Changing Function Values at Finitely Many Points

Why finitely many values can be changed

A definite Riemann integral records accumulated behavior over an interval. A function’s value at one point, or at any finite collection of points, does not occupy an interval of positive length. This suggests that changing finitely many values should not change the integral. The proof must nevertheless account for the fact that a Riemann sum or Darboux sum can be affected by values at partition boundaries.

The key is to surround each changed point with a small interval. The total length of these intervals can be made as small as desired. Since the functions are bounded, their contributions on those intervals are correspondingly small. Away from them, the functions agree.

Darboux sums and the proof strategy

A partition \(P\) of \([a,b]\) is a finite list \(a=x_0<x_1<\cdots<x_m=b\). Its subintervals are \([x_{j-1},x_j]\), with lengths \(\Delta x_j=x_j-x_{j-1}\). For a bounded function \(u\), let \(M_j(u)\) and \(m_j(u)\) denote its supremum and infimum on the \(j\)-th subinterval. The upper and lower Darboux sums are

The Darboux criterion says that a bounded function is Riemann integrable if, for every \(\varepsilon>0\), there is a partition \(P\) such that \(U(u,P)-L(u,P)<\varepsilon\). For an integrable function, its integral lies between the lower and upper sums for every partition. We will use these facts to prove both integrability and equality of integrals.

Write the distinct points where \(f\) and \(g\) differ as \(c_1,\ldots,c_n\). Choose small, disjoint intervals around these points, using one-sided intervals if a changed point is an endpoint. We can make their total length as small as we wish. Refine a partition by adding the endpoints of these intervals and the points \(c_i\). Then every subinterval whose closure contains a changed point lies inside one of the small intervals. Call such subintervals affected. All other subintervals are unaffected, and \(f=g\) everywhere on each of them.

Proof of the theorem

Because both functions are bounded, choose \(M\geq 0\) such that \(|f(x)|\leq M\) and \(|g(x)|\leq M\) for every \(x\in[a,b]\). On any affected subinterval, the oscillation of either function is at most \(2M\). Here the oscillation means the supremum minus the infimum. The total length of affected subintervals can be made arbitrarily small by the choice of the surrounding intervals.

If \(M=0\), both functions are identically zero and the result is immediate. If \(M>0\), choose \(\eta<\varepsilon/(4M)\). The bound just obtained is then less than \(\varepsilon\), so the Darboux criterion proves that \(g\) is integrable.

It remains to prove the equality of integrals. Use the same refined partition \(Q\). On unaffected subintervals, the upper contribution for \(g\) minus the lower contribution for \(f\) is exactly the corresponding Darboux-gap contribution for \(f\). On affected subintervals, the supremum of \(g\) minus the infimum of \(f\) is at most \(2M\). Consequently,

Since \(\int_a^b g(x)\,dx\leq U(g,Q)\) and \(\int_a^b f(x)\,dx\geq L(f,Q)\), this bounds \(\int_a^b g(x)\,dx-\int_a^b f(x)\,dx\) from above. Interchanging \(f\) and \(g\) gives the same bound for the difference in the other direction. Therefore

We can choose \(\eta\) arbitrarily small, as well as choose the original Darboux gap arbitrarily small. Thus the absolute difference between the two integrals is less than every positive tolerance. It must be zero, proving that the integrals are equal.

Worked examples

A common partition mistake

It is tempting to think that placing a changed point at a partition point makes its new value irrelevant. That is not generally true: the supremum and infimum on a partition subinterval include its endpoints. For example, let \(f(x)=0\) on \([0,1]\), and change only \(g(1/2)\) to \(100\). The partition \(P=\{0,1/2,1\}\) places the changed point at a boundary, but both subintervals include it. On each subinterval, the supremum of \(g\) is \(100\) and the infimum is \(0\). Thus

This partition does not demonstrate integrability. The correction is to put boundaries on both sides of \(1/2\), close enough that the total length of the affected intervals is small. The changed value can still influence their suprema, but its contribution to the Darboux gap is bounded by \(100\) times that small total length.

Why the result matters

The theorem explains why the integral is insensitive to isolated alterations in a bounded function. A graph can have a few points moved far above or below its usual values without changing its accumulated integral. Boundedness matters: it supplies a uniform limit on the size of the effect inside the small intervals. The finiteness of the changed set matters too: finitely many points can be enclosed in intervals with arbitrarily small total length.

The proof does not say that changed values are ignored by every partition. Rather, it shows that their influence can be confined to intervals whose total length is as small as needed. This distinction is the reason the partition must be designed around the changed points.