What Interval Integrals Can Determine
An indefinite Lebesgue integral records accumulated area from a fixed starting point. Its increments over subintervals contain enough information to determine the integrand almost everywhere, but not its values at every point: changing an integrable function on a set of measure zero does not change any of its integrals. A related freedom appears for primitives: adding a constant changes their values but changes neither their increments nor their derivatives.
Throughout, let \(a<b\) be finite real numbers and use the finite measure space \(([a,b],\mathcal{L}_{[a,b]},m)\), where \(\mathcal{L}_{[a,b]}\) is the Lebesgue sigma-algebra restricted to \([a,b]\), and \(m\) is Lebesgue measure. Functions \(f,g\) are real-valued and in \(L^1(m)\), meaning they are measurable and \(\int_a^b |f|\,dm<\infty\), with the same condition for \(g\). Integrals over subintervals are Lebesgue integrals; including or excluding an endpoint does not affect them, since a singleton has measure zero.
The requirement \(a<b\) matters. On a degenerate interval \([a,a]\), every interval integral is zero, regardless of the functions, so interval integrals cannot determine them there. For \(a<b\), it is enough to check intervals of the form \([a,x]\): if the integrals agree on all such intervals, additivity gives agreement on \([u,v]\) by subtracting the values at \(v\) and \(u\).
Proof of the Recovery Theorem
We use the differentiation theorem for indefinite Lebesgue integrals: if \(h\in L^1(m)\) on a finite interval and \(H(x)=\int_a^x h\,dm\), then \(H\) is absolutely continuous and \(H'(x)=h(x)\) for \(m\)-almost every \(x\in(a,b)\). This standard result was established in the earlier material on differentiating indefinite Lebesgue integrals. Its hypotheses require integrability of \(h\); it does not assert a derivative identity at every point.
Proof. First suppose the integrals of \(f\) and \(g\) agree on every subinterval. Set \(h=f-g\), which belongs to \(L^1(m)\) because $$ \int_a^b |f-g|\,dm\le \int_a^b |f|\,dm+\int_a^b |g|\,dm<\infty. $$ For each \(x\in[a,b]\), the assumed equality on \([a,x]\) gives \(H(x)=\int_a^x h\,dm=0\). Thus \(H\) is the zero function, and its derivative is zero at every interior point. By the differentiation theorem, \(H'=h\) almost everywhere. Consequently \(h=0\) almost everywhere, or \(f=g\) almost everywhere.
Conversely, suppose \(f=g\) almost everywhere. Then \(h=f-g\) vanishes outside a Lebesgue null set. The integral of an integrable function over a null set is zero, so for every \(a\le u\le v\le b\), $$ \int_u^v f\,dm-\int_u^v g\,dm=\int_u^v h\,dm=0. $$ This proves both implications. \(\square\)
The proof also explains the mechanism: the anchored integral \(H(x)\) is a function whose derivative recovers \(f-g\) almost everywhere. If every interval increment vanishes, then \(H\) is constant (in this anchored case, identically zero), so its derivative vanishes almost everywhere.
Worked Example: A Step Integrand and a Null-Set Change
On \([0,1]\), define \(f(x)=1\) for \(0\le x<\tfrac12\) and \(f(x)=3\) for \(\tfrac12\le x\le1\). Let \(g(x)=f(x)\) except that \(g(\tfrac12)=100\). Both functions are measurable and integrable; their only difference is at the singleton \(\{\tfrac12\}\), which has measure zero.
For \(x\le\tfrac12\), the anchored integral is \(P(x)=\int_0^x f\,dm=x\). For \(x\ge\tfrac12\), the first half contributes \(\tfrac12\) and the remaining interval contributes \(3(x-\tfrac12)\), so $$ P(x)=\frac12+3\left(x-\frac12\right)=3x-1. $$ Since \(g=f\) almost everywhere, its anchored integral is the same \(P\). Therefore, for any \(0\le u\le v\le1\), both interval integrals equal \(P(v)-P(u)\). For instance, on \([\tfrac14,\tfrac34]\), the integral is \(P(\tfrac34)-P(\tfrac14)=(\tfrac94-1)-\tfrac14=\tfrac54-\tfrac14=1\). Directly, the interval has length \(\tfrac14\) below \(\tfrac12\) and length \(\tfrac14\) at or above it, giving \(1\cdot\tfrac14+3\cdot\tfrac14=1\). The value assigned at the single point does not affect this calculation.
Check: At \(x=\tfrac12\), the formula from the right gives \(3(\tfrac12)-1=\tfrac12\), agreeing with the formula from the left. The theorem concludes \(f=g\) almost everywhere, not everywhere: here their values differ at \(x=\tfrac12\).
Primitive Functions and the Constant of Integration
For \(f\in L^1(m)\) and \(C\in\mathbb{R}\), define $$ F_C(x)=C+\int_a^x f\,dm,\qquad a\le x\le b. $$ The indefinite-integral differentiation theorem says \(F_C\) is absolutely continuous and \(F_C'=f\) almost everywhere. The constant \(C\) changes the value of \(F_C\) at every point by the same amount, but it cancels from every increment \(F_C(v)-F_C(u)\).
Proof. Set \(Q=F-G\). The difference of absolutely continuous functions is absolutely continuous, and \(Q'=F'-G'=0\) almost everywhere. By the integral representation theorem for absolutely continuous functions, \(Q(x)=Q(a)+\int_a^x Q'\,dm\) for every \(x\in[a,b]\). Since \(Q'=0\) almost everywhere, each of these integrals is zero, so \(Q(x)=Q(a)\) everywhere. Conversely, if \(Q\) is constant, its derivative is zero at every interior point; wherever \(F\) and \(G\) are differentiable, \(F'-G'=Q'=0\). Absolute continuity ensures their derivatives exist almost everywhere, proving the converse. \(\square\)
More precisely, if \(F\) is absolutely continuous and \(F'=f\) almost everywhere for some \(f\in L^1(m)\), the integral representation theorem gives \(F(x)=F(a)+\int_a^x f\,dm\) for every \(x\). Thus every absolutely continuous primitive of \(f\) has the displayed form, with the sole freedom being its starting value \(F(a)\). The almost-everywhere condition on the derivative is essential to this representation; it is not a claim that \(F'\) exists at every point.
Worked Example: Two Primitives with Different Constants
On \([0,1]\), take \(f(x)=2x\), \(F(x)=x^2+4\), and \(G(x)=x^2-3\). The integrand is integrable, and both functions are continuously differentiable, hence absolutely continuous. Their derivatives are \(F'(x)=2x\) and \(G'(x)=2x\) at every \(x\in[0,1]\) where the endpoint derivative is interpreted one-sided; in particular the derivatives agree almost everywhere on \((0,1)\).
For every \(0\le u\le v\le1\), direct integration gives $$ \int_u^v 2x\,dm=x^2\big|_u^v=v^2-u^2. $$ The increments of \(F\) and \(G\) are respectively \((v^2+4)-(u^2+4)=v^2-u^2\) and \((v^2-3)-(u^2-3)=v^2-u^2\). For \(u=\tfrac14\) and \(v=\tfrac34\), the integral is \(\tfrac9{16}-\tfrac1{16}=\tfrac8{16}=\tfrac12\), and each function's increment is also \(\tfrac12\). Finally, \(F(x)-G(x)=7\) for every \(x\), confirming the constant difference.
Check: Differentiating either expression gives \(2x\), and subtracting them gives \(7\) at every point. The two checks confirm respectively the shared integrand and the freedom to add a constant.
Why Equality of One Integral Is Not Enough
The recovery theorem requires equality on every subinterval, not merely equality of the total integrals over \([a,b]\). Positive and negative differences can cancel over the full interval. The following calculation shows exactly how that cancellation can conceal distinct integrands.
Worked Example: Equal Total Integrals but Different Integrands
On \([0,1]\), let \(f(x)=1\) and \(g(x)=2x\). Both are real-valued, measurable, and integrable. Their total integrals agree: $$ \int_0^1 f\,dm=1,\qquad \int_0^1 g\,dm=\int_0^1 2x\,dm=1^2-0^2=1. $$ But on \([0,\tfrac12]\), the respective integrals are $$ \int_0^{1/2} f\,dm=\frac12,\qquad \int_0^{1/2} g\,dm=\left(\frac12\right)^2-0^2=\frac14. $$ Thus their interval integrals do not all agree. The functions themselves agree only where \(1=2x\), namely at \(x=\tfrac12\); they differ almost everywhere on \([0,1]\). As an additional check, their anchored integrals at \(x=\tfrac12\) are \(\tfrac12\) and \(\tfrac14\), matching the subinterval calculation.
A common mistake is to infer equality of integrands from equality of total integrals. The example shows why that inference fails: a single number records only the net accumulated difference, whereas the family of all interval increments detects changes at every scale. In applications, it is often enough to verify equality of the anchored integrals \(x\mapsto\int_a^x f\,dm\) and \(x\mapsto\int_a^x g\,dm\) for every \(x\), since subtracting their values at \(u\) and \(v\) recovers the integrals over \([u,v]\).
The conclusion remains almost everywhere rather than pointwise. Lebesgue integration identifies functions that differ only on null sets, and the differentiation theorem recovers the integrand only almost everywhere. These are not defects: an \(L^1\) integrand is naturally treated up to equality almost everywhere. By contrast, a primitive is an actual function defined at every point, and once its integrand is fixed almost everywhere, its values are determined everywhere after one starting value is specified.
Check Your Understanding
Use interval increments and the hypotheses of the recovery theorem to answer each question.
- Why is it enough to know that the integrals of two \(L^1\) functions agree on every interval \([a,x]\) to conclude agreement on every \([u,v]\)?
- If two integrable functions differ only at one point, what can be concluded about their interval integrals, and what cannot be concluded about their pointwise values?
- Suppose \(F\) and \(G\) are absolutely continuous and \(F'=G'\) almost everywhere. Which theorem gives the representation needed to conclude that \(F-G\) is constant?
- Give a pair of distinct integrable functions on \([0,1]\) whose total integrals agree, and identify a subinterval on which their integrals differ.
- Why does the recovery theorem require a nondegenerate interval \([a,b]\) with \(a<b\)?