Deriving Additivity From Riemann Sums

Why Additivity Can Be Seen in the Riemann Sums

If an interval is divided into two adjacent pieces, the total accumulation over the whole interval should equal the accumulation over the first piece plus the accumulation over the second. The additivity rule expresses this simple idea. Here we derive it from the Riemann sums that define the integral.

The key construction is to take a partition of each piece and join them at their common endpoint. Every subinterval in the joined partition belongs to exactly one of the two pieces. Consequently, the sum for the joined partition is exactly the sum of the two separate sums. As the partitions become fine, those sums approach the corresponding integrals.

Tagged Partitions and Riemann Sums

For an interval \([u,v]\), a partition is a finite increasing list of points \(u=x_0<x_1<\cdots<x_n=v\). Its subintervals are \([x_{i-1},x_i]\), with widths \(\Delta x_i=x_i-x_{i-1}\). The mesh of the partition is the largest subinterval width. A tagged partition also chooses a tag \(\xi_i\in[x_{i-1},x_i]\) for each subinterval.

The number \(S(f,P,\xi)\) is a tagged Riemann sum. If \(f\) is Riemann integrable on \([u,v]\), then its Riemann sums approach \(\int_u^v f(x)\,dx\) as the mesh approaches zero: for every \(\varepsilon>0\), there is a \(\delta>0\) such that every tagged partition with \(\|P\|<\delta\) has \(\left|S(f,P,\xi)-\int_u^v f(x)\,dx\right|<\varepsilon\). This convergence holds for every choice of tags.

We will use the standard restriction property: if \(f\) is Riemann integrable on \([a,b]\), then it is Riemann integrable on each subinterval. One justification uses the Darboux criterion: a bounded function is Riemann integrable if, for every \(\varepsilon>0\), some partition has upper sum minus lower sum less than \(\varepsilon\). Add a chosen interior endpoint \(c\) to such a partition. Refining a partition cannot increase that difference, and the parts of the refined partition on either side of \(c\) give the criterion on each subinterval. Thus the restricted functions are integrable.

The Additivity Theorem

The central case is \(a<c<b\). Choose any partition of \([a,c]\) and any partition of \([c,b]\). Their union, with the common point \(c\) written only once, is a partition of \([a,b]\). The tagged sums on the two pieces combine exactly into the tagged sum on the whole interval.

Derivation by Joining Partitions

Take sequences of partitions \(P_n\) of \([a,c]\) and \(Q_n\) of \([c,b]\) whose meshes both tend to zero. Such partitions can, for example, be obtained by dividing each interval into \(n\) equal parts. Choose tags freely in each subinterval. Join \(P_n\) and \(Q_n\) at \(c\), and call the resulting partition \(R_n\). Its mesh is the larger of the two original meshes, so it too tends to zero.

For each \(n\), the identity of finite sums is exact:

There is no approximation in this equality: the terms on the left are the same terms as those on the right, simply collected into one list. Since \(f\) is integrable on \([a,b]\), and its restrictions are integrable, the tagged sums converge as the meshes tend to zero. Taking limits in the finite-sum identity gives

which is exactly

If \(c=a\), the first interval has zero length and its integral is zero; if \(c=b\), the second interval has zero length and its integral is zero. In either case the same formula follows directly.

Why the Shared Endpoint Adds No Separate Area

A Riemann sum has one term for each subinterval, and each term is a function value multiplied by that subinterval's width. The point \(c\) is a boundary between subintervals, not an extra subinterval. The joined partition therefore has no separate term of the form \(f(c)\times\text{(a width belonging only to }c\text{)}\). A single point has zero width.

There is an important qualification: a tag for a subinterval may be chosen to equal \(c\). Then that subinterval's Riemann-sum term is \(f(c)\Delta x_i\), which need not be zero for a finite partition. It is not an additional contribution assigned to the point alone; it is the term for an interval of positive width. If \(f\) is bounded by \(M\) in absolute value, the magnitude of this term is at most \(M\Delta x_i\), and \(\Delta x_i\) is at most the mesh. Thus any such endpoint-tag term tends to zero as the mesh shrinks. Tags at \(c\) do not obstruct the additivity proof, which works for every choice of tags.

Worked Example: Joining Two Partitions for \(f(x)=x\)

Worked Example: Limits of Left-Endpoint Sums

Worked Example: A Special Value at the Common Endpoint

A Common Mistake

A tempting but incorrect calculation is to treat the shared endpoint as a third piece. For the last example, one might write

The value \(100=f(1)\) has been added as if the point \(1\) occupied an interval of width \(1\). It does not. The Riemann sum uses height times width, and the point by itself has width zero. The correct result is

The same distinction prevents a subtler error: a finite tagged sum may contain a term using \(f(c)\) when \(c\) is chosen as a tag, but that term belongs to an adjacent interval and is controlled by that interval's shrinking width. It is not an extra term for the shared point.

Why the Partition Argument Matters

The additivity formula is not based on a geometric picture alone. Riemann sums encode accumulation by adding contributions from subintervals; when a partition is joined, those contributions are simply grouped into one list. The limit preserves that exact finite-sum identity. This explains both why additivity holds and why no separate area is assigned to the boundary point.

The proof also identifies its assumptions. The integral over the full interval must exist, so that its tagged sums converge; the restrictions to the two pieces are then integrable as well. The argument does not require \(f\) to be positive or continuous, and the sums may contain positive and negative terms. What matters here is integrability and the fact that the partitions become fine.