_1=f_2\) almost everywhere, then \(gf_1=gf_2\) almost everywhere because \(g\) is finite-valued. It is linear, and its size is controlled by \(\|g\|_\infty\).
In particular, if \(g\) and \(h\) are measurable and both satisfy finite almost-everywhere bounds, then \(g-h\) is essentially bounded. Applying the theorem to \(g-h\) and \(f\) gives $$ \|(g-h)f\|_1\le\|g-h\|_\infty\|f\|_1. $$ Thus if two multipliers are close in essential supremum, their products with the same integrable function are close in \(L^1\). This estimate follows from the same product bound; it does not require an interchange of limits and integrals.
Worked Example: A Finite Counting-Measure Calculation
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) contain all subsets of \(X\), and let \(\mu\) be counting measure, so the integral is the sum over the three points. Define \(f(1)=2\), \(f(2)=-1\), \(f(3)=3\), and \(g(1)=\frac12\), \(g(2)=-\frac14\), \(g(3)=\frac13\). The largest multiplier magnitude is \(C=\frac12\).
The input norm is $$ \|f\|_1=|2|+|-1|+|3|=2+1+3=6. $$ Compute every product value: \((gf)(1)=\frac12\cdot2=1\), \((gf)(2)=(-\frac14)(-1)=\frac14\), and \((gf)(3)=\frac13\cdot3=1\). Therefore $$ \int_Xgf\,d\mu=1+\frac14+1=\frac94, \qquad \int_X|gf|\,d\mu=1+\frac14+1=\frac94. $$ The theorem's upper bound is \(C\|f\|_1=\frac12\cdot6=3\). Check: \(\left|\int_Xgf\,d\mu\right|=\frac94\le\int_X|gf|\,d\mu=\frac94\le3\). The finite sum also verifies the product and both integrals directly.
There is a useful order-based refinement for nonnegative functions. Suppose \(f\ge0\) is integrable and \(g\) is real-valued and measurable with \(a\le g\le b\) almost everywhere, where \(a,b\) are finite real constants. Then \(gf\) is integrable and $$ a\int_Xf\,d\mu\le\int_Xgf\,d\mu\le b\int_Xf\,d\mu. $$ Indeed, \((g-a)f\ge0\) and \((b-g)f\ge0\) almost everywhere, so monotonicity gives nonnegative integrals for both products. Linearity then gives the two displayed inequalities. Integrability of \(gf\) follows from the bounded-multiplier theorem, since \(|g|\le\max(|a|,|b|)\) almost everywhere. No such order conclusion is asserted for a general signed or complex \(f\).
Worked Example: An Almost-Everywhere Bound
On \([0,1]\) with Lebesgue measure, let \(f(x)=1\) and set \(g(0)=1000\), while \(g(x)=\frac12\) for \(0<x\le1\). This is a measurable real-valued function. The singleton \(\{0\}\) has measure zero, so \(|g|\le\frac12\) almost everywhere, although \(g\) is not bounded by \(\frac12\) at every point.
Since \(\int_0^1|f|\,dx=1\), the theorem with the almost-everywhere constant \(C=\frac12\) gives \(\int_0^1|gf|\,dx\le\frac12\). Directly, \(gf\) equals \(\frac12\) except at \(0\), so $$ \int_0^1gf\,dx=\frac12,\qquad \int_0^1|gf|\,dx=\frac12. $$ Check: the bound reads \(\frac12\le\frac12\cdot1=\frac12\), with equality. The exceptional value \(1000\) contributes zero to the Lebesgue integral because it occurs on a null set.
Why Boundedness Cannot Be Dropped
The condition is a bound on the multiplier, not merely measurability of the multiplier. A large multiplier can turn an integrable function into a nonintegrable one. The following calculation also shows why a finite measure space does not by itself remove the need for a bound.
Worked Example: An Unbounded Multiplier Can Break Integrability
Take \([0,1]\) with Lebesgue measure and define \(f(x)=x\). Let \(g(0)=0\) and \(g(x)=x^{-2}\) for \(0<x\le1\). Both functions are measurable and finite-valued at every point, but \(g\) is not essentially bounded. The input is integrable: $$ \int_0^1|f(x)|\,dx=\int_0^1x\,dx=\frac12. $$ For \(x>0\), \(g(x)f(x)=1/x\), while the product is zero at \(x=0\). For each integer \(n\ge1\), monotonicity of the nonnegative integral gives $$ \int_0^1|gf|\,dx\ge\int_{1/n}^1\frac{dx}{x} =[\ln x]_{1/n}^1=\ln n. $$ As \(n\to\infty\), \(\ln n\to\infty\); hence \(\int_0^1|gf|\,dx=\infty\), and \(gf\) is not integrable. Check: the input integral is the finite value \(\frac12\), but the product integral is at least \(\ln n\) for every \(n\), so it cannot be finite.
A common mistake is to bound the product's integral using a value of \(g\) that is not a valid almost-everywhere bound. For the last example, one might incorrectly write \(|gf|\le 1\cdot|f|\) because \(f\le1\) on \([0,1]\). That compares \(gf\) with \(f\), not with a valid multiplier bound: for \(0<x<1\), \(|g(x)|=x^{-2}>1\). In fact, at \(x=\frac12\), \(|g(x)f(x)|=2\), while \(|f(x)|=\frac12\), so the proposed pointwise inequality already fails there. The correction is to verify \(|g|\le C\) almost everywhere before applying the theorem; here no finite \(C\) works, and the product is indeed nonintegrable.
Check Your Understanding
In each problem, identify the measure space and check that the multiplier bound is valid before using the theorem.
- If \(f\in L^1(\mu)\) and \(|g|\le3\) \(\mu\)-a.e. on an arbitrary measure space, what bounds follow for \(\|gf\|_1\) and \(\left|\int_Xgf\,d\mu\right|\)?
- On \([0,1]\) with Lebesgue measure, take \(f(x)=1\) and \(g(x)=x-\frac12\). Find an almost-everywhere bound \(C\), then calculate \(\int_0^1gf\,dx\) and verify the theorem's integral bound.
- For counting measure on \(\{1,2\}\), let \(f(1)=1,\ f(2)=-2,\ g(1)=-1,\ g(2)=\frac12\). Compute \(\|f\|_1\), \(\|gf\|_1\), and \(\int gf\,d\mu\); check the bound using \(C=1\).
- Why does a multiplier taking a very large value at one point still satisfy an almost-everywhere bound on \([0,1]\) with Lebesgue measure if it equals \(\frac12\) everywhere else?
- Give an integrable function and a measurable unbounded multiplier whose product is not integrable, and identify which hypothesis of the theorem fails.