Why the Sign of a Rate Matters
An accumulation function records the net amount contributed by a rate over an interval. If the rate is nonnegative, each interval contributes a nonnegative amount, so moving the input to the right cannot decrease the accumulation. If the rate is nonpositive, each such contribution is nonpositive, so moving to the right cannot increase it.
The key step is to compare two values of the accumulation directly. The interval-increment identity expresses their difference as an integral over the interval between the inputs. The sign of that integral then determines the direction of change.
Accumulation and Monotonicity
Let \(I\) be an interval, let \(c\in I\) be a fixed basepoint, and suppose \(f\) is Riemann integrable on every closed subinterval of \(I\). Define the accumulation function \(A\) by \(A(x)=\int_c^x f(t)\,dt\), with the usual orientation convention for integrals. The variable \(t\) is the integration variable; the input \(x\) determines the endpoint.
For \(u<v\) in \(I\), the interval-increment identity gives
This identity is the bridge between the rate \(f\) and the monotonicity of \(A\). The basepoint \(c\) does not appear on the right-hand side. Thus, the direction of change between \(u\) and \(v\) is controlled by the accumulation of \(f\) on \([u,v]\), regardless of where the basepoint was chosen.
The Sign Theorem
The hypotheses apply throughout the interval, not merely at the basepoint or at the two inputs being compared. The conclusion concerns every pair \(u<v\) in the domain.
Why the Sign Theorem Works
For a Riemann integrable function that is nonnegative throughout an interval, its integral over that interval is nonnegative. One way to see the sign is to form Riemann sums: each is a sum of terms \(f(\xi_i)\Delta x_i\), where each sampled value \(f(\xi_i)\) and each subinterval width \(\Delta x_i\) are nonnegative. Thus every such sum is nonnegative, and its limit, the integral, is also nonnegative. For a nonpositive integrand, the same reasoning gives a nonpositive integral.
Let \(u<v\) be any two points in \(I\). The aim is to compare \(A(u)\) and \(A(v)\).
Write \(A(v)-A(u)=\int_u^v f(t)\,dt\).
If \(f\ge 0\) throughout \(I\), then \(f\ge 0\) on \([u,v]\), so \(\int_u^v f(t)\,dt\ge 0\).
Since \(A(v)-A(u)\ge 0\), it follows that \(A(u)\le A(v)\). As \(u<v\) was arbitrary, \(A\) is nondecreasing.
For the nonpositive case, the same argument gives \(\int_u^v f(t)\,dt\le 0\), hence \(A(v)-A(u)\le 0\) and \(A(u)\ge A(v)\). This proves that \(A\) is nonincreasing.
Worked Example: A Nonnegative Rate
Worked Example: A Nonnegative Rate
Let \(f(t)=3t^2\) on \(I=[0,2]\), and define \(A(x)=\int_0^x 3t^2\,dt\). Since \(3t^2\ge 0\) for every \(t\in[0,2]\), the theorem predicts that \(A\) is nondecreasing. Compute \(A\), then verify the comparison for arbitrary inputs.
First, evaluate the accumulation:
Take any \(u,v\in[0,2]\) with \(u<v\). The change in the accumulation is
The last inequality follows because the integrand is nonnegative throughout \([u,v]\). Therefore \(A(u)\le A(v)\), as required. As a numerical check, \(A(1)=1\) and \(A(2)=8\), and the interval increment is \(8-1=7\). Direct integration gives \(\int_1^2 3t^2\,dt=2^3-1^3=7\), in agreement with the identity.
Worked Example: A Nonpositive Rate
Worked Example: A Nonpositive Rate
Let \(f(t)=-2(t+1)\) on \(I=[0,2]\), with \(A(x)=\int_0^x f(t)\,dt\). Since \(t+1\ge 1\) on this interval, \(f(t)\le 0\) throughout. We expect \(A\) to be nonincreasing.
Compute the accumulation:
For arbitrary \(0\le u<v\le 2\), the change is
Here \(v-u>0\) and \(v+u+2>0\), so the product with the leading minus sign is negative. Equivalently, by the interval-increment identity, this difference is \(\int_u^v -2(t+1)\,dt\le 0\). Hence \(A(v)\le A(u)\), confirming that \(A\) is nonincreasing. For a numerical check, \(A(0)=0\) and \(A(1)=-3\); direct integration gives \(\int_0^1 -2(t+1)\,dt=-3\), matching \(A(1)-A(0)\).
The Exact Condition for Monotonicity
The sign theorem gives useful sufficient conditions, but it does not characterize every way an accumulation can be nondecreasing or nonincreasing. The exact test comes from the interval-increment identity itself.
To prove the first equivalence, suppose first that \(A\) is nondecreasing. For any \(u<v\), its definition gives \(A(u)\le A(v)\). Subtracting \(A(u)\) shows \(A(v)-A(u)\ge 0\). By the interval-increment identity, this is exactly \(\int_u^v f(t)\,dt\ge 0\). Conversely, suppose every such interval integral is nonnegative. Then for every \(u<v\), the identity gives \(A(v)-A(u)=\int_u^v f(t)\,dt\ge 0\). Therefore \(A(u)\le A(v)\), so \(A\) is nondecreasing. The nonincreasing equivalence follows by reversing each inequality: \(A(u)\ge A(v)\) is equivalent to \(A(v)-A(u)\le 0\), which is equivalent to \(\int_u^v f(t)\,dt\le 0\).
This distinction matters: pointwise nonnegativity implies nonnegative integrals, but nonnegative integrals over all intervals do not require \(f\) to be nonnegative at every individual point.
Worked Example: A Negative Value Without a Decrease
Worked Example: A Negative Value Without a Decrease
On \(I=[0,1]\), define \(f(0)=-1\) and \(f(t)=1\) for \(0<t\le 1\). This function is Riemann integrable: it is bounded and has only one point of discontinuity. It is not pointwise nonnegative, because \(f(0)=-1\). Nevertheless, its accumulation from \(0\) is nondecreasing.
For any \(x\in[0,1]\), the value at the single endpoint \(t=0\) does not change the Riemann integral, so
In particular, \(A(0)=0\). For any \(0\le u<v\le 1\), the same endpoint observation gives
Thus \(A(u)\le A(v)\) for every \(u<v\), and \(A\) is nondecreasing even though \(f(0)<0\). As a direct check, \(A(0)=0\), \(A(1)=1\), and \(\int_0^1 f(t)\,dt=1=A(1)-A(0)\). This example shows why \(f\ge 0\) everywhere is sufficient, but not necessary. The exact condition is nonnegativity of every interval integral.
A Common Sign-Checking Mistake
A common error is to notice that a rate is positive somewhere, or that its total integral over the whole domain is nonnegative, and then conclude that the accumulation is nondecreasing everywhere. Monotonicity requires the appropriate sign for the increment on every interval between two inputs.
Worked Example: Why the Sign Must Be Checked on Every Interval
Let \(f(t)=t-1\) on \([0,2]\), and let \(A(x)=\int_0^x (t-1)\,dt\). A mistaken argument might say: “The rate is positive for \(t>1\), and the total accumulation at \(2\) is zero, so \(A\) is nondecreasing.” Neither observation establishes that conclusion.
Compute the accumulation and compare values:
Then \(A(0)=0\), \(A(1)=\frac12-1=-\frac12\), and \(A(2)=2-2=0\). Since \(0<1\) but \(A(0)=0>A(1)=-\frac12\), \(A\) is not nondecreasing. Since \(1<2\) but \(A(1)=-\frac12<A(2)=0\), it is not nonincreasing either. The relevant increments confirm this: \(\int_0^1(t-1)\,dt=-\frac12\), while \(\int_1^2(t-1)\,dt=\frac12\). They do not have one consistent sign across all intervals. The total \(\int_0^2(t-1)\,dt=0\) alone cannot determine monotonicity.
Using the Result in Accumulation Models
In a model, \(f\) may represent a rate and \(A(x)\) the net amount accumulated up to input \(x\). A nonnegative rate guarantees that the accumulated amount cannot go down as the input increases. A nonpositive rate guarantees that it cannot go up. This conclusion follows from interval contributions and does not require comparing values only to the basepoint.
When the rate changes sign, do not infer global monotonicity from a single value, from the sign at the basepoint, or from the total accumulation over the whole domain. Instead, use the identity for arbitrary \(u<v\). For a general integrable \(f\), the precise criterion is whether every interval integral has the required sign. The pointwise sign test is often easy to apply, but it is a sufficient shortcut rather than the exact characterization.
Check Your Understanding
Use the interval-increment identity in answering these questions.
- State the definition of a nondecreasing function using inputs \(u<v\).
- If \(f(t)\le 0\) throughout \(I\), what inequality does the identity give for \(A(v)-A(u)\), and what does that imply?
- State the exact condition on interval integrals that is equivalent to \(A\) being nondecreasing.
- Why does the example \(f(0)=-1\), \(f(t)=1\) for \(t>0\) show that pointwise nonnegativity is not necessary?
- For \(f(t)=t-1\) on \([0,2]\), compute \(\int_0^1 f(t)\,dt\) and explain what it says about nondecreasing behavior.
