Lifting Integral Bounds from Simple Functions

From a Simple-Function Estimate to a General One

Many integral inequalities are easiest to verify when a function takes only finitely many values. For a nonnegative simple function, the integral is a finite sum, so estimates can be checked on its level sets. The challenge is to pass from that calculation to a general measurable function without assuming that the desired inequality already holds in the general case.

The standard route is to choose nonnegative simple functions increasing pointwise to the function of interest. If the two sides of the estimate are integrals that also increase along this approximation, the Monotone Convergence Theorem transfers the inequality to the limit. The direction of approximation matters: pointwise convergence on its own does not provide this conclusion.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Functions in this tutorial are measurable and take values in \([0,\infty]\). All pointwise statements are meant everywhere unless an almost-everywhere qualification is given. Integrals may equal \(+\infty\), so inequalities are interpreted in the extended order on \([0,\infty]\).

A simple function is a measurable function with finite range. It can be written as \(\varphi=\sum_{j=1}^r c_j\chi_{E_j}\), where \(c_j\geq0\) are finite constants and the measurable sets \(E_j\) are disjoint. Its integral is \(\int\varphi\,d\mu=\sum_{j=1}^r c_j\mu(E_j)\). We use the earlier result on simple-function approximation: for every measurable \(f:X\to[0,\infty]\), there are nonnegative simple functions \(\varphi_n\) with \(\varphi_n\uparrow f\) pointwise. One standard choice is \(\varphi_n=2^{-n}\lfloor 2^n\min(f,n)\rfloor\), with the value at \(f=\infty\) interpreted using \(\min(\infty,n)=n\). The Monotone Convergence Theorem says that if measurable \(u_n:X\to[0,\infty]\) increase pointwise to \(u\), then \(\int u_n\,d\mu\uparrow\int u\,d\mu\).

The Simple-Function Lifting Principle

Proof. Fix a measurable \(f:X\to[0,\infty]\) and choose nonnegative simple \(\varphi_n\uparrow f\). For each \(n\), the hypothesis applied to \(\varphi_n\) gives \(\int a\varphi_n\,d\mu\leq\int b\varphi_n\,d\mu\). We check that the Monotone Convergence Theorem applies to both sequences of products. At a point where \(a(x)<\infty\), \(\varphi_n(x)\uparrow f(x)\) implies \(a(x)\varphi_n(x)\uparrow a(x)f(x)\). At a point where \(a(x)=\infty\), use the convention \(0\cdot\infty=0\): if \(f(x)=0\), every \(\varphi_n(x)=0\) and the products remain zero; if \(f(x)>0\), some \(\varphi_n(x)>0\), after which the products equal \(+\infty\). Thus \(a\varphi_n\uparrow af\) everywhere. The same argument gives \(b\varphi_n\uparrow bf\). These products are measurable and nonnegative, so monotone convergence gives \[ \int a\varphi_n\,d\mu\uparrow\int af\,d\mu, \qquad \int b\varphi_n\,d\mu\uparrow\int bf\,d\mu. \] Taking limits in the inequalities for each \(n\) preserves the order, including when a limit is infinite. Hence \(\int af\,d\mu\leq\int bf\,d\mu\). \(\square\)

The condition on all simple functions can often be checked just on indicators. This is useful because an indicator tests the estimate on a single measurable set.

Proof. If the inequality holds for every simple function, take \(\varphi=\chi_E\). Since \(a\chi_E\) equals \(a\) on \(E\) and zero on \(E^c\), its integral is \(\int_Ea\,d\mu\); the corresponding identity holds for \(b\). This proves the set inequality. Conversely, suppose the set inequality holds for every measurable \(E\). Write a nonnegative simple function as \(\varphi=\sum_{j=1}^r c_j\chi_{E_j}\), with disjoint \(E_j\) and \(c_j\geq0\). Additivity of the nonnegative integral gives \[ \int a\varphi\,d\mu=\sum_{j=1}^r c_j\int_{E_j}a\,d\mu \leq\sum_{j=1}^r c_j\int_{E_j}b\,d\mu =\int b\varphi\,d\mu. \] Each sum has finitely many nonnegative terms, so the calculation remains valid if a term or sum is infinite. This proves the reverse implication. \(\square\)

Combining the proposition with the theorem gives a practical version: if the estimate is known on every measurable set, then it holds after multiplying both integrands by any nonnegative measurable \(f\). The key is that the set estimate first yields the simple-function estimate; increasing approximation and monotone convergence then handle \(f\).

Worked Applications

Why Increasing Approximation Is Essential

The lifting proof uses more than the fact that simple functions approach \(f\) pointwise. It needs them to increase, so that the products increase and the Monotone Convergence Theorem applies. The theorem permits infinite integrals and requires no integrable dominating function, but its monotonicity hypothesis cannot be discarded.

For example, on \((0,1)\) with Lebesgue measure, define \(f_n(x)=n\chi_{(0,1/n)}(x)\). For each fixed \(x\in(0,1)\), once \(n>1/x\), one has \(x\notin(0,1/n)\), so \(f_n(x)=0\) thereafter and \(f_n(x)\to0\). Nevertheless, \(\int_0^1f_n\,dm=n\cdot(1/n)=1\) for every \(n\), whereas \(\int_0^1\lim_n f_n\,dm=0\). This sequence is not increasing: for instance, at \(x=3/4\), \(f_1(x)=1\) and \(f_2(x)=0\). Thus pointwise convergence does not justify replacing an integral of a limit by a limit of integrals.

A second error is to pass to limits without checking what happens to each side of the inequality. In the lifting theorem, the check is explicit: the approximants increase, multiplication by each fixed nonnegative factor preserves that increase (including the zero and infinite cases), and monotone convergence identifies both limits with the desired integrals. Without those facts, an inequality at every finite stage need not give the claimed inequality for the target functions.