Building an Integral from Values Below a Function
For a nonnegative measurable function, the integral can be defined without first finding an antiderivative or taking a limit of integrals. The starting objects are simple functions: measurable functions taking only finitely many finite values. A simple function lying below the given function gives a lower estimate; taking the supremum of all such estimates defines the integral, including when its value is infinite.
Throughout the definition and the results below, \((X,\mathcal{M},\mu)\) is an arbitrary measure space. No assumption that \(\mu\) is finite, complete, or sigma-finite is needed. The function \(f:X\to[0,\infty]\) is extended nonnegative and \(\mathcal{M}\)-measurable. A nonnegative simple function is a measurable function with finite range contained in \([0,\infty)\); in particular, its values are finite, although its integral can be infinite if it is nonzero on a set of infinite measure.
The inequality \(\varphi\le f\) in this definition is pointwise on all of \(X\), not merely almost everywhere. The definition itself makes no finiteness assertion: the supremum may be any value in \([0,\infty]\). It also does not require the supremum to be represented by one particular simple function.
Two Facts Used in the Construction
Proof. Partition \(X\) into the finitely many measurable sets on which the pair \((u,v)\) has a fixed value \((c,d)\). On each such set, \(0\le c\le d\), so \(c\mu(A)\le d\mu(A)\), including when \(\mu(A)=\infty\); when \(d=0\), both contributions are zero. Adding the finitely many inequalities gives \(\int u\,d\mu\le\int v\,d\mu\). \(\square\)
We also use continuity from below for measures: if \(A_1\subseteq A_2\subseteq\cdots\) are measurable, then \(\mu(\bigcup_n A_n)=\lim_n\mu(A_n)\), with infinity allowed. Indeed, the disjoint sets \(D_1=A_1\) and \(D_j=A_j\setminus A_{j-1}\) for \(j\ge2\) have union \(\bigcup_n A_n\); countable additivity gives \(\mu(\bigcup_n A_n)=\sum_j\mu(D_j)=\lim_n\sum_{j=1}^n\mu(D_j)=\lim_n\mu(A_n)\).
A Canonical Sequence of Simple Minorants
There are many possible simple functions below \(f\). The next result shows that a particular sequence, formed by truncating and rounding down on progressively finer grids, is enough to recover the supremum. This also supplies concrete minorants when \(f\) is unbounded or takes the value \(+\infty\).
Proof. The values of \(q_n\) belong to the finite set \(\{k2^{-n}:k=0,1,\ldots,n2^n\}\). For \(k<n2^n\), the set where \(q_n=k2^{-n}\) is \(\{k2^{-n}\le \min(f,n)<(k+1)2^{-n}\}\); the top value \(n\) occurs on \(\{f\ge n\}\). These sets are measurable because \(f\) is measurable, so \(q_n\) is simple and measurable. Rounding down gives \(q_n\le f\).
To compare successive terms, fix \(x\) and write \(t=f(x)\). If \(t\le n\), rounding down to the finer grid with spacing \(2^{-(n+1)}\) gives a value at least as large as rounding down to the grid with spacing \(2^{-n}\): every multiple of \(2^{-n}\) is also a multiple of \(2^{-(n+1)}\). If \(t>n\), including \(t=+\infty\), then \(q_n(x)=n\), while \(\min\{t,n+1\}\ge n\), so \(q_{n+1}(x)\ge n\). Thus \(q_n\le q_{n+1}\) everywhere.
If \(t=f(x)<\infty\), then for every integer \(n>t\), the truncation has no effect and rounding down gives \(t-2^{-n}<q_n(x)\le t\). Hence \(q_n(x)\to t\). If \(t=+\infty\), then \(q_n(x)=n\) for every \(n\), so \(q_n(x)\to+\infty\).
Set \(S=\sup_n\int q_n\,d\mu\). Since every \(q_n\) is an eligible simple minorant, the definition gives \(S\le\int f\,d\mu\). For the reverse inequality, let \(\psi\) be any nonnegative simple function with \(\psi\le f\), and fix a real number \(s\) with \(0<s<1\). Define \(E_n=\{x:q_n(x)\ge s\psi(x)\}\). These measurable sets increase with \(n\). Their union is \(X\): if \(\psi(x)=0\), then \(x\in E_n\) for every \(n\); if \(\psi(x)>0\), then \(s\psi(x)<\psi(x)\le f(x)\), and \(q_n(x)\to f(x)\), so \(q_n(x)\ge s\psi(x)\) for some \(n\).
The simple function \(s\psi\chi_{E_n}\) is bounded above by \(q_n\) everywhere: on \(E_n\) this follows from its definition, and outside \(E_n\) it is zero. The lemma therefore gives \(\int q_n\,d\mu\ge s\int\psi\chi_{E_n}\,d\mu\). Write \(\psi=\sum_{j=1}^k a_j\chi_{A_j}\) on disjoint measurable sets \(A_j\). Since \(E_n\) increases to \(X\), continuity from below gives \(\mu(A_j\cap E_n)\to\mu(A_j)\) for each \(j\). Consequently, \(\int\psi\chi_{E_n}\,d\mu=\sum_{j=1}^k a_j\mu(A_j\cap E_n)\to\sum_{j=1}^k a_j\mu(A_j)=\int\psi\,d\mu\), where the finite sum and its limit are interpreted in \([0,\infty]\). Taking \(n\to\infty\) in the inequality yields \(S\ge s\int\psi\,d\mu\). Letting \(s\) increase to \(1\) gives \(S\ge\int\psi\,d\mu\); if this integral is infinite, the inequality for any fixed \(s>0\) already forces \(S=\infty\). This holds for every eligible \(\psi\), so taking their supremum gives \(S\ge\int f\,d\mu\). Together with the opposite inequality, this proves the formula. \(\square\)
The proof uses continuity from below for sets and comparison of simple-function integrals; it does not exchange a limit and an integral. In particular, the conclusion is an identity between the defining supremum and the supremum of a specified sequence of simple integrals.
Worked Example: The Function f(x) = x on [0,1]
Take \(X=[0,1]\), \(\mathcal{M}\) the Lebesgue sigma-algebra restricted to \([0,1]\), and \(\mu=m\). The function \(f(x)=x\) is finite, nonnegative, and measurable. For \(n\ge1\), the canonical minorant is \(q_n(x)=2^{-n}\lfloor 2^n x\rfloor\), since the cap at \(n\) does not affect values on \([0,1]\). For \(k=0,\ldots,2^n-1\), it equals \(k/2^n\) on \([k/2^n,(k+1)/2^n)\); at \(x=1\) it equals \(1\), which contributes zero measure to the integral.
Each interval in the displayed partition has measure \(2^{-n}\). Therefore \[ \int q_n\,dm =\sum_{k=0}^{2^n-1}\frac{k}{2^n}\,2^{-n} =\frac{1}{4^n}\frac{(2^n-1)2^n}{2} =\frac{2^n-1}{2^{n+1}}. \] As \(n\to\infty\), this tends to \(1/2\). The theorem identifies the supremum of the integrals of all simple minorants with this limit, so \(\int_{[0,1]}x\,dm=1/2\). Check: at \(n=1\), the formula gives \((2-1)/2^2=1/4\); at \(n=2\), it gives \((4-1)/2^3=3/8\). Independently, the Riemann integral is \([x^2/2]_0^1=1/2\).
Worked Example: An Infinite Function on a Set of Positive Measure
On \((\mathbb{R},\mathcal{L},m)\), let \(E=(0,2)\) and define \(f=+\infty\) on \(E\) and \(f=0\) on \(E^c\). This is an extended nonnegative measurable function. For every finite \(c\ge0\), \(\varphi_c=c\chi_E\) is a nonnegative simple function and satisfies \(\varphi_c\le f\) everywhere. Since \(m(E)=2\), its simple integral is \[ \int\varphi_c\,dm=c\,m(E)=2c. \] As \(c\) ranges over finite nonnegative values, \(2c\) is unbounded above. The defining supremum is therefore \(+\infty\).
The canonical sequence gives the same result directly: for every \(n\ge1\), \(q_n=n\chi_E\), so \(\int q_n\,dm=n\,m(E)=2n\). These values are unbounded. Check: choosing \(c=3\) gives \(3\cdot 2=6\), and the canonical term at \(n=3\) also has integral \(3\cdot 2=6\). No finite value of \(f\) on \(E\) is required for these finite-valued minorants.
Worked Example: A Geometric Sequence under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mu\) be counting measure, and set \(f(k)=2^{-k}\). For each \(N\ge1\), the simple function \(\psi_N=\sum_{k=1}^N2^{-k}\chi_{\{k\}}\) lies below \(f\) everywhere, and \[ \int\psi_N\,d\mu=\sum_{k=1}^N2^{-k}=1-2^{-N}. \] These values tend to \(1\), so the supremum in the definition is at least \(1\).
For the upper bound, let \(\psi\) be any nonnegative simple minorant. Under counting measure its integral is the sum of its point values, with infinity allowed. For each \(K\), \(\sum_{k=1}^K\psi(k)\le\sum_{k=1}^K2^{-k}=1-2^{-K}\le1\). Taking the supremum over \(K\) shows \(\int\psi\,d\mu\le1\). Thus every eligible simple minorant has integral at most \(1\), and the supremum is exactly \(1\). Check: at \(N=3\), the partial sum is \(1/2+1/4+1/8=7/8\), while \(1-2^{-3}=1-1/8=7/8\). The upper bound and these lower estimates agree in the limit.
A Common Mistake: Using Only Constant Minorants
A minorant must be simple, but it need not be constant. Restricting the supremum to constant functions can lose almost all the information about a function. For example, on \([0,1]\) with Lebesgue measure, if a constant \(c\ge0\) satisfies \(c\le x\) at every \(x\in[0,1]\), then evaluating at \(x=0\) forces \(c=0\). The incorrect calculation “the only constant minorant is zero, so the integral is zero” ignores nonconstant simple minorants.
The correction is to use the dyadic minorants from the worked example. They are not constant, they satisfy \(q_n(x)\le x\) at every point, and their integrals are \((2^n-1)/2^{n+1}\), whose supremum is \(1/2\). The definition allows every nonnegative measurable simple minorant, not just constants.
The supremum definition is useful because it reduces the construction of the nonnegative integral to measurable sets and finite-valued functions. Its full generality matters: the measure space need not be finite, the function may be unbounded or infinite at some points, and the resulting integral is permitted to be \(+\infty\). The dyadic construction gives a systematic family to use, while the supremum remains over all simple minorants.
Check Your Understanding
Use the pointwise minorant condition and the supremum definition in each question.
- For a nonnegative simple \(\psi\le f\), what does the definition guarantee about \(\int\psi\,d\mu\) and \(\int f\,d\mu\)?
- Why does the canonical formula \(q_n(x)=2^{-n}\lfloor2^n\min\{f(x),n\}\rfloor\) give a finite-valued function even when \(f(x)=+\infty\)?
- On \([0,1]\) with Lebesgue measure, compute the integral of the first dyadic minorant of \(f(x)=x\), corresponding to \(n=1\).
- For \(f=+\infty\) on a measurable set \(E\) with \(m(E)=3\) and \(f=0\) outside \(E\), what integrals result from the minorants \(c\chi_E\), with finite \(c\ge0\)?
- Under counting measure on \(\mathbb{N}\), why does \(\sum_{k=1}^N2^{-k}\) provide a lower estimate for the integral of \(f(k)=2^{-k}\)?