Integral Bounds from Pointwise Estimates

From Pointwise Information to Integral Bounds

A pointwise estimate can give useful information about an integral even when the function is not integrated exactly. If a nonnegative function is at least a fixed amount on a measurable set, that set contributes at least its measure multiplied by that amount. Conversely, a uniform upper bound on a set limits the integral over that set. The size of the set matters: a bound on a small set does not by itself give an upper bound for the integral over the whole space.

Throughout, \((X,\mathcal{M},\mu)\) is any measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions \(f:X\to[0,\infty]\) are measurable, and their integrals may equal \(+\infty\). We use the established facts that the nonnegative integral is monotone and that, for \(E\in\mathcal{M}\) and finite \(c\ge0\), \(\int c\chi_E\,d\mu=c\mu(E)\). In the product \(c\mu(E)\), the value is \(0\) when \(c=0\), including when \(\mu(E)=\infty\). Inequalities stated \(\mu\)-a.e. may be used in integral bounds because changing a nonnegative measurable function on a null set does not change its integral.

Proof. First suppose the inequalities hold everywhere on \(E\). The function \(f\chi_E\) is nonnegative and measurable. The pointwise estimate \(f\chi_E\ge a\chi_E\) and monotonicity give \(\int_E f\,d\mu=\int_X f\chi_E\,d\mu\ge\int_X a\chi_E\,d\mu=a\mu(E)\). Since \(f\ge0\) on \(X\), we also have \(f\ge a\chi_E\) on all of \(X\), so monotonicity gives \(\int_X f\,d\mu\ge a\mu(E)\). For the upper bound, \(f\chi_E\le b\chi_E\) everywhere, and therefore \(\int_E f\,d\mu=\int_X f\chi_E\,d\mu\le\int_X b\chi_E\,d\mu=b\mu(E)\). If either estimate holds only \(\mu\)-a.e. on \(E\), its failure is confined to a null set; changing \(f\chi_E\) on that set does not affect its integral, so the same inequalities follow. \(\square\)

The lower estimate on \(X\) uses the nonnegativity of \(f\) outside \(E\). The upper estimate is only for the integral over \(E\): to bound \(\int_X f\,d\mu\) from above, one needs suitable control over the rest of \(X\) as well. If \(E\) has infinite measure and \(a>0\), the lower bound says \(\int_E f\,d\mu=\infty\). If \(E\) has measure zero, both estimates on \(E\) give no positive lower bound, and \(\int_E f\,d\mu=0\).

Upper Estimates and the Role of the Set

An upper estimate \(f\le b\) on \(E\) yields \(\int_E f\,d\mu\le b\mu(E)\). It does not imply \(\int_X f\,d\mu\le b\mu(E)\) unless the contribution outside \(E\) is zero or is separately bounded. A useful whole-space version is immediate: if \(f\le b\) \(\mu\)-a.e. on \(X\) and \(\mu(X)<\infty\), then \(\int_X f\,d\mu\le b\mu(X)\). The finite-measure hypothesis makes this a finite numerical upper bound when \(b\) is finite; if \(\mu(X)=\infty\), the inequality remains true but may say only that the integral is at most \(+\infty\).

Sets Where the Function Exceeds a Threshold

For a measurable \(f:X\to[0,\infty]\) and a finite threshold \(t\ge0\), the set \(E_t=\{x\in X:f(x)>t\}\) is measurable. Since \(f\ge t\) on \(E_t\), the lower-bound theorem gives \(\int_X f\,d\mu\ge t\mu(E_t)\). The set is defined using the strict condition \(f>t\), but the resulting integral estimate is non-strict. This bound is useful when the size of a level set is known, even if the values of \(f\) elsewhere are not.

Strict pointwise inequality can yield a strict integral bound when the set has positive finite measure. Merely replacing \(f\ge a\) by \(f>a\) does not always justify a strict bound: if \(E\) is null, its integral is zero. The next result states precisely a setting where strictness does follow.

Proof. For each positive integer \(n\), define the measurable set \(H_n=\{x\in E:f(x)-a\ge1/n\}\), interpreting \(\infty-a=\infty\). Because \(f>a\) a.e. on \(E\), the union of the sets \(H_n\) covers \(E\) up to a null set: for every point where \(f>a\), the positive difference \(f-a\), including the value \(+\infty\), is at least \(1/n\) for some positive integer \(n\). If every \(H_n\) had measure zero, their countable union would have measure zero, contradicting \(\mu(E)>0\). Thus some \(H_n\) has \(\mu(H_n)>0\). On \(E\setminus H_n\), \(f\ge a\); on \(H_n\), \(f\ge a+1/n\). Hence \(f\chi_E\ge a\chi_E+(1/n)\chi_{H_n}\) a.e. Integrating and using additivity for nonnegative simple functions and monotonicity gives \(\int_E f\,d\mu\ge a\mu(E)+(1/n)\mu(H_n)>a\mu(E)\). The final inequality is strict because \(1/n>0\) and \(\mu(H_n)>0\). \(\square\)

Almost-Everywhere Estimates and a Common Pitfall

Pointwise bounds need not hold at every point to control an integral. For example, if \(f\le b\) except on a measurable null set, the upper bound on \(\int_E f\,d\mu\) is unchanged: the exceptional set contributes zero to the integral. This does not mean that the function is bounded everywhere. It means only that changes on a null set do not affect its integral. In particular, a function may take the value \(+\infty\) on a null set and still have a finite integral.

Common mistake: using the measure of the whole space instead of the level set. On \(X=[0,3]\) with Lebesgue measure, let \(f=2\chi_{[0,1]}\) and \(t=1\). The strict level set is \(E_t=\{x:f(x)>1\}=[0,1]\), so \(m(E_t)=1\). The valid estimate is \(\int_X f\,dm\ge1\cdot1=1\), and the exact integral is \(2\cdot1=2\). A wrong calculation would use \(m(X)=3\) and claim \(\int_X f\,dm\ge1\cdot3=3\). That contradicts the exact value \(2\). The correction is to use the set on which the threshold condition holds: the function is not greater than \(1\) throughout \(X\), only on \(E_t\). The endpoints cause no ambiguity here: \(f(0)=f(1)=2>1\), while \(f(x)=0\) for \(1<x\le3\).

These estimates are elementary consequences of monotonicity, but they are useful throughout integration theory. They turn information about the size of a function on a set into information about the integral, without requiring an exact integral calculation. When applying them, identify the domain of the pointwise estimate, use that domain's measure, and check whether the conclusion is intended for the integral over that set or over the entire space.