Testing Direct Riemann Integrability

Why Mesh Sums Matter in Renewal Theory

In renewal equations, a forcing function describes the contribution made at a given time, while renewals average that contribution over a growing collection of renewal epochs. The key renewal theorem uses a regularity condition on the forcing function: direct Riemann integrability. This condition is designed for the half-line and checks not only the area under a function, but also how much its values can vary across small mesh cells. A function may be integrable and still fail this stronger test.

The renewal process itself is defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, but direct Riemann integrability is a deterministic property of a function $f:[0,\infty)\to[0,\infty)$. No probability or conditioning is needed to test it. The mesh convention matters: cells are half-open intervals, so a point on a cell boundary belongs to the cell starting at that boundary.

Definition: For mesh width $h>0$ and integer $k\ge0$, let $I_{k,h}=[kh,(k+1)h)$. For a nonnegative measurable function $f$, define the upper and lower mesh sums by $$ U_h(f)=h\sum_{k=0}^{\infty}\sup_{x\in I_{k,h}}f(x), \qquad L_h(f)=h\sum_{k=0}^{\infty}\inf_{x\in I_{k,h}}f(x). $$ We call $f$ directly Riemann integrable (DRI) if $U_h(f)<\infty$ for every $h>0$ and $U_h(f)-L_h(f)\to0$ as $h\downarrow0$. For a signed function, we use the convention that $f$ is DRI when $|f|$ is DRI.

The sums use the supremum and infimum on each actual cell, not just values at selected sample points. If $f$ is integrable, then its integral is bounded by the lower and upper sums whenever those sums are finite: on every cell, $\inf_{I_{k,h}}f\le f(x)\le\sup_{I_{k,h}}f$. Summing the cellwise integral bounds gives $L_h(f)\le\int_0^\infty f(x)\,dx\le U_h(f)$. Thus the DRI condition makes the upper and lower approximations squeeze together as the mesh shrinks.

In the renewal setting, DRI is a check on the forcing term before applying a key renewal limit. It is not a substitute for the other hypotheses of that limit, such as the relevant assumptions on the interarrival distribution. Here the task is narrower: determine whether a function meets the mesh-sum condition, and understand useful sufficient tests.

Indicators and Step Functions

Worked Example: An Interval Indicator on a Boundary-Aligned Mesh

Let $f(x)=\mathbf{1}_{[1,3)}(x)$ for $x\ge0$. Use mesh width $h=\tfrac12$ and compute both sums. Membership is inclusive at $1$ and strict at $3$, as specified by the half-open interval $[1,3)$.

The cells are $I_{k,1/2}=[k/2,(k+1)/2)$. In particular, $[1/2,1)$ lies outside the interval, the four cells $[1,3/2)$, $[3/2,2)$, $[2,5/2)$, and $[5/2,3)$ lie inside it, and $[3,7/2)$ lies outside it. All earlier and later cells also lie outside. Because both endpoints of the indicated interval are mesh boundaries, every cell has constant function value: zero outside and one inside. There are four inside cells, each of width $\tfrac12$, so $$ U_{1/2}(f)=L_{1/2}(f)=4\cdot\tfrac12=2. $$

More generally, an interval indicator $\mathbf{1}_{[a,b)}$, with $0\le a<b<\infty$, is DRI. Only cells containing $a$ or $b$ can have both zero and one values; all other cells have zero oscillation. There are at most two such cells, each with oscillation at most one. Hence $0\le U_h-L_h\le2h$. Also, the number of cells meeting $[a,b)$ is finite for each fixed $h$, so $U_h<\infty$. Letting $h\downarrow0$ proves the claim.

Proposition: A bounded step function with finitely many jumps and compact support is DRI, provided its values are assigned consistently on the half-open intervals between successive jump locations.

Proof. Let the finitely many jump locations be $a_1,\ldots,a_m$, and let $C$ be the difference between the largest and smallest values of the function, including its zero value outside the support. A mesh cell not containing a jump location has a constant value and therefore contributes zero to $U_h-L_h$. Each jump location belongs to exactly one half-open mesh cell, so at most $m$ cells can contribute a positive amount. The oscillation on each such cell is at most $C$, and its contribution to the gap is at most $hC$. Thus $0\le U_h-L_h\le mhC$, which tends to zero. Compact support and boundedness imply that only finitely many cells have nonzero supremum for each $h>0$, so $U_h<\infty$. Therefore the function is DRI. $\square$

Worked Example: A Two-Level Step Function

Let $f(x)=2$ on $[0,1)$, $f(x)=1$ on $[1,2)$, and $f(x)=0$ on $[2,\infty)$. At mesh width $h=\tfrac12$, the first two cells have supremum and infimum $2$; the next two cells have supremum and infimum $1$; all remaining cells have supremum and infimum $0$. Consequently, $$ U_{1/2}(f)=L_{1/2}(f) =\tfrac12(2+2+1+1)=3. $$

The integral is also $2\cdot1+1\cdot1=3$, which checks the mesh sums. The only jump locations are $1$ and $2$; the proposition proves that the function is DRI for arbitrary mesh widths, including widths for which a jump lies inside a cell rather than at its boundary.

Continuous Functions with Compact Support

Theorem: Every nonnegative continuous function with compact support on $[0,\infty)$ is DRI.

Proof. Suppose $f(x)=0$ for $x>A$, where $A<\infty$. Fix $h\le1$. The function is uniformly continuous on the compact interval $[0,A+1]$. Define its modulus of continuity there by $$ \omega(h)=\sup\{|f(x)-f(y)|:x,y\in[0,A+1],\ |x-y|\le h\}. $$ Uniform continuity says $\omega(h)\to0$ as $h\downarrow0$. Any mesh cell that does not meet $[0,A]$ has $f=0$ throughout it, so its contribution to the upper-minus-lower gap is zero. There are at most $A/h+2$ cells that meet $[0,A]$. On each such cell, the difference between the supremum and infimum is at most $\omega(h)$, since the distance between any two points in a cell is at most $h$. Therefore $$ 0\le U_h(f)-L_h(f) \le h(A/h+2)\omega(h) =(A+2h)\omega(h)\longrightarrow0. $$ For any fixed $h>0$, compact support means only finitely many cells have nonzero supremum, and continuity makes those suprema finite. Thus $U_h(f)<\infty$. Both conditions in the definition hold, proving the theorem. $\square$

Worked Example: A Continuous Triangle

Consider $f(x)=1-x$ on $[0,1]$ and $f(x)=0$ for $x\ge1$. This function is continuous, nonnegative, and compactly supported, so the theorem applies. The following mesh calculation gives a direct check. Use $h=\tfrac14$; on the four nonzero-range cells the endpoint values decrease as follows: $$ [0,\tfrac14):\ 1\text{ to }\tfrac34,\quad [\tfrac14,\tfrac12):\ \tfrac34\text{ to }\tfrac12,\quad [\tfrac12,\tfrac34):\ \tfrac12\text{ to }\tfrac14,\quad [\tfrac34,1):\ \tfrac14\text{ to }0. $$ The function is decreasing on these cells, so the listed left values are the suprema and the right-limit values are the infima. Cells starting at or beyond $1$ contribute zero. Hence $$ U_{1/4}(f)=\tfrac14(1+\tfrac34+\tfrac12+\tfrac14)=\tfrac58, \qquad L_{1/4}(f)=\tfrac14(\tfrac34+\tfrac12+\tfrac14+0)=\tfrac38. $$ The gap is $\tfrac58-\tfrac38=\tfrac14$. The integral, computed as the area of a triangle of base $1$ and height $1$, is $\tfrac12$, and indeed $\tfrac38\le\tfrac12\le\tfrac58$.

Monotone Integrable Tails

Proposition: If $f$ is nonnegative, nonincreasing, and integrable on $[0,\infty)$, then $f$ is DRI.

Proof. Write $I=\int_0^\infty f(x)\,dx<\infty$. Since $f$ is nonincreasing, the supremum on $[kh,(k+1)h)$ is $f(kh)$. For $k\ge1$, every $x\in[(k-1)h,kh)$ has $f(x)\ge f(kh)$, so $$ h f(kh)\le\int_{(k-1)h}^{kh} f(x)\,dx. $$ It follows that $U_h(f)\le hf(0)+I<\infty$. Also, integrability and monotonicity imply $f(x)\to0$: if its nonnegative limit were $c>0$, then $\int_0^R f(x)\,dx\ge cR$ for every $R$, contradicting finite $I$. The infimum on $[kh,(k+1)h)$ is at least $f((k+1)h)$. For each finite $N$, $$ h\sum_{k=0}^{N}\bigl(\sup_{I_{k,h}}f-\inf_{I_{k,h}}f\bigr) \le h\sum_{k=0}^{N}\bigl(f(kh)-f((k+1)h)\bigr) =h\bigl(f(0)-f((N+1)h)\bigr) \le hf(0). $$ Taking $N\to\infty$ gives $0\le U_h-L_h\le hf(0)$. This bound tends to zero as $h\downarrow0$, so $f$ is DRI. $\square$

Worked Example: An Integrable Power Tail

Take $f(x)=(1+x)^{-2}$ for $x\ge0$. This function is nonnegative and nonincreasing. Its integral is $$ \int_0^\infty(1+x)^{-2}\,dx =\left[-(1+x)^{-1}\right]_{0}^{\infty}=1. $$ The proposition therefore applies. More explicitly, for every $h>0$, $$ U_h(f)\le h f(0)+1=h+1,\qquad 0\le U_h(f)-L_h(f)\le h f(0)=h, $$ because $f(0)=1$. At $h=1$, these bounds give $U_1(f)\le2$ and $U_1(f)-L_1(f)\le1$; as $h\downarrow0$, the gap bound tends to zero. This verifies both parts of the DRI condition without needing a closed form for the infinite mesh sum.

Why Integrability Alone Is Not Enough

A finite integral controls area, but it does not control the supremum on every mesh cell. For example, define $f(x)=1$ when $x$ is rational and $f(x)=0$ when $x$ is irrational. This is a bounded measurable function with Lebesgue integral zero. Yet every cell $[kh,(k+1)h)$ contains both rational and irrational numbers, so its supremum is $1$ and its infimum is $0$. Thus $U_h(f)=h\sum_{k=0}^{\infty}1=\infty$ and $L_h(f)=0$ for every $h>0$. The function is not DRI. A concrete mistake would be to conclude “integral zero, therefore DRI”; the correction is to compute the cellwise extrema, which show that every upper sum is infinite.

For practical renewal applications, the preceding results give useful sufficient tests. Indicators of bounded intervals and finite step functions with compact support pass the test because only finitely many boundary cells can oscillate. Continuous compact-support functions pass because uniform continuity makes oscillations small. Nonincreasing integrable functions pass because their upper-minus-lower gap is at most $hf(0)$. These are sufficient criteria, not a claim that every DRI function must have one of these forms.

Common pitfall. Do not group cells with the wrong width or treat a mesh-boundary point as belonging to both neighboring cells. For $h=\tfrac12$, the cells are $[k/2,(k+1)/2)$, each of width $\tfrac12$. For the indicator of $[1,3)$, the cells $[1/2,1)$ and $[3,7/2)$ are outside, while $[1,3/2)$, $[3/2,2)$, $[2,5/2)$, and $[5/2,3)$ are inside. Every cell has zero oscillation, giving $U_{1/2}=L_{1/2}=2$. Endpoint conventions and cell width must be checked before summing.

Check Your Understanding

For each question, state the cell convention you use and justify the conclusion from the upper and lower sums.

For $f=\mathbf{1}_{[0,2)}$ and $h=1$, list the nonzero cells and compute $U_1(f)$ and $L_1(f)$.
Why does a step function with finitely many jumps and compact support have only finitely many cells that can contribute to $U_h-L_h$ at a fixed mesh width?
For a nonnegative, nonincreasing, integrable function, what bound on $U_h-L_h$ proves that the gap vanishes with the mesh?
For $f(x)=(1+x)^{-2}$, compute the integral and the bound on $U_h(f)$ at mesh width $h=\tfrac12$.
Why is the function equal to $1$ on the rationals and $0$ on the irrationals not DRI, despite having integral zero?