Why Bounded Rates Give Continuous Accumulation
An accumulation function records the net effect of a rate between a fixed basepoint and a variable input. If the rate stays bounded, then over a very short interval it cannot add or remove an arbitrarily large amount. The length of that interval limits how much the accumulated value can change.
This observation gives a direct proof of continuity. It does not require the rate itself to be continuous. The essential ingredients are that the rate is integrable and that its size is bounded near the input where continuity is being considered.
Accumulation and the Interval-Increment Identity
Let \(I\) be an interval, let \(a\in I\) be a fixed basepoint, and let \(f\) be integrable on every compact subinterval of \(I\). The accumulation function with basepoint \(a\) is
Integrals with reversed bounds use the convention \(\int_u^v f(t)\,dt=-\int_v^u f(t)\,dt\). For any \(x,c\in I\), splitting the integral at \(c\) gives the interval-increment identity
This identity says that the difference between two accumulated values is exactly the accumulation over the interval between their inputs. We will combine it with a bound on that interval’s accumulation.
If \(u\leq v\) and \(f\) is integrable on \([u,v]\), the integral estimate for a bounded function is
The first inequality is the standard absolute-value estimate for an integral; the second uses \(|f(t)|\leq M\). If the bounds are reversed, changing their order changes the sign of the integral, not its absolute value. Thus, for either ordering of \(u\) and \(v\),
The Continuity Theorem
The proof uses the interval-increment identity and the integral estimate. Fix any \(c\in I\). For \(x\in I\), those two results give
To verify continuity at \(c\), let \(\varepsilon>0\) be given and choose
This is positive because \(\varepsilon>0\) and \(M+1>0\). If \(x\in I\) and \(|x-c|<\delta\), then
The middle inequality follows from \(M\geq0\) and \(|x-c|<\delta\); writing it as a non-strict inequality also covers \(M=0\). The final inequality holds because \(M/(M+1)<1\) for every finite \(M\geq0\). We have therefore shown that for every \(\varepsilon>0\), there is a \(\delta>0\) such that inputs within \(\delta\) of \(c\) have accumulation values within \(\varepsilon\) of \(F(c)\). That is precisely continuity at \(c\). Since \(c\) was arbitrary, \(F\) is continuous throughout \(I\).
Worked Examples
Worked Example: A Constant Rate
Let \(f(t)=2\) on \(I=[0,4]\), and define \(F(x)=\int_0^x 2\,dt\). The rate is bounded by \(M=2\). Evaluate the accumulation function and use the bound to verify continuity at \(c=1\) for \(\varepsilon=0.3\).
First compute the accumulation:
In particular, \(F(1)=2\). Choose \(\delta=\varepsilon/(M+1)=0.3/3=0.1\). Whenever \(x\in[0,4]\) and \(|x-1|<0.1\), the increment estimate gives
As a direct check, \(F(x)-F(1)=2x-2=2(x-1)\), so its absolute value is exactly \(2|x-1|\), as the estimate predicts. Also, differentiating \(2x\) gives \(2\), the stated rate.
Worked Example: A Rate That Changes Sign
Let \(f(t)=t-1\) on \(I=[0,2]\), with basepoint \(a=0\). Prove continuity at \(c=1.5\) using the bound \(M=1\) and \(\varepsilon=0.04\).
On this interval, \(t-1\) ranges from \(-1\) to \(1\), so \(|f(t)|\leq1\). Compute \(F\):
At \(c=1.5\),
Choose \(\delta=\varepsilon/(M+1)=0.04/2=0.02\). If \(x\in[0,2]\) and \(|x-1.5|<0.02\), then
For a calculation check, the antiderivative \(\frac{x^2}{2}-x\) differentiates to \(x-1\), and its value at \(0\) is \(0\), as required by the basepoint. The rate changes sign, but the absolute-value bound still controls the change in accumulated value.
Worked Example: A Zero Rate Bound
Suppose \(f(t)=0\) on \(I=[-1,1]\) and \(F(x)=\int_0^x f(t)\,dt\). This example makes explicit how the continuity estimate works when the bound is \(M=0\). For \(\varepsilon=0.05\), the theorem’s choice is
The accumulation is
For every \(x\in[-1,1]\) with \(|x-0|<0.05\), the estimate gives
Checking directly, \(F(x)-F(0)=0-0=0\), so the required conclusion holds. The derivative of the constant antiderivative \(F(x)=0\) is also \(0\), matching the rate.
Bounded Nearby Is Enough
The theorem states a convenient global condition: one bound works everywhere on \(I\). Continuity at a particular point \(c\) requires less. It is enough that \(f\) be integrable between nearby inputs and bounded on some interval around \(c\). Indeed, if \(|f(t)|\leq M\) whenever \(t\) is sufficiently close to \(c\), then for inputs \(x\) close enough to \(c\), the whole interval between \(c\) and \(x\) lies in that region. The same estimate \(|F(x)-F(c)|\leq M|x-c|\) proves continuity there.
This local version explains the role of the assumptions. The integral over the interval from \(c\) to \(x\) must be defined, and a finite bound on the rate over that interval must be available. Continuity of \(f\) is not required for this argument: boundedness and integrability suffice.
A Common Pitfall: Strict Inequalities When \(M=0\)
When choosing \(\delta=\varepsilon/(M+1)\), it is tempting to write the estimate as
That strict inequality does not always hold. For example, if \(M=0\), then \(M|x-c|=0\) and \(M\delta=0\), so the claimed inequality would say \(0<0\), which is false. This is a flaw in the proof chain even though the continuity conclusion is true.
The correction is to use a non-strict inequality at that step. Since \(|x-c|<\delta\) and \(M\geq0\), we have \(M|x-c|\leq M\delta\). Then, including when \(M=0\),
When \(M=0\), the middle expression is \(0\), and \(0<\varepsilon\) because \(\varepsilon\) was chosen positive. Checking boundary cases is part of making an epsilon–delta proof valid for every value allowed by its assumptions.
Why This Result Matters
The theorem turns a qualitative condition on a rate—having a finite bound—into a precise conclusion about the behavior of its accumulation. The estimate also explains the mechanism: over an interval of length \(|x-c|\), a rate of size at most \(M\) can contribute at most \(M|x-c|\) in absolute accumulated value.
This is useful even when an exact accumulation formula is unavailable. If a model supplies a bound on the rate, the estimate gives control over how much the accumulated quantity can change between nearby inputs. The conclusion concerns the accumulation function; it does not assert that the rate has no discontinuities or that the accumulation has any particular derivative. The continuity proof needs only the interval-increment identity and the bound.
Check Your Understanding
Use the increment estimate and the assumptions of the theorem to answer these questions.
- What identity expresses \(F(x)-F(c)\) as an integral over the interval between \(c\) and \(x\)?
- If \(|f(t)|\leq 3\), what bound does the theorem give for \(|F(x)-F(c)|\)?
- For a given \(\varepsilon>0\), what positive \(\delta\) does the proof choose when the rate bound is \(M\geq0\)?
- Why must the proof avoid claiming \(M|x-c|<M\delta\) when \(M=0\)?
- Does this continuity argument require \(f\) to be continuous? State the conditions it uses instead.
